[{"Name":"Linear Transformation Definition","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Lesson 1 - What is a Linear Transformation","Duration":"17m 48s","ChapterTopicVideoID":9678,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.404","Text":"In this clip, we\u0027re introducing a new concept,"},{"Start":"00:03.404 ","End":"00:06.820","Text":"that of linear transformation."},{"Start":"00:06.980 ","End":"00:10.860","Text":"The word transformation is just another word for function,"},{"Start":"00:10.860 ","End":"00:15.150","Text":"but in linear algebra we use transformation rather than function."},{"Start":"00:15.150 ","End":"00:19.230","Text":"The question is, what makes a transformation linear?"},{"Start":"00:19.230 ","End":"00:23.970","Text":"It turns out that we have 2 conditions or axioms"},{"Start":"00:23.970 ","End":"00:29.730","Text":"that this function or transformation has to satisfy and then it will be called linear."},{"Start":"00:29.730 ","End":"00:33.100","Text":"That\u0027s a bit vague. Let\u0027s get more concrete."},{"Start":"00:33.100 ","End":"00:35.645","Text":"I\u0027ll explain through an example."},{"Start":"00:35.645 ","End":"00:39.200","Text":"Let\u0027s take a transformation and we use"},{"Start":"00:39.200 ","End":"00:44.215","Text":"a T typically just like we would use f for function,"},{"Start":"00:44.215 ","End":"00:46.230","Text":"T for transformation,"},{"Start":"00:46.230 ","End":"00:48.960","Text":"and linear algebra we use that."},{"Start":"00:48.960 ","End":"00:54.180","Text":"Here are two vector spaces, well it’s the same one, from R^2 to R^2."},{"Start":"00:54.180 ","End":"00:59.175","Text":"And it\u0027s defined by sending the vector x,"},{"Start":"00:59.175 ","End":"01:02.670","Text":"y to the vector y, x."},{"Start":"01:02.670 ","End":"01:05.700","Text":"This is in R^2 and this is in R^2."},{"Start":"01:05.700 ","End":"01:12.185","Text":"Small remark and notation often don\u0027t bother putting these extra brackets in."},{"Start":"01:12.185 ","End":"01:18.825","Text":"It\u0027s pretty clear when we write it this way that T is applying to the whole vector x, y."},{"Start":"01:18.825 ","End":"01:23.125","Text":"We typically write it this way without the extra brackets."},{"Start":"01:23.125 ","End":"01:27.770","Text":"If I apply the transformation T to the vector 1,"},{"Start":"01:27.770 ","End":"01:31.295","Text":"2, I get the vector 2, 1."},{"Start":"01:31.295 ","End":"01:33.260","Text":"If I apply T to 3,"},{"Start":"01:33.260 ","End":"01:35.800","Text":"4 I get 4, 3."},{"Start":"01:35.800 ","End":"01:41.325","Text":"This third example is special little which pretty much picked at random,"},{"Start":"01:41.325 ","End":"01:42.990","Text":"but the third 1,"},{"Start":"01:42.990 ","End":"01:45.135","Text":"where I sent 4, 6 to 6,"},{"Start":"01:45.135 ","End":"01:47.470","Text":"4 is more interesting."},{"Start":"01:47.470 ","End":"01:50.855","Text":"Notice that if I give this vector 1,"},{"Start":"01:50.855 ","End":"01:52.865","Text":"2 a name u,"},{"Start":"01:52.865 ","End":"01:59.120","Text":"and I give this vector a name v. If I take u plus v,"},{"Start":"01:59.120 ","End":"02:00.500","Text":"1 plus 3 is 4,"},{"Start":"02:00.500 ","End":"02:01.895","Text":"2 plus 4 is 6."},{"Start":"02:01.895 ","End":"02:10.785","Text":"This vector here is u plus v. Now also notice that,"},{"Start":"02:10.785 ","End":"02:13.800","Text":"this is T of u, write it like that,"},{"Start":"02:13.800 ","End":"02:17.690","Text":"and this is T of v and if I add them,"},{"Start":"02:17.690 ","End":"02:22.305","Text":"this plus this, I get 2 plus 4 is 6,"},{"Start":"02:22.305 ","End":"02:24.855","Text":"1 plus 3 is 4."},{"Start":"02:24.855 ","End":"02:27.530","Text":"I get T of u plus T of v,"},{"Start":"02:27.530 ","End":"02:36.585","Text":"but it\u0027s the same thing as T of u plus v. Now I\u0027m going to go with this again."},{"Start":"02:36.585 ","End":"02:39.905","Text":"If I take 2 vectors,"},{"Start":"02:39.905 ","End":"02:41.750","Text":"this will work in general,"},{"Start":"02:41.750 ","End":"02:49.715","Text":"u and v and add them and then I apply T to this and to this and to this."},{"Start":"02:49.715 ","End":"02:52.715","Text":"This plus this will be this."},{"Start":"02:52.715 ","End":"03:01.385","Text":"It doesn\u0027t matter if I first apply T to u and T to v and then add,"},{"Start":"03:01.385 ","End":"03:07.950","Text":"or if I first add u and v to get this and then take T of that."},{"Start":"03:08.090 ","End":"03:13.990","Text":"The transformation of the sum is the sum of the transformations."},{"Start":"03:13.990 ","End":"03:16.685","Text":"Another way of saying it in words."},{"Start":"03:16.685 ","End":"03:19.165","Text":"I\u0027m going to write it another way."},{"Start":"03:19.165 ","End":"03:22.990","Text":"T of 1, 2 plus 3,"},{"Start":"03:22.990 ","End":"03:26.130","Text":"4, this bit here is T of 4,"},{"Start":"03:26.130 ","End":"03:30.280","Text":"6, is the same as T of 1, 2,"},{"Start":"03:30.280 ","End":"03:32.140","Text":"which is this plus T of 3,"},{"Start":"03:32.140 ","End":"03:35.175","Text":"4, which is the 4, 3."},{"Start":"03:35.175 ","End":"03:38.440","Text":"Other words, take the sum of the vectors and apply"},{"Start":"03:38.440 ","End":"03:41.950","Text":"T or apply T to each 1 and then take the sum."},{"Start":"03:41.950 ","End":"03:44.305","Text":"Same thing in this case."},{"Start":"03:44.305 ","End":"03:48.040","Text":"The first axiom or condition that we want for"},{"Start":"03:48.040 ","End":"03:51.640","Text":"linearity will be that this should work all the time."},{"Start":"03:51.640 ","End":"03:53.420","Text":"That if I take any 2 vectors,"},{"Start":"03:53.420 ","End":"03:54.530","Text":"u and v,"},{"Start":"03:54.530 ","End":"03:57.020","Text":"and add them and then apply T,"},{"Start":"03:57.020 ","End":"04:02.350","Text":"it should be the same thing as if I apply T to each of them and then do the addition."},{"Start":"04:02.350 ","End":"04:05.280","Text":"That\u0027s axiom number 1."},{"Start":"04:05.280 ","End":"04:07.635","Text":"Now there are actually 2 axioms."},{"Start":"04:07.635 ","End":"04:10.965","Text":"We\u0027ve done 1. I\u0027ll show you what the other 1 is."},{"Start":"04:10.965 ","End":"04:13.220","Text":"Again we\u0027ll start with an example."},{"Start":"04:13.220 ","End":"04:15.725","Text":"Suppose I want to know what is 4,"},{"Start":"04:15.725 ","End":"04:19.870","Text":"the scalar, times T of 1, 2."},{"Start":"04:19.870 ","End":"04:21.900","Text":"This would be our u,"},{"Start":"04:21.900 ","End":"04:24.340","Text":"in fact, from here."},{"Start":"04:24.530 ","End":"04:28.305","Text":"What we\u0027ll do is we take T of 1,"},{"Start":"04:28.305 ","End":"04:30.990","Text":"2 and we get 2, 1."},{"Start":"04:30.990 ","End":"04:33.125","Text":"Then 4 times that,"},{"Start":"04:33.125 ","End":"04:35.270","Text":"we take a scalar times a vector,"},{"Start":"04:35.270 ","End":"04:36.965","Text":"we multiply by each component,"},{"Start":"04:36.965 ","End":"04:38.450","Text":"so 4 times 2 is 8,"},{"Start":"04:38.450 ","End":"04:39.620","Text":"4 times 1 is 4,"},{"Start":"04:39.620 ","End":"04:42.120","Text":"we get 8, 4."},{"Start":"04:42.650 ","End":"04:47.210","Text":"On the other hand, let\u0027s do another computation"},{"Start":"04:47.210 ","End":"04:49.420","Text":"where we first take the vector 1,"},{"Start":"04:49.420 ","End":"04:54.175","Text":"2 and multiply by 4 and then we want to apply T to that."},{"Start":"04:54.175 ","End":"04:58.890","Text":"4 times 1, 2 is 4, 8."},{"Start":"04:58.890 ","End":"05:01.100","Text":"When we apply T to it,"},{"Start":"05:01.100 ","End":"05:02.795","Text":"T switches the order,"},{"Start":"05:02.795 ","End":"05:07.980","Text":"we get 8, 4 and these 2 are the same."},{"Start":"05:09.230 ","End":"05:13.790","Text":"The second axiom is to require that this should always happen."},{"Start":"05:13.790 ","End":"05:15.395","Text":"This shouldn\u0027t just be a fluke,"},{"Start":"05:15.395 ","End":"05:16.520","Text":"because we chose 1,"},{"Start":"05:16.520 ","End":"05:18.140","Text":"2 and we chose 4."},{"Start":"05:18.140 ","End":"05:20.030","Text":"It should be that for any scalar,"},{"Start":"05:20.030 ","End":"05:21.710","Text":"we choose and for any vector,"},{"Start":"05:21.710 ","End":"05:24.020","Text":"this thing should work out."},{"Start":"05:24.020 ","End":"05:26.210","Text":"Summarizing this particular case,"},{"Start":"05:26.210 ","End":"05:30.700","Text":"we say 4 times T of this is T of 4 times this,"},{"Start":"05:30.700 ","End":"05:34.755","Text":"and in general, we require this."},{"Start":"05:34.755 ","End":"05:40.940","Text":"4 is a scalar and the typical scalar will be k and then 1,"},{"Start":"05:40.940 ","End":"05:43.075","Text":"2 is our typical vector."},{"Start":"05:43.075 ","End":"05:47.850","Text":"This is k, this is a typical vector u. K times T of u,"},{"Start":"05:47.850 ","End":"05:52.185","Text":"is T of k times u."},{"Start":"05:52.185 ","End":"05:54.500","Text":"It doesn\u0027t make any difference."},{"Start":"05:54.500 ","End":"06:01.810","Text":"If we first multiply by the scalar and then take T of that,"},{"Start":"06:01.810 ","End":"06:07.400","Text":"or apply T to the vector and then multiply the scalar by that."},{"Start":"06:07.400 ","End":"06:09.920","Text":"Either way you do it, should come out the same."},{"Start":"06:09.920 ","End":"06:12.510","Text":"This, together with this,"},{"Start":"06:12.510 ","End":"06:17.360","Text":"are the 2 axioms that make a transformation linear."},{"Start":"06:17.360 ","End":"06:22.320","Text":"Next I want to just make it slightly more formal."},{"Start":"06:22.700 ","End":"06:25.320","Text":"I don\u0027t want it in general."},{"Start":"06:25.320 ","End":"06:27.120","Text":"Instead of just R^2 and R^2,"},{"Start":"06:27.120 ","End":"06:32.110","Text":"we take any 2 vector spaces called them U and V,"},{"Start":"06:32.110 ","End":"06:38.090","Text":"but they have to be over the same field K. Let\u0027s say both of them over the real numbers."},{"Start":"06:38.090 ","End":"06:41.570","Text":"A transformation, a function,"},{"Start":"06:41.570 ","End":"06:46.250","Text":"call it T from V to U,"},{"Start":"06:46.250 ","End":"06:50.900","Text":"is called linear if it satisfies the following 2 conditions."},{"Start":"06:50.900 ","End":"06:55.280","Text":"That\u0027s going to be the 2 conditions above that we had in general."},{"Start":"06:55.280 ","End":"06:59.460","Text":"The first 1 was the first 1 we had above is that,"},{"Start":"06:59.460 ","End":"07:02.290","Text":"if you apply the transformation to a sum,"},{"Start":"07:02.290 ","End":"07:05.440","Text":"it\u0027s the same as the sum of applying the transformation"},{"Start":"07:05.440 ","End":"07:08.740","Text":"to each of the vectors and here I\u0027ve written it in formal language."},{"Start":"07:08.740 ","End":"07:14.800","Text":"T of u plus v is T of u plus T of v for all u and v in"},{"Start":"07:14.800 ","End":"07:23.110","Text":"the vector space V. That\u0027s in the source domain vector space,"},{"Start":"07:23.110 ","End":"07:25.505","Text":"that\u0027s where u and v come from."},{"Start":"07:25.505 ","End":"07:29.305","Text":"The second condition is that if we take a scalar,"},{"Start":"07:29.305 ","End":"07:32.830","Text":"multiplied by a vector and take T of that,"},{"Start":"07:32.830 ","End":"07:36.340","Text":"it\u0027s the same as the first applying T to the vector,"},{"Start":"07:36.340 ","End":"07:38.335","Text":"then multiplying by k, In other words,"},{"Start":"07:38.335 ","End":"07:42.255","Text":"T of k, u is k times T of u."},{"Start":"07:42.255 ","End":"07:44.410","Text":"For all k in K,"},{"Start":"07:44.410 ","End":"07:46.160","Text":"and is very scalar."},{"Start":"07:46.160 ","End":"07:52.700","Text":"K is the field of scalars like real numbers or complex numbers."},{"Start":"07:52.700 ","End":"07:57.225","Text":"For all u in V,"},{"Start":"07:57.225 ","End":"08:02.495","Text":"again, the source domain vector space."},{"Start":"08:02.495 ","End":"08:05.530","Text":"1 more definition."},{"Start":"08:05.530 ","End":"08:09.650","Text":"If V and U where the same vector space,"},{"Start":"08:09.650 ","End":"08:11.570","Text":"which is what we had in the example,"},{"Start":"08:11.570 ","End":"08:14.000","Text":"this was R^2 and this was R^2,"},{"Start":"08:14.000 ","End":"08:20.370","Text":"then the linear transformation is also called a linear operator."},{"Start":"08:20.370 ","End":"08:30.240","Text":"Linear operator means that the domain and target vector spaces are the same."},{"Start":"08:30.410 ","End":"08:35.120","Text":"I mean, just highlight the word operator."},{"Start":"08:35.120 ","End":"08:39.530","Text":"Now, there\u0027s something else I want to show you."},{"Start":"08:39.530 ","End":"08:43.130","Text":"There is an alternative definition for linear,"},{"Start":"08:43.130 ","End":"08:47.120","Text":"which combines the 2 rules or 2 axioms,"},{"Start":"08:47.120 ","End":"08:50.425","Text":"a and b into 1 single axiom."},{"Start":"08:50.425 ","End":"08:54.020","Text":"The alternative definition for linear transformation."},{"Start":"08:54.020 ","End":"08:55.640","Text":"Again, same beginning,"},{"Start":"08:55.640 ","End":"08:57.590","Text":"U and V are vector spaces over"},{"Start":"08:57.590 ","End":"09:02.860","Text":"the same field and we have a transformation T from V to U,"},{"Start":"09:02.860 ","End":"09:06.420","Text":"and it\u0027s called linear if it satisfies."},{"Start":"09:06.420 ","End":"09:10.470","Text":"This time I\u0027m going to give you 1 condition instead of 2."},{"Start":"09:10.470 ","End":"09:20.090","Text":"Here it is, that if you apply a transformation to au plus bv,"},{"Start":"09:20.090 ","End":"09:28.960","Text":"and that\u0027s equal to a times T of u plus b times T of v for all a and b in k,"},{"Start":"09:28.960 ","End":"09:34.065","Text":"a and b are scalars and all u and v in V,"},{"Start":"09:34.065 ","End":"09:43.175","Text":"u and v are vectors from the source or domain."},{"Start":"09:43.175 ","End":"09:47.635","Text":"This is sometimes more convenient than the 2 separate axioms."},{"Start":"09:47.635 ","End":"09:53.695","Text":"Let me give you an idea of where this comes from or why it might be true."},{"Start":"09:53.695 ","End":"10:00.895","Text":"If we let the scalar a be 1, let\u0027s say,"},{"Start":"10:00.895 ","End":"10:04.375","Text":"and the scalar b is 1,"},{"Start":"10:04.375 ","End":"10:09.970","Text":"then what we\u0027ll get is T of u plus v. This will be 1 and 1,"},{"Start":"10:09.970 ","End":"10:11.815","Text":"will be T of u plus T of v,"},{"Start":"10:11.815 ","End":"10:14.530","Text":"that would be like the first axiom."},{"Start":"10:14.530 ","End":"10:22.780","Text":"On the other hand, if we let a replace it by k and replace b by 0,"},{"Start":"10:22.780 ","End":"10:29.095","Text":"it will say that T of k u is k times T of u,"},{"Start":"10:29.095 ","End":"10:31.930","Text":"which will be the second axiom."},{"Start":"10:31.930 ","End":"10:37.600","Text":"It gives you an idea why this is equivalent to the 2 axioms."},{"Start":"10:37.600 ","End":"10:40.690","Text":"Like I said, occasionally we\u0027ll be using this form."},{"Start":"10:40.690 ","End":"10:48.350","Text":"They usually will be using the 2 conditions or 2 axioms to prove linearity."},{"Start":"10:48.360 ","End":"10:52.839","Text":"We\u0027ve talked a lot about when transformation is linear."},{"Start":"10:52.839 ","End":"10:59.300","Text":"I would also like to give you an example of a transformation that is not linear."},{"Start":"10:59.570 ","End":"11:02.265","Text":"Here it is. Once again,"},{"Start":"11:02.265 ","End":"11:07.905","Text":"I\u0027m taking u and v both to be the vector space R^2."},{"Start":"11:07.905 ","End":"11:15.025","Text":"We\u0027ll define T by what it does to the vector x, y."},{"Start":"11:15.025 ","End":"11:19.315","Text":"It sends it to the vector x plus 1 comma y."},{"Start":"11:19.315 ","End":"11:24.290","Text":"Second component unchanged and we add 1 to the first component."},{"Start":"11:24.360 ","End":"11:28.645","Text":"For example, t of 1 comma 2,"},{"Start":"11:28.645 ","End":"11:32.800","Text":"we just add 1 to the first component is 2 comma 2."},{"Start":"11:32.800 ","End":"11:35.890","Text":"T of 3, 4 is 4,"},{"Start":"11:35.890 ","End":"11:39.460","Text":"4, this is the same numbers I used earlier."},{"Start":"11:39.460 ","End":"11:46.630","Text":"This we called it U and this we called V. The third example where 4,"},{"Start":"11:46.630 ","End":"11:48.010","Text":"6 goes to 5, 6,"},{"Start":"11:48.010 ","End":"11:52.750","Text":"I chose it because I chose this 1 to be u plus v,"},{"Start":"11:52.750 ","End":"11:53.950","Text":"1, 2 plus 3,"},{"Start":"11:53.950 ","End":"11:56.455","Text":"4 is 4, 6, just like before."},{"Start":"11:56.455 ","End":"12:00.850","Text":"The thing is that if I take this plus this,"},{"Start":"12:00.850 ","End":"12:03.280","Text":"it does not equal this 2,"},{"Start":"12:03.280 ","End":"12:06.805","Text":"2 plus 4, 4 would be 6,"},{"Start":"12:06.805 ","End":"12:12.520","Text":"6, which is not the same as 5, 6."},{"Start":"12:12.520 ","End":"12:15.400","Text":"Rewriting that. T of 1,"},{"Start":"12:15.400 ","End":"12:16.810","Text":"2 plus 3, 4,"},{"Start":"12:16.810 ","End":"12:18.295","Text":"which is T of this,"},{"Start":"12:18.295 ","End":"12:21.190","Text":"is not the same as taking T of 1,"},{"Start":"12:21.190 ","End":"12:24.175","Text":"2 and adding that to T of 3, 4."},{"Start":"12:24.175 ","End":"12:31.090","Text":"Notice that this 1 here is T of u plus v, and it\u0027s T of this,"},{"Start":"12:31.090 ","End":"12:37.060","Text":"and this is T of u plus T of v,"},{"Start":"12:37.060 ","End":"12:38.875","Text":"because this is T of u,"},{"Start":"12:38.875 ","End":"12:42.910","Text":"and this is T of v. In general,"},{"Start":"12:42.910 ","End":"12:51.505","Text":"we have violated the axiom that T of u plus v should equal T of u plus T of v. This time,"},{"Start":"12:51.505 ","End":"12:53.230","Text":"they are not equal."},{"Start":"12:53.230 ","End":"13:01.465","Text":"All you need is 1 counterexample to say that this rule is violated."},{"Start":"13:01.465 ","End":"13:03.520","Text":"I only need 1 example."},{"Start":"13:03.520 ","End":"13:05.890","Text":"I don\u0027t mean that this is always different from this"},{"Start":"13:05.890 ","End":"13:08.590","Text":"could sometimes by chance be the same."},{"Start":"13:08.590 ","End":"13:12.700","Text":"But it\u0027s not always true and we have 1 counterexample."},{"Start":"13:12.700 ","End":"13:15.610","Text":"Already we know that this T is not linear,"},{"Start":"13:15.610 ","End":"13:18.310","Text":"but just for the sake of exercise,"},{"Start":"13:18.310 ","End":"13:23.140","Text":"let\u0027s also show that the other axiom fails also."},{"Start":"13:23.140 ","End":"13:26.020","Text":"I\u0027m using the same numbers I used earlier,"},{"Start":"13:26.020 ","End":"13:33.679","Text":"take the constant k. This 1 is the k and this 1 is the vector u."},{"Start":"13:34.710 ","End":"13:39.230","Text":"This is k times T of u,"},{"Start":"13:39.870 ","End":"13:44.050","Text":"which is, comes out to be this."},{"Start":"13:44.050 ","End":"13:49.915","Text":"The other hand, if I first multiply u by k and then apply T,"},{"Start":"13:49.915 ","End":"13:51.415","Text":"I get T of 4,"},{"Start":"13:51.415 ","End":"13:53.845","Text":"8, which is 5, 8."},{"Start":"13:53.845 ","End":"13:55.270","Text":"Remember we add 1 to this,"},{"Start":"13:55.270 ","End":"13:57.685","Text":"and this certainly is not equal to this."},{"Start":"13:57.685 ","End":"14:04.060","Text":"This is k times T of u,"},{"Start":"14:04.060 ","End":"14:09.054","Text":"and this is T of K times u,"},{"Start":"14:09.054 ","End":"14:12.130","Text":"and they are not the same in this case."},{"Start":"14:12.130 ","End":"14:15.055","Text":"This axiom does not hold."},{"Start":"14:15.055 ","End":"14:18.775","Text":"In our case, this doesn\u0027t hold."},{"Start":"14:18.775 ","End":"14:24.310","Text":"In general, this means that in general we cannot say"},{"Start":"14:24.310 ","End":"14:34.240","Text":"that K times T of u is the same as T of k u."},{"Start":"14:34.240 ","End":"14:36.055","Text":"They\u0027re not the same here."},{"Start":"14:36.055 ","End":"14:40.000","Text":"Actually both axioms fail the first and the second."},{"Start":"14:40.000 ","End":"14:43.669","Text":"T is definitely not linear."},{"Start":"14:43.669 ","End":"14:48.030","Text":"Now I want to show 1 more example of linearity."},{"Start":"14:48.030 ","End":"14:52.845","Text":"In fact, this is the very first example we had only we just"},{"Start":"14:52.845 ","End":"14:58.220","Text":"illustrated it with a particular example with certain numbers."},{"Start":"14:58.220 ","End":"15:01.690","Text":"But I want to do it in general, more formally,"},{"Start":"15:01.690 ","End":"15:07.990","Text":"I commend the preparation phase with stuff that you\u0027re going to use."},{"Start":"15:07.990 ","End":"15:11.710","Text":"We certainly need a u and a v because we\u0027re"},{"Start":"15:11.710 ","End":"15:15.250","Text":"going to need u plus v. We\u0027ll let u be little a,"},{"Start":"15:15.250 ","End":"15:17.800","Text":"little b and v be capital A capital B."},{"Start":"15:17.800 ","End":"15:20.635","Text":"We\u0027ll need u plus v to say plus a,"},{"Start":"15:20.635 ","End":"15:25.360","Text":"b plus b and we\u0027ll need a scalar times u k u."},{"Start":"15:25.360 ","End":"15:31.360","Text":"We\u0027ll use the 2 axiom version, not the all-in-1."},{"Start":"15:31.360 ","End":"15:34.630","Text":"The 2 axioms, the first 1 is that"},{"Start":"15:34.630 ","End":"15:38.080","Text":"the transformation of the sum is the sum of the transformations."},{"Start":"15:38.080 ","End":"15:39.550","Text":"Let\u0027s see if that holds."},{"Start":"15:39.550 ","End":"15:43.270","Text":"True. Make some computations."},{"Start":"15:43.270 ","End":"15:45.100","Text":"First of all, I need T of u,"},{"Start":"15:45.100 ","End":"15:48.520","Text":"which is T of a, b."},{"Start":"15:48.520 ","End":"15:53.365","Text":"Remember we just switch the coordinates so that\u0027s b comma a."},{"Start":"15:53.365 ","End":"15:56.545","Text":"Similarly with u and v,"},{"Start":"15:56.545 ","End":"16:00.550","Text":"both come from the source or domain from,"},{"Start":"16:00.550 ","End":"16:04.105","Text":"they\u0027re both the same but from the left 1."},{"Start":"16:04.105 ","End":"16:08.800","Text":"Then we do u plus v and apply T to that."},{"Start":"16:08.800 ","End":"16:10.660","Text":"Well, u plus v is a plus a,"},{"Start":"16:10.660 ","End":"16:15.230","Text":"b plus b apply T to that is switching the order."},{"Start":"16:15.420 ","End":"16:19.340","Text":"I need a bit more space here."},{"Start":"16:19.410 ","End":"16:23.185","Text":"Notice that this right-hand side,"},{"Start":"16:23.185 ","End":"16:25.600","Text":"b a plus this big B,"},{"Start":"16:25.600 ","End":"16:29.095","Text":"big A is equal to this."},{"Start":"16:29.095 ","End":"16:32.845","Text":"In other words, this plus this is this,"},{"Start":"16:32.845 ","End":"16:35.725","Text":"which is this, and that\u0027s the first axiom."},{"Start":"16:35.725 ","End":"16:42.835","Text":"We\u0027ve got the first axiom proved and now we just go on and do the second axiom."},{"Start":"16:42.835 ","End":"16:48.250","Text":"We want to show this time that this equality always holds."},{"Start":"16:48.250 ","End":"16:53.860","Text":"T of u, this bit here we can just copy from here, we\u0027ve done it."},{"Start":"16:53.860 ","End":"16:59.200","Text":"It\u0027s b comma a multiply both sides by"},{"Start":"16:59.200 ","End":"17:05.560","Text":"k. I have k T of u and I multiply this by k. It\u0027s component-wise that kb, ka."},{"Start":"17:05.560 ","End":"17:09.220","Text":"On the other hand, and this is the left-hand side here."},{"Start":"17:09.220 ","End":"17:11.125","Text":"T of ku,"},{"Start":"17:11.125 ","End":"17:12.655","Text":"ku is ka,"},{"Start":"17:12.655 ","End":"17:16.450","Text":"kb and T of that is switch the order,"},{"Start":"17:16.450 ","End":"17:18.520","Text":"we get kb, ka."},{"Start":"17:18.520 ","End":"17:23.695","Text":"Now, this is certainly equal to this,"},{"Start":"17:23.695 ","End":"17:26.470","Text":"which means that k times T u,"},{"Start":"17:26.470 ","End":"17:29.500","Text":"which is, this, is equal."},{"Start":"17:29.500 ","End":"17:34.120","Text":"Sorry, this is the right-hand side,"},{"Start":"17:34.120 ","End":"17:38.005","Text":"and this is the left-hand side anyway, they\u0027re equal."},{"Start":"17:38.005 ","End":"17:41.905","Text":"We\u0027ve proven both the first and the second axioms."},{"Start":"17:41.905 ","End":"17:48.140","Text":"T is indeed a linear transformation and we\u0027re done."}],"ID":10223},{"Watched":false,"Name":"Exercise 1","Duration":"4m 15s","ChapterTopicVideoID":9694,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"In this exercise,"},{"Start":"00:01.890 ","End":"00:10.365","Text":"is a transformation T from R^2 to R^2 given by T of the vector x,y,"},{"Start":"00:10.365 ","End":"00:15.450","Text":"is the vector x plus y, x minus y."},{"Start":"00:15.450 ","End":"00:20.230","Text":"The question is, is T linear?"},{"Start":"00:20.360 ","End":"00:23.760","Text":"I start by doing some preparations,"},{"Start":"00:23.760 ","End":"00:27.900","Text":"taking some general vectors from the source,"},{"Start":"00:27.900 ","End":"00:32.970","Text":"or domain, the 1 that T comes out of."},{"Start":"00:32.970 ","End":"00:37.380","Text":"Let\u0027s take u and v. Since there are now 2,"},{"Start":"00:37.380 ","End":"00:39.420","Text":"they will be a,"},{"Start":"00:39.420 ","End":"00:42.205","Text":"b and capital A,B."},{"Start":"00:42.205 ","End":"00:44.450","Text":"I\u0027m also going to need u plus v,"},{"Start":"00:44.450 ","End":"00:45.800","Text":"and I\u0027m also going to need ku,"},{"Start":"00:45.800 ","End":"00:48.810","Text":"so this is just a bit of preparation."},{"Start":"00:48.860 ","End":"00:52.534","Text":"Remember, there are 2 rules that we have to check."},{"Start":"00:52.534 ","End":"00:56.150","Text":"The first 1 is written like this,"},{"Start":"00:56.150 ","End":"01:01.495","Text":"basically that the transformation of a sum is the sum of the transformations,"},{"Start":"01:01.495 ","End":"01:06.530","Text":"and I just copied this here because we\u0027re going to be scrolling."},{"Start":"01:06.530 ","End":"01:10.320","Text":"There we are. Let\u0027s start with this 1."},{"Start":"01:10.320 ","End":"01:13.545","Text":"T of u is T of a b,"},{"Start":"01:13.545 ","End":"01:15.585","Text":"and according to the formula,"},{"Start":"01:15.585 ","End":"01:18.000","Text":"it\u0027s the sum, the difference,"},{"Start":"01:18.000 ","End":"01:20.865","Text":"a plus b, a minus b."},{"Start":"01:20.865 ","End":"01:26.420","Text":"Similarly with v, only this time it\u0027s just with capital letters,"},{"Start":"01:26.420 ","End":"01:28.490","Text":"but same thing, T of A B,"},{"Start":"01:28.490 ","End":"01:30.950","Text":"A plus B, A minus B."},{"Start":"01:30.950 ","End":"01:34.760","Text":"Now we need T of u plus v. We"},{"Start":"01:34.760 ","End":"01:38.810","Text":"have u plus v precomputed already, should have mentioned that."},{"Start":"01:38.810 ","End":"01:43.430","Text":"Of course we take the components, add them."},{"Start":"01:43.430 ","End":"01:45.680","Text":"The first component to the first component,"},{"Start":"01:45.680 ","End":"01:48.200","Text":"and the second component with the second component,"},{"Start":"01:48.200 ","End":"01:52.700","Text":"and likewise with the scalar multiplier,"},{"Start":"01:52.700 ","End":"01:54.105","Text":"we multiply by each."},{"Start":"01:54.105 ","End":"01:56.260","Text":"We\u0027ll need that in a moment."},{"Start":"01:57.080 ","End":"02:01.005","Text":"T of u plus v is T of this,"},{"Start":"02:01.005 ","End":"02:03.320","Text":"and by the rule here,"},{"Start":"02:03.320 ","End":"02:05.840","Text":"we take the sum of these 2,"},{"Start":"02:05.840 ","End":"02:07.865","Text":"which is this plus this,"},{"Start":"02:07.865 ","End":"02:11.285","Text":"and then the difference, this minus this."},{"Start":"02:11.285 ","End":"02:15.470","Text":"If that\u0027s not clear, just think of the a plus A as an x,"},{"Start":"02:15.470 ","End":"02:19.250","Text":"b plus B as the y. I\u0027m looking here,"},{"Start":"02:19.250 ","End":"02:22.430","Text":"I need x plus y,"},{"Start":"02:22.430 ","End":"02:27.930","Text":"and here I need x minus y."},{"Start":"02:31.100 ","End":"02:39.260","Text":"Now, notice that if I take this and add it to this,"},{"Start":"02:39.260 ","End":"02:44.360","Text":"then the first component plus the first component is this, and the second component,"},{"Start":"02:44.360 ","End":"02:51.300","Text":"and the second component gives this because it\u0027s a plus A minus b plus B,"},{"Start":"02:51.300 ","End":"02:53.915","Text":"so this plus this equals this,"},{"Start":"02:53.915 ","End":"02:58.160","Text":"which means that T of u plus T of v is equal to T of u plus"},{"Start":"02:58.160 ","End":"03:06.220","Text":"v. That\u0027s the first rule from the 2 rules that we have to prove, rules or axioms."},{"Start":"03:06.320 ","End":"03:09.575","Text":"The second 1 was this,"},{"Start":"03:09.575 ","End":"03:12.755","Text":"is if we apply T to a scalar times u,"},{"Start":"03:12.755 ","End":"03:15.860","Text":"we can take the scalar outside."},{"Start":"03:15.860 ","End":"03:20.555","Text":"Here again is the definition of T, which scrolled off."},{"Start":"03:20.555 ","End":"03:23.075","Text":"This part T of u,"},{"Start":"03:23.075 ","End":"03:24.950","Text":"since u is little a,"},{"Start":"03:24.950 ","End":"03:27.830","Text":"little b, is just a plus b,"},{"Start":"03:27.830 ","End":"03:30.220","Text":"a minus b from the definition."},{"Start":"03:30.220 ","End":"03:34.505","Text":"I want to multiply this by k to get the right-hand side here."},{"Start":"03:34.505 ","End":"03:39.470","Text":"K times this is k times the first component."},{"Start":"03:39.470 ","End":"03:42.890","Text":"I skipped this stage, it\u0027s k brackets a plus b,"},{"Start":"03:42.890 ","End":"03:45.695","Text":"but I opened the brackets already, and similarly here,"},{"Start":"03:45.695 ","End":"03:49.585","Text":"k times a minus b is ka minus kb."},{"Start":"03:49.585 ","End":"03:54.510","Text":"T of ku, we already computed ku as ka,kb,"},{"Start":"03:54.510 ","End":"03:57.300","Text":"and then T of that is the sum,"},{"Start":"03:57.300 ","End":"04:01.325","Text":"the difference ka plus kb, ka minus kb."},{"Start":"04:01.325 ","End":"04:04.750","Text":"If you look, this is exactly equal to this,"},{"Start":"04:04.750 ","End":"04:10.400","Text":"and so we\u0027ve proved that the second axiom also holds in general."},{"Start":"04:10.400 ","End":"04:16.170","Text":"That means that T is indeed linear. We\u0027re done."}],"ID":10224},{"Watched":false,"Name":"Exercise 2","Duration":"4m 27s","ChapterTopicVideoID":9695,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.450","Text":"In this exercise, we have a transformation from R^3 to R^3,"},{"Start":"00:06.450 ","End":"00:08.010","Text":"it\u0027s called T."},{"Start":"00:08.010 ","End":"00:10.450","Text":"What it does to the vector x,"},{"Start":"00:10.450 ","End":"00:12.480","Text":"y, z is it sends it to,"},{"Start":"00:12.480 ","End":"00:15.105","Text":"well, what\u0027s written here."},{"Start":"00:15.105 ","End":"00:20.770","Text":"We have to check if this T is linear."},{"Start":"00:20.920 ","End":"00:26.915","Text":"I routinely start these exercises with writing an expression for u,"},{"Start":"00:26.915 ","End":"00:29.690","Text":"v, u plus v and ku."},{"Start":"00:29.690 ","End":"00:34.350","Text":"In this case, since it\u0027s R^3, then a, b,"},{"Start":"00:34.350 ","End":"00:38.780","Text":"c would be a typical little u and use capitals for v,"},{"Start":"00:38.780 ","End":"00:47.085","Text":"u plus v component-wise addition and a scalar times u just multiply component-wise."},{"Start":"00:47.085 ","End":"00:51.270","Text":"We have to check if the 2 axioms hold."},{"Start":"00:51.270 ","End":"00:54.315","Text":"This is our first axiom."},{"Start":"00:54.315 ","End":"00:58.905","Text":"This has to hold for all u and v in R^3,"},{"Start":"00:58.905 ","End":"01:03.120","Text":"the source vector space."},{"Start":"01:03.120 ","End":"01:06.845","Text":"Question marks because that\u0027s what we have to show."},{"Start":"01:06.845 ","End":"01:10.980","Text":"I just copied it because I want to scroll."},{"Start":"01:11.990 ","End":"01:17.190","Text":"T of u and u is a, b, c."},{"Start":"01:17.190 ","End":"01:19.115","Text":"Using the formula,"},{"Start":"01:19.115 ","End":"01:20.600","Text":"just replace the x with a,"},{"Start":"01:20.600 ","End":"01:22.930","Text":"y with b and z with c,"},{"Start":"01:22.930 ","End":"01:25.910","Text":"and we\u0027ll get this expression."},{"Start":"01:25.910 ","End":"01:28.640","Text":"Practically the same thing for v,"},{"Start":"01:28.640 ","End":"01:30.800","Text":"except that we have capital A, B, C."},{"Start":"01:30.800 ","End":"01:35.490","Text":"Then we apply T to u plus v."},{"Start":"01:35.490 ","End":"01:37.805","Text":"We have u plus v over here,"},{"Start":"01:37.805 ","End":"01:40.115","Text":"but it didn\u0027t all fit in 1 line."},{"Start":"01:40.115 ","End":"01:41.915","Text":"This is the first component,"},{"Start":"01:41.915 ","End":"01:44.630","Text":"the second and the third."},{"Start":"01:44.630 ","End":"01:48.400","Text":"Just using this formula, each time,"},{"Start":"01:48.400 ","End":"01:51.195","Text":"x plus y minus 2z,"},{"Start":"01:51.195 ","End":"01:53.340","Text":"this would be our x."},{"Start":"01:53.340 ","End":"01:54.645","Text":"The a plus A,"},{"Start":"01:54.645 ","End":"01:56.480","Text":"b plus B would be the y,"},{"Start":"01:56.480 ","End":"02:00.180","Text":"and this would be the z, and so on."},{"Start":"02:00.290 ","End":"02:02.555","Text":"If you look at it,"},{"Start":"02:02.555 ","End":"02:05.015","Text":"this equation does hold."},{"Start":"02:05.015 ","End":"02:07.220","Text":"Because what we have here,"},{"Start":"02:07.220 ","End":"02:09.155","Text":"that\u0027s the T of u plus v,"},{"Start":"02:09.155 ","End":"02:11.750","Text":"and it is equal to this plus this."},{"Start":"02:11.750 ","End":"02:14.795","Text":"If you add this to this component-wise,"},{"Start":"02:14.795 ","End":"02:18.080","Text":"you\u0027ll get this possibly in a different order."},{"Start":"02:18.080 ","End":"02:21.865","Text":"We\u0027d get a plus b minus 2c plus capital A,"},{"Start":"02:21.865 ","End":"02:24.535","Text":"capital B minus capital 2C."},{"Start":"02:24.535 ","End":"02:26.600","Text":"If you slightly rearrange it,"},{"Start":"02:26.600 ","End":"02:28.440","Text":"then you will get this."},{"Start":"02:28.440 ","End":"02:30.930","Text":"Similarly for the other 2 components."},{"Start":"02:30.930 ","End":"02:33.360","Text":"This plus this equals this."},{"Start":"02:33.360 ","End":"02:35.565","Text":"That\u0027s the first axiom."},{"Start":"02:35.565 ","End":"02:39.080","Text":"Now we need to move on to the second axiom."},{"Start":"02:39.080 ","End":"02:42.180","Text":"Here it is."},{"Start":"02:42.180 ","End":"02:46.770","Text":"T of ku is the same as k times T of u."},{"Start":"02:46.770 ","End":"02:50.090","Text":"We have to check. Once again,"},{"Start":"02:50.090 ","End":"02:51.890","Text":"I copied the original formula."},{"Start":"02:51.890 ","End":"02:54.215","Text":"I keep scrolling off screen."},{"Start":"02:54.215 ","End":"02:57.140","Text":"Now, T of u, we\u0027ve actually done above,"},{"Start":"02:57.140 ","End":"03:00.330","Text":"so I just copy pasted it."},{"Start":"03:00.430 ","End":"03:08.990","Text":"Scalar k times this would just be putting k times this,"},{"Start":"03:08.990 ","End":"03:11.870","Text":"which is multiplying each component by k,"},{"Start":"03:11.870 ","End":"03:15.865","Text":"a plus b minus 2c multiplied by k,"},{"Start":"03:15.865 ","End":"03:17.870","Text":"you put it in the brackets first of course,"},{"Start":"03:17.870 ","End":"03:23.405","Text":"you get ka plus kb minus 2kc and similarly the other 2 components."},{"Start":"03:23.405 ","End":"03:27.965","Text":"On the other hand, if I apply T to ku,"},{"Start":"03:27.965 ","End":"03:33.960","Text":"then I will use this formula."},{"Start":"03:33.960 ","End":"03:41.370","Text":"We already have ku here."},{"Start":"03:41.370 ","End":"03:43.790","Text":"We figured out before, we\u0027ll figure it out again,"},{"Start":"03:43.790 ","End":"03:46.070","Text":"ku is ka, kb, kc."},{"Start":"03:46.070 ","End":"03:53.370","Text":"Now we apply this formula here with this being x, y, and z."},{"Start":"03:53.370 ","End":"03:56.625","Text":"This is the expression we get."},{"Start":"03:56.625 ","End":"03:58.985","Text":"If we examine it,"},{"Start":"03:58.985 ","End":"04:03.275","Text":"this expression is the same as this expression."},{"Start":"04:03.275 ","End":"04:07.655","Text":"Possibly with a change of the order,"},{"Start":"04:07.655 ","End":"04:11.700","Text":"maybe even exactly the same here."},{"Start":"04:12.440 ","End":"04:17.250","Text":"The second axiom is verified also."},{"Start":"04:17.250 ","End":"04:22.550","Text":"We conclude that our transformation T is indeed linear."},{"Start":"04:22.550 ","End":"04:27.620","Text":"We\u0027re done."}],"ID":10225},{"Watched":false,"Name":"Exercise 3","Duration":"4m 14s","ChapterTopicVideoID":9696,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.685","Text":"In this exercise, we have a transformation."},{"Start":"00:03.685 ","End":"00:06.910","Text":"This time is from R_3 to R_2."},{"Start":"00:06.910 ","End":"00:08.830","Text":"This is the formula for it."},{"Start":"00:08.830 ","End":"00:14.310","Text":"It takes x, y, z to the pair 2x plus z comma,"},{"Start":"00:14.310 ","End":"00:16.320","Text":"the absolute value of y."},{"Start":"00:16.320 ","End":"00:19.330","Text":"We want to see if this transformation is linear."},{"Start":"00:19.330 ","End":"00:22.805","Text":"To be linear, it has to satisfy 2 axioms."},{"Start":"00:22.805 ","End":"00:24.990","Text":"I\u0027m getting ahead of myself."},{"Start":"00:24.990 ","End":"00:26.970","Text":"Before we get to the 2 axioms,"},{"Start":"00:26.970 ","End":"00:28.030","Text":"this is what I always do."},{"Start":"00:28.030 ","End":"00:30.399","Text":"I write out u, v,"},{"Start":"00:30.399 ","End":"00:34.225","Text":"u plus v and ku in the more expanded form,"},{"Start":"00:34.225 ","End":"00:35.830","Text":"like if you as a 3D vector,"},{"Start":"00:35.830 ","End":"00:39.980","Text":"give names to each of the components and so on."},{"Start":"00:40.070 ","End":"00:42.855","Text":"Here\u0027s the first axiom,"},{"Start":"00:42.855 ","End":"00:45.980","Text":"the question marks because that\u0027s what we have to verify."},{"Start":"00:45.980 ","End":"00:48.970","Text":"We don\u0027t know if it\u0027s true or not yet."},{"Start":"00:48.970 ","End":"00:52.595","Text":"Now, I\u0027ll just copy the formula for T over here,"},{"Start":"00:52.595 ","End":"00:53.810","Text":"because as you see,"},{"Start":"00:53.810 ","End":"00:59.885","Text":"I\u0027m going to scroll enough screen. Let\u0027s see."},{"Start":"00:59.885 ","End":"01:04.755","Text":"T of u is T of a b, c,"},{"Start":"01:04.755 ","End":"01:07.335","Text":"and just substituting in here,"},{"Start":"01:07.335 ","End":"01:09.600","Text":"replace x by a,"},{"Start":"01:09.600 ","End":"01:11.280","Text":"y is b, z is c,"},{"Start":"01:11.280 ","End":"01:13.450","Text":"this is what we get."},{"Start":"01:13.880 ","End":"01:17.355","Text":"Very similarly with T of v,"},{"Start":"01:17.355 ","End":"01:19.155","Text":"this is what we get."},{"Start":"01:19.155 ","End":"01:22.655","Text":"Then we need T of u plus v,"},{"Start":"01:22.655 ","End":"01:25.650","Text":"u plus v is written here,"},{"Start":"01:25.650 ","End":"01:29.210","Text":"you have to apply T to it using this formula."},{"Start":"01:29.210 ","End":"01:33.020","Text":"Instead of x, we need to put a plus A and so on,"},{"Start":"01:33.020 ","End":"01:36.990","Text":"and so what we get is this."},{"Start":"01:37.550 ","End":"01:41.030","Text":"Particularly interesting is the second component,"},{"Start":"01:41.030 ","End":"01:42.800","Text":"absolute value of b plus B,"},{"Start":"01:42.800 ","End":"01:45.245","Text":"and you\u0027ll see why in a moment."},{"Start":"01:45.245 ","End":"01:51.155","Text":"The thing is that this plus this is not equal to this."},{"Start":"01:51.155 ","End":"01:53.420","Text":"The first component is okay,"},{"Start":"01:53.420 ","End":"01:58.955","Text":"2a plus C and then capital 2A plus C will give us this."},{"Start":"01:58.955 ","End":"02:00.590","Text":"But the question is,"},{"Start":"02:00.590 ","End":"02:02.600","Text":"is this plus this equal to this?"},{"Start":"02:02.600 ","End":"02:05.180","Text":"We would get the same thing as this almost,"},{"Start":"02:05.180 ","End":"02:12.355","Text":"but we would get absolute value of b plus absolute value of capital B,"},{"Start":"02:12.355 ","End":"02:17.140","Text":"and is that the same as capital value of b plus B?"},{"Start":"02:17.140 ","End":"02:19.329","Text":"Well, the answer is no."},{"Start":"02:19.329 ","End":"02:22.870","Text":"When I say no, I mean it can be equal for some values,"},{"Start":"02:22.870 ","End":"02:24.610","Text":"but it\u0027s not an identity;"},{"Start":"02:24.610 ","End":"02:26.410","Text":"its not always true."},{"Start":"02:26.410 ","End":"02:28.765","Text":"To show that it\u0027s not always true,"},{"Start":"02:28.765 ","End":"02:31.760","Text":"we need a counter example."},{"Start":"02:31.830 ","End":"02:39.005","Text":"What I\u0027m going to do is take u to be this,"},{"Start":"02:39.005 ","End":"02:41.835","Text":"and I took v to be this vector."},{"Start":"02:41.835 ","End":"02:48.820","Text":"The idea I had was the only interesting thing is the B, the middle component."},{"Start":"02:48.820 ","End":"02:51.995","Text":"I made them to be of opposite signs,"},{"Start":"02:51.995 ","End":"02:55.510","Text":"because this plus this will not equal to this."},{"Start":"02:55.510 ","End":"02:58.760","Text":"If little b and capital B are of opposite signs,"},{"Start":"02:58.760 ","End":"03:04.100","Text":"the plus and the minus will partially cancel each other when I add them here."},{"Start":"03:04.100 ","End":"03:06.935","Text":"Let\u0027s see it in this case."},{"Start":"03:06.935 ","End":"03:11.420","Text":"If we add just the u plus the v,"},{"Start":"03:11.420 ","End":"03:14.450","Text":"then u plus v comes out this."},{"Start":"03:14.450 ","End":"03:17.240","Text":"Minus 2 plus the 2 is the 0,"},{"Start":"03:17.240 ","End":"03:18.830","Text":"or the middle term is 0."},{"Start":"03:18.830 ","End":"03:21.805","Text":"Then if you apply T to that,"},{"Start":"03:21.805 ","End":"03:25.950","Text":"it\u0027s scrolled off but you get 10 comma 0."},{"Start":"03:25.950 ","End":"03:34.650","Text":"The second component was the absolute value of B here."},{"Start":"03:34.840 ","End":"03:38.615","Text":"On the other hand, if we add this to this,"},{"Start":"03:38.615 ","End":"03:40.775","Text":"T of u plus T of v,"},{"Start":"03:40.775 ","End":"03:47.990","Text":"we get 10 comma 4 which is not completely different,"},{"Start":"03:47.990 ","End":"03:50.840","Text":"but half different from 10 comma 0."},{"Start":"03:50.840 ","End":"03:55.055","Text":"Because the absolute values made it 2 plus 2,"},{"Start":"03:55.055 ","End":"03:59.220","Text":"but here we added minus 2 and 2,"},{"Start":"03:59.220 ","End":"04:01.770","Text":"and so we didn\u0027t get the same thing."},{"Start":"04:01.770 ","End":"04:07.470","Text":"All you need is 1 counterexample to say that this rule doesn\u0027t hold."},{"Start":"04:07.490 ","End":"04:09.670","Text":"T is not linear,"},{"Start":"04:09.670 ","End":"04:14.130","Text":"and we don\u0027t need to check the other axiom. We\u0027re done."}],"ID":10226},{"Watched":false,"Name":"Exercise 4","Duration":"2m 54s","ChapterTopicVideoID":9697,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.890","Text":"In this exercise, our a transformation is from R^2-R^3 and here\u0027s the formula."},{"Start":"00:07.890 ","End":"00:13.290","Text":"As usual, we have to decide if this is linear or not."},{"Start":"00:13.290 ","End":"00:16.410","Text":"After a while, you get a feeling for these things"},{"Start":"00:16.410 ","End":"00:19.305","Text":"and the xy doesn\u0027t look very linear."},{"Start":"00:19.305 ","End":"00:22.830","Text":"Usually linear means some constant times x plus"},{"Start":"00:22.830 ","End":"00:25.050","Text":"some constant times y, constant times z."},{"Start":"00:25.050 ","End":"00:30.305","Text":"A product of 2 variables or an exponent usually means not linear."},{"Start":"00:30.305 ","End":"00:34.860","Text":"Anyway, let\u0027s say you don\u0027t have any intuition, let\u0027s just proceed."},{"Start":"00:34.990 ","End":"00:39.380","Text":"I\u0027m going to skip the preparation phase that I usually do"},{"Start":"00:39.380 ","End":"00:44.045","Text":"and in previous exercises because I\u0027m expecting it not to be linear."},{"Start":"00:44.045 ","End":"00:50.570","Text":"Let\u0027s go right ahead and check if the first axiom holds true always."},{"Start":"00:50.570 ","End":"00:56.670","Text":"For convenience, I just copied the T so I can scroll a bit."},{"Start":"00:58.100 ","End":"01:02.150","Text":"I\u0027m going to use the technique of just guessing,"},{"Start":"01:02.150 ","End":"01:04.669","Text":"just trying different values."},{"Start":"01:04.669 ","End":"01:06.740","Text":"Usually if this is not true,"},{"Start":"01:06.740 ","End":"01:09.365","Text":"I mean, could by chance be true,"},{"Start":"01:09.365 ","End":"01:11.750","Text":"but let\u0027s just plug in some values"},{"Start":"01:11.750 ","End":"01:15.409","Text":"until maybe we get something where it doesn\u0027t hold the equality."},{"Start":"01:15.409 ","End":"01:20.795","Text":"Let\u0027s try, u is 1, 2, and then T of that,"},{"Start":"01:20.795 ","End":"01:23.305","Text":"if you compute it comes out this."},{"Start":"01:23.305 ","End":"01:26.385","Text":"Then for v, let\u0027s take 3, 4,"},{"Start":"01:26.385 ","End":"01:31.600","Text":"and then apply T under this formula, and we get this."},{"Start":"01:32.030 ","End":"01:37.845","Text":"Then u plus v, 1 plus 3 is 4, 2 plus 4 is 6."},{"Start":"01:37.845 ","End":"01:39.140","Text":"On the other hand,"},{"Start":"01:39.140 ","End":"01:42.050","Text":"if we add T of u plus T of v,"},{"Start":"01:42.050 ","End":"01:48.120","Text":"we get completely something different."},{"Start":"01:49.250 ","End":"01:51.705","Text":"Well, sorry, not completely different."},{"Start":"01:51.705 ","End":"01:52.919","Text":"In the first component,"},{"Start":"01:52.919 ","End":"01:55.660","Text":"different, because if I added this to this,"},{"Start":"01:55.660 ","End":"02:00.985","Text":"what I would get would be 2 plus 12 is 14,"},{"Start":"02:00.985 ","End":"02:05.560","Text":"2 plus 4 is 6, 1 plus 3 is 4."},{"Start":"02:05.560 ","End":"02:09.010","Text":"Yeah, we\u0027re okay with the second and third components"},{"Start":"02:09.010 ","End":"02:11.400","Text":"because just y and just x are linear,"},{"Start":"02:11.400 ","End":"02:13.890","Text":"but the xy messed things up,"},{"Start":"02:13.890 ","End":"02:17.310","Text":"14 is not equal to 24."},{"Start":"02:17.310 ","End":"02:20.875","Text":"Here I just wrote it out in more detail."},{"Start":"02:20.875 ","End":"02:23.080","Text":"T of this vector plus"},{"Start":"02:23.080 ","End":"02:26.880","Text":"this vector is not equal to T of this vector"},{"Start":"02:26.880 ","End":"02:29.095","Text":"plus T of the other vector."},{"Start":"02:29.095 ","End":"02:30.910","Text":"Because if you compute this,"},{"Start":"02:30.910 ","End":"02:32.180","Text":"it comes out that,"},{"Start":"02:32.180 ","End":"02:33.965","Text":"this right-hand side comes out this,"},{"Start":"02:33.965 ","End":"02:38.050","Text":"and they\u0027re not the same because 24 is not equal to 14."},{"Start":"02:38.050 ","End":"02:40.035","Text":"So T is not linear,"},{"Start":"02:40.035 ","End":"02:46.010","Text":"all you need is 1 counterexample and destroys the linearity."},{"Start":"02:46.010 ","End":"02:49.850","Text":"This is true, has to be true for all u and v, 1 exception,"},{"Start":"02:49.850 ","End":"02:54.150","Text":"and this is false, and we\u0027re done."}],"ID":10227},{"Watched":false,"Name":"Exercise 5","Duration":"3m 34s","ChapterTopicVideoID":9698,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.790","Text":"In this exercise, we have to check if this transformation from"},{"Start":"00:05.790 ","End":"00:12.090","Text":"R3 to R3 given by this formula is linear."},{"Start":"00:12.090 ","End":"00:16.260","Text":"In this exercise, I\u0027m going to show you a trick,"},{"Start":"00:16.260 ","End":"00:20.740","Text":"not really a trick, it\u0027s a proposition that can help sometimes."},{"Start":"00:20.740 ","End":"00:27.605","Text":"What the proposition says is that if we have a linear transformation T,"},{"Start":"00:27.605 ","End":"00:31.670","Text":"then necessarily T of 0 is 0."},{"Start":"00:31.670 ","End":"00:34.865","Text":"It takes the 0 vector to the 0 vector."},{"Start":"00:34.865 ","End":"00:39.950","Text":"Now, in logic, there\u0027s something called the contrapositive."},{"Start":"00:39.950 ","End":"00:44.115","Text":"If A implies B,"},{"Start":"00:44.115 ","End":"00:49.870","Text":"then not B implies not A."},{"Start":"00:51.170 ","End":"00:57.540","Text":"In this case, it means that if T0 is not 0,"},{"Start":"00:57.540 ","End":"01:01.040","Text":"then T is not linear because if it was linear,"},{"Start":"01:01.040 ","End":"01:03.110","Text":"then T of 0 would be 0."},{"Start":"01:03.110 ","End":"01:04.850","Text":"This is just the logical equivalent,"},{"Start":"01:04.850 ","End":"01:06.470","Text":"it\u0027s not really conversely,"},{"Start":"01:06.470 ","End":"01:10.175","Text":"it\u0027s what we call the contrapositive in logic."},{"Start":"01:10.175 ","End":"01:12.965","Text":"We\u0027ll all just accept it,"},{"Start":"01:12.965 ","End":"01:14.540","Text":"if T of 0 is not 0,"},{"Start":"01:14.540 ","End":"01:16.325","Text":"then t is not linear."},{"Start":"01:16.325 ","End":"01:22.790","Text":"In our case, I just copy the formula for T cause I\u0027m going to be scrolling."},{"Start":"01:22.790 ","End":"01:28.164","Text":"Now we\u0027re going to plug in the 0 vector."},{"Start":"01:28.164 ","End":"01:31.635","Text":"The 0 vector is 0, 0, 0,"},{"Start":"01:31.635 ","End":"01:33.440","Text":"and if you compute it here,"},{"Start":"01:33.440 ","End":"01:35.315","Text":"it comes out to 1, 0, 0,"},{"Start":"01:35.315 ","End":"01:37.730","Text":"which is not the 0 vector,"},{"Start":"01:37.730 ","End":"01:41.320","Text":"so T of 0 is not equal to 0,"},{"Start":"01:41.320 ","End":"01:44.390","Text":"and so T is not linear."},{"Start":"01:44.390 ","End":"01:46.250","Text":"Now, if you want,"},{"Start":"01:46.250 ","End":"01:49.415","Text":"we could consider this exercise solved,"},{"Start":"01:49.415 ","End":"01:52.530","Text":"but there\u0027s a very simple proof of the proposition,"},{"Start":"01:52.530 ","End":"01:54.275","Text":"and if you\u0027d like to stay,"},{"Start":"01:54.275 ","End":"01:58.260","Text":"then I\u0027ll show you why the proposition is true."},{"Start":"02:00.560 ","End":"02:05.000","Text":"We suppose that T is linear,"},{"Start":"02:05.000 ","End":"02:10.325","Text":"and we need to show that T of the 0 vector is the 0 vector."},{"Start":"02:10.325 ","End":"02:12.440","Text":"Now since T is linear,"},{"Start":"02:12.440 ","End":"02:14.650","Text":"it satisfies the 2nd axiom,"},{"Start":"02:14.650 ","End":"02:22.515","Text":"which is that T of k.u equals k.T of u for any scalar k and for any vector u."},{"Start":"02:22.515 ","End":"02:26.210","Text":"If k is 0,"},{"Start":"02:26.210 ","End":"02:31.415","Text":"the scalar 0, then k times u is 0 times u."},{"Start":"02:31.415 ","End":"02:33.905","Text":"Here I have 0 times this,"},{"Start":"02:33.905 ","End":"02:38.450","Text":"and this gives us that T of 0 equals 0,"},{"Start":"02:38.450 ","End":"02:39.920","Text":"which is what we wanted."},{"Start":"02:39.920 ","End":"02:43.250","Text":"Just going to say that it doesn\u0027t matter what you choose as u,"},{"Start":"02:43.250 ","End":"02:47.484","Text":"just choose any u in the vector space."},{"Start":"02:47.484 ","End":"02:50.805","Text":"It doesn\u0027t matter, we get that."},{"Start":"02:50.805 ","End":"02:52.085","Text":"Now, this was so short."},{"Start":"02:52.085 ","End":"02:54.835","Text":"I\u0027ll show you another proof if you want."},{"Start":"02:54.835 ","End":"02:56.825","Text":"In the alternate proof,"},{"Start":"02:56.825 ","End":"03:03.904","Text":"I start off with saying that because 0 plus 0 is 0 for vectors also,"},{"Start":"03:03.904 ","End":"03:07.765","Text":"then T of this equals T of this,"},{"Start":"03:07.765 ","End":"03:10.700","Text":"and since we\u0027re assuming our T is linear,"},{"Start":"03:10.700 ","End":"03:15.815","Text":"the left-hand side can be broken up into T of 0 plus T of 0,"},{"Start":"03:15.815 ","End":"03:18.105","Text":"so we get this."},{"Start":"03:18.105 ","End":"03:24.350","Text":"Now we can cancel from both sides or subtract T of 0 from both sides."},{"Start":"03:24.350 ","End":"03:28.855","Text":"Once again, we get that T of 0 equals 0."},{"Start":"03:28.855 ","End":"03:31.080","Text":"Useful little theorem,"},{"Start":"03:31.080 ","End":"03:34.900","Text":"and an easy proof, and we\u0027re done."}],"ID":10228},{"Watched":false,"Name":"Exercise 6","Duration":"4m 45s","ChapterTopicVideoID":9679,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Before I read the exercise,"},{"Start":"00:01.800 ","End":"00:04.090","Text":"I just want to remind you what is this,"},{"Start":"00:04.090 ","End":"00:08.490","Text":"M sub n [R.] This is"},{"Start":"00:08.490 ","End":"00:15.165","Text":"the space of n by n matrices over the field of real numbers."},{"Start":"00:15.165 ","End":"00:21.765","Text":"It\u0027s a vector space and its dimension is n squared, but that\u0027s irrelevant."},{"Start":"00:21.765 ","End":"00:25.784","Text":"We\u0027re given a transformation from this space to itself,"},{"Start":"00:25.784 ","End":"00:28.125","Text":"and it\u0027s defined as follows."},{"Start":"00:28.125 ","End":"00:31.890","Text":"What the transformation does to a matrix A,"},{"Start":"00:31.890 ","End":"00:34.140","Text":"A, is variable is,"},{"Start":"00:34.140 ","End":"00:37.215","Text":"it sends it to BA plus AB,"},{"Start":"00:37.215 ","End":"00:42.205","Text":"where B is a fixed matrix from this space."},{"Start":"00:42.205 ","End":"00:47.720","Text":"In other words, for each A we get T of A different each time,"},{"Start":"00:47.720 ","End":"00:49.235","Text":"B is the same."},{"Start":"00:49.235 ","End":"00:55.870","Text":"Our question as usual is to decide if this transformation is linear or not."},{"Start":"00:55.870 ","End":"00:58.280","Text":"By the way, if it is a linear transformation,"},{"Start":"00:58.280 ","End":"01:00.560","Text":"it\u0027s also called a linear operator because"},{"Start":"01:00.560 ","End":"01:03.050","Text":"when it\u0027s from a space to itself, it\u0027s an operator."},{"Start":"01:03.050 ","End":"01:05.860","Text":"Oh, well, let\u0027s begin."},{"Start":"01:05.860 ","End":"01:12.830","Text":"Now the first rule we\u0027ve gotten very used to saying T of u plus v is T of u plus T of v,"},{"Start":"01:12.830 ","End":"01:16.400","Text":"but we could choose different letters for matrices,"},{"Start":"01:16.400 ","End":"01:22.270","Text":"the letters u and v don\u0027t suit me and we\u0027ll use capital A and capital C instead,"},{"Start":"01:22.270 ","End":"01:24.350","Text":"so this the question."},{"Start":"01:24.350 ","End":"01:29.089","Text":"Does this equal this for all A and C in this space?"},{"Start":"01:29.089 ","End":"01:34.110","Text":"I just copied the formula for T because I\u0027m going to scroll."},{"Start":"01:35.420 ","End":"01:37.890","Text":"This is a general A,"},{"Start":"01:37.890 ","End":"01:40.265","Text":"like the variable X and here we have"},{"Start":"01:40.265 ","End":"01:44.225","Text":"our own particular choice of A and C and at the same formula,"},{"Start":"01:44.225 ","End":"01:46.190","Text":"copied this here, but with C,"},{"Start":"01:46.190 ","End":"01:51.405","Text":"I replace A by C. T of C is BC plus CB."},{"Start":"01:51.405 ","End":"01:54.395","Text":"Then if I apply the formula,"},{"Start":"01:54.395 ","End":"01:58.230","Text":"replacing the variable with A plus C,"},{"Start":"01:58.230 ","End":"02:03.720","Text":"then it\u0027s just B times A plus C plus A plus C times"},{"Start":"02:03.720 ","End":"02:08.970","Text":"B. T of thingy is B times thingy plus thingy times B,"},{"Start":"02:08.970 ","End":"02:10.665","Text":"doesn\u0027t matter what that thing is,"},{"Start":"02:10.665 ","End":"02:12.730","Text":"it\u0027s A plus C here."},{"Start":"02:12.730 ","End":"02:17.735","Text":"Now it\u0027s not obvious if this plus this is equal to this,"},{"Start":"02:17.735 ","End":"02:20.930","Text":"but let\u0027s expand using"},{"Start":"02:20.930 ","End":"02:26.615","Text":"the distributive law and now I think it is clear that if we take this plus this,"},{"Start":"02:26.615 ","End":"02:33.680","Text":"it will be equal to this because all the terms are here, possibly reversed."},{"Start":"02:33.680 ","End":"02:35.765","Text":"Let\u0027s see, BA is BA,"},{"Start":"02:35.765 ","End":"02:39.300","Text":"AB is AB, BC is BC,"},{"Start":"02:39.300 ","End":"02:40.620","Text":"and CB is CB, yeah,"},{"Start":"02:40.620 ","End":"02:42.735","Text":"exactly as is nothing, yeah,"},{"Start":"02:42.735 ","End":"02:45.705","Text":"1 for 1 accounted for so yes."},{"Start":"02:45.705 ","End":"02:52.950","Text":"Number 1, axiom is being demonstrated but remember there\u0027s also axiom 2."},{"Start":"02:52.950 ","End":"02:54.690","Text":"Wait before axiom 2,"},{"Start":"02:54.690 ","End":"02:56.090","Text":"we have 1 more line."},{"Start":"02:56.090 ","End":"02:59.990","Text":"I just wanted to stress that T of A plus C is"},{"Start":"02:59.990 ","End":"03:03.905","Text":"equal to T of A plus T of C. This highlighted,"},{"Start":"03:03.905 ","End":"03:06.620","Text":"is this highlighted plus this highlighted."},{"Start":"03:06.620 ","End":"03:10.560","Text":"Now onto axiom 2 and once again,"},{"Start":"03:10.560 ","End":"03:18.500","Text":"the letter u doesn\u0027t suit me for matrices and I\u0027ll replace u with letter A."},{"Start":"03:18.500 ","End":"03:27.180","Text":"What we have to show is T of k times A equals k times T of A,"},{"Start":"03:27.180 ","End":"03:28.695","Text":"that\u0027s what we have to check."},{"Start":"03:28.695 ","End":"03:31.930","Text":"So that\u0027s the second axiom."},{"Start":"03:32.180 ","End":"03:35.565","Text":"Here I\u0027m reminding you of the formula,"},{"Start":"03:35.565 ","End":"03:37.410","Text":"this is for a general A,"},{"Start":"03:37.410 ","End":"03:39.630","Text":"whatever A is,"},{"Start":"03:39.630 ","End":"03:43.940","Text":"so T of A from here is equal to,"},{"Start":"03:43.940 ","End":"03:45.660","Text":"well, I\u0027m just copying it really."},{"Start":"03:45.660 ","End":"03:47.060","Text":"Here it\u0027s a variable and here,"},{"Start":"03:47.060 ","End":"03:57.190","Text":"it\u0027s specific T of A equals BA plus AB and multiplying by k is k times BA plus AB."},{"Start":"03:57.190 ","End":"04:00.620","Text":"On the other hand, if I look at the left-hand side here,"},{"Start":"04:00.620 ","End":"04:02.510","Text":"I want T of kA,"},{"Start":"04:02.510 ","End":"04:04.510","Text":"that\u0027s equal to this,"},{"Start":"04:04.510 ","End":"04:09.555","Text":"B times this plus this times B and that\u0027s what I wrote here."},{"Start":"04:09.555 ","End":"04:12.935","Text":"Now the question is, does this line equal this line?"},{"Start":"04:12.935 ","End":"04:15.280","Text":"Well the right-hand sides, I mean."},{"Start":"04:15.280 ","End":"04:17.865","Text":"If I just pull k out,"},{"Start":"04:17.865 ","End":"04:23.090","Text":"there\u0027s one of those rules that the constants can be pulled anywhere in front so we"},{"Start":"04:23.090 ","End":"04:29.670","Text":"get k times BA plus AB and now we really do have equality."},{"Start":"04:29.670 ","End":"04:36.990","Text":"In general, for any scalar k and any n by n matrix A we have, this is true."},{"Start":"04:37.220 ","End":"04:40.850","Text":"Finally, because both axioms are satisfied,"},{"Start":"04:40.850 ","End":"04:46.110","Text":"we conclude that T is indeed linear. We\u0027re done."}],"ID":10229},{"Watched":false,"Name":"Exercise 7","Duration":"3m 35s","ChapterTopicVideoID":9680,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.200","Text":"In this exercise, we\u0027re given a transformation T from this space to itself."},{"Start":"00:07.200 ","End":"00:08.280","Text":"What is this space?"},{"Start":"00:08.280 ","End":"00:14.230","Text":"It\u0027s the n by n matrices over the real numbers."},{"Start":"00:15.260 ","End":"00:24.720","Text":"We define it by T of A is equal to A plus A transpose. This makes sense."},{"Start":"00:24.720 ","End":"00:26.500","Text":"This is a square n by n matrix."},{"Start":"00:26.500 ","End":"00:29.565","Text":"A transpose is also a square n by n matrix,"},{"Start":"00:29.565 ","End":"00:31.230","Text":"same size, we can add them."},{"Start":"00:31.230 ","End":"00:34.455","Text":"No problem. Let\u0027s see if it\u0027s linear."},{"Start":"00:34.455 ","End":"00:38.620","Text":"I\u0027m going to check the 2 axioms or the 2 conditions."},{"Start":"00:38.620 ","End":"00:41.435","Text":"I don\u0027t think letters,"},{"Start":"00:41.435 ","End":"00:44.120","Text":"lowercase, u and v are suitable for matrices."},{"Start":"00:44.120 ","End":"00:49.865","Text":"I prefer uppercase and closer to the beginning of the alphabet."},{"Start":"00:49.865 ","End":"00:55.350","Text":"Let\u0027s rephrase rule 1 as T of A plus C equals T of"},{"Start":"00:55.350 ","End":"01:01.010","Text":"A plus T of C for all A and C in this space."},{"Start":"01:01.010 ","End":"01:03.185","Text":"Let\u0027s check if this is so."},{"Start":"01:03.185 ","End":"01:08.580","Text":"I just copied the formula before I scroll it off-screen."},{"Start":"01:09.230 ","End":"01:11.975","Text":"This is the general formula,"},{"Start":"01:11.975 ","End":"01:14.330","Text":"just a dummy variable A,"},{"Start":"01:14.330 ","End":"01:17.680","Text":"but we can apply it to our particular A and C. T"},{"Start":"01:17.680 ","End":"01:21.210","Text":"of A is A plus A-transpose, basically a copy."},{"Start":"01:21.210 ","End":"01:24.810","Text":"T of C equals C plus C transpose."},{"Start":"01:24.810 ","End":"01:30.200","Text":"T of something, is that something plus the something transpose."},{"Start":"01:30.200 ","End":"01:32.150","Text":"That will work for A plus C also."},{"Start":"01:32.150 ","End":"01:35.850","Text":"A plus C and A plus C transpose."},{"Start":"01:37.400 ","End":"01:43.265","Text":"Drop the brackets here and use the property of the transpose."},{"Start":"01:43.265 ","End":"01:46.640","Text":"The transpose of a sum is the sum of the transposes."},{"Start":"01:46.640 ","End":"01:52.175","Text":"Now we can see that this plus this is"},{"Start":"01:52.175 ","End":"01:57.605","Text":"equal to this which means that this equals this,"},{"Start":"01:57.605 ","End":"02:01.250","Text":"which is exactly what we were trying to prove up here."},{"Start":"02:01.250 ","End":"02:03.320","Text":"Rule number 1,"},{"Start":"02:03.320 ","End":"02:05.690","Text":"axiom number 1, we\u0027re okay with."},{"Start":"02:05.690 ","End":"02:08.970","Text":"We still have the second axiom."},{"Start":"02:09.290 ","End":"02:11.690","Text":"Here it is and once again,"},{"Start":"02:11.690 ","End":"02:16.415","Text":"I\u0027ve replaced the letter U with the capital A."},{"Start":"02:16.415 ","End":"02:23.355","Text":"It\u0027s more suitable for a matrix notation."},{"Start":"02:23.355 ","End":"02:31.690","Text":"To remind you, this was the formula T of anything is that anything plus its transpose."},{"Start":"02:31.690 ","End":"02:34.440","Text":"Let\u0027s work on the right-hand side first,"},{"Start":"02:34.440 ","End":"02:38.400","Text":"k times T of A. T of A is this."},{"Start":"02:38.400 ","End":"02:42.745","Text":"A plus A transpose times k is k times this."},{"Start":"02:42.745 ","End":"02:45.140","Text":"I didn\u0027t expand here, I left it like this."},{"Start":"02:45.140 ","End":"02:47.015","Text":"Let\u0027s go to the left-hand side,"},{"Start":"02:47.015 ","End":"02:55.090","Text":"which is this T of kA is that kA plus its transpose."},{"Start":"02:55.090 ","End":"02:57.740","Text":"By the rules of transpose,"},{"Start":"02:57.740 ","End":"02:59.225","Text":"if you have a constant,"},{"Start":"02:59.225 ","End":"03:01.250","Text":"a scalar times a matrix,"},{"Start":"03:01.250 ","End":"03:02.480","Text":"when you take the transpose,"},{"Start":"03:02.480 ","End":"03:05.840","Text":"you can just take the scalar outside."},{"Start":"03:05.840 ","End":"03:07.855","Text":"This is what we get."},{"Start":"03:07.855 ","End":"03:13.310","Text":"At this point, I think it\u0027s easy to see that this is equal to this."},{"Start":"03:13.310 ","End":"03:20.110","Text":"Just have to expand using the distributive law."},{"Start":"03:20.110 ","End":"03:25.590","Text":"We\u0027ve proven this which is what we wanted to prove up here."},{"Start":"03:26.030 ","End":"03:29.415","Text":"Now we\u0027ve proven everything we need."},{"Start":"03:29.415 ","End":"03:35.850","Text":"We conclude that T is linear and we\u0027re done."}],"ID":10230},{"Watched":false,"Name":"Exercise 8","Duration":"3m 10s","ChapterTopicVideoID":9681,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.070","Text":"In this exercise, we have a transformation T which goes"},{"Start":"00:05.070 ","End":"00:10.185","Text":"from the M by n matrices over R to the same space."},{"Start":"00:10.185 ","End":"00:12.720","Text":"It\u0027s given as follows;"},{"Start":"00:12.720 ","End":"00:21.210","Text":"transformation sends a matrix A to its determinant times the identity matrix."},{"Start":"00:21.210 ","End":"00:25.980","Text":"Let\u0027s see if the 2 axioms are satisfied."},{"Start":"00:25.980 ","End":"00:29.310","Text":"The first one, which, well,"},{"Start":"00:29.310 ","End":"00:32.820","Text":"we used to say that T of u plus v is T of u plus T of v,"},{"Start":"00:32.820 ","End":"00:36.600","Text":"but I prefer to use different letters for matrices,"},{"Start":"00:36.600 ","End":"00:43.710","Text":"uppercase letters, we\u0027ll call them A and C instead of u and v. Question is,"},{"Start":"00:43.710 ","End":"00:46.470","Text":"is this always true?"},{"Start":"00:46.470 ","End":"00:50.235","Text":"I just copied the formula."},{"Start":"00:50.235 ","End":"00:53.085","Text":"Now we can plug in, first of all,"},{"Start":"00:53.085 ","End":"00:55.350","Text":"A equals A and secondly,"},{"Start":"00:55.350 ","End":"00:57.045","Text":"A equals C,"},{"Start":"00:57.045 ","End":"00:59.325","Text":"so we get this,"},{"Start":"00:59.325 ","End":"01:02.625","Text":"and that will give us the right-hand side when we add them."},{"Start":"01:02.625 ","End":"01:08.690","Text":"The left-hand side will give us the determinant of"},{"Start":"01:08.690 ","End":"01:15.485","Text":"this times I as A could be replaced by A plus C. Now,"},{"Start":"01:15.485 ","End":"01:20.315","Text":"the question is, is this plus this equal this?"},{"Start":"01:20.315 ","End":"01:21.980","Text":"It doesn\u0027t look like it."},{"Start":"01:21.980 ","End":"01:23.930","Text":"I mean, if you ignore the I part,"},{"Start":"01:23.930 ","End":"01:26.570","Text":"it really basically says that determinant of"},{"Start":"01:26.570 ","End":"01:29.540","Text":"a plus determinant of C equals determinant of"},{"Start":"01:29.540 ","End":"01:35.870","Text":"A plus C. This is definitely not true in general,"},{"Start":"01:35.870 ","End":"01:38.240","Text":"it could be true in a specific case,"},{"Start":"01:38.240 ","End":"01:40.450","Text":"it\u0027s not true in general."},{"Start":"01:40.450 ","End":"01:43.940","Text":"It\u0027s customary to produce a counterexample,"},{"Start":"01:43.940 ","End":"01:46.500","Text":"not just to leave it at that."},{"Start":"01:46.540 ","End":"01:50.500","Text":"Here are some 2 by 2 examples."},{"Start":"01:50.500 ","End":"01:53.655","Text":"We\u0027ll take A as 1, 2, 3, 4,"},{"Start":"01:53.655 ","End":"01:55.530","Text":"I played with some numbers,"},{"Start":"01:55.530 ","End":"01:59.040","Text":"I took C as 1, 2, 3, 5."},{"Start":"01:59.040 ","End":"02:01.380","Text":"Then A plus C,"},{"Start":"02:01.380 ","End":"02:04.305","Text":"just add each component."},{"Start":"02:04.305 ","End":"02:06.705","Text":"We get 2, 4, 7, 9."},{"Start":"02:06.705 ","End":"02:10.420","Text":"Now, let\u0027s check the determinants."},{"Start":"02:10.690 ","End":"02:16.805","Text":"The determinant of A is this diagonal minus this diagonal in the product,"},{"Start":"02:16.805 ","End":"02:20.875","Text":"4 minus 6 is minus 2."},{"Start":"02:20.875 ","End":"02:22.820","Text":"I take the minus 2,"},{"Start":"02:22.820 ","End":"02:24.670","Text":"I multiply by I."},{"Start":"02:24.670 ","End":"02:27.720","Text":"This is the identity matrix for 2 by 2,"},{"Start":"02:27.720 ","End":"02:30.855","Text":"which is when we get this."},{"Start":"02:30.855 ","End":"02:33.690","Text":"Next, for the case of C,"},{"Start":"02:33.690 ","End":"02:36.630","Text":"1 times 5 minus 2 times 3,"},{"Start":"02:36.630 ","End":"02:43.250","Text":"5 minus 6 is minus 1 times identity matrix and we get this."},{"Start":"02:43.400 ","End":"02:46.400","Text":"Then if we figure this,"},{"Start":"02:46.400 ","End":"02:49.655","Text":"we need the determinant of A plus C, which is what,"},{"Start":"02:49.655 ","End":"02:58.260","Text":"18 minus 28, which is minus 10 times I, which is this."},{"Start":"02:58.670 ","End":"03:03.875","Text":"Well, certainly this plus this does not equal this."},{"Start":"03:03.875 ","End":"03:10.590","Text":"We are not linear because already we failed the first axiom. Okay, we\u0027re done."}],"ID":10231},{"Watched":false,"Name":"Exercise 9","Duration":"3m 52s","ChapterTopicVideoID":9682,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.530","Text":"In this exercise, we have a transformation T,"},{"Start":"00:04.530 ","End":"00:09.000","Text":"which goes from M to the space of n"},{"Start":"00:09.000 ","End":"00:13.995","Text":"by n matrices over the real numbers, and it\u0027s defined."},{"Start":"00:13.995 ","End":"00:16.305","Text":"What it does to A,"},{"Start":"00:16.305 ","End":"00:24.130","Text":"A is a square matrix is it takes A and multiplies it by a transpose."},{"Start":"00:24.290 ","End":"00:29.040","Text":"Now to show that this is a linear transformation,"},{"Start":"00:29.040 ","End":"00:33.160","Text":"we have to satisfy the 2 axioms."},{"Start":"00:34.010 ","End":"00:38.980","Text":"I\u0027m actually going to start with the second one for a change."},{"Start":"00:38.980 ","End":"00:41.255","Text":"As I usually do,"},{"Start":"00:41.255 ","End":"00:45.890","Text":"instead of using u and v like in the formula,"},{"Start":"00:45.890 ","End":"00:48.800","Text":"I like to use A and C. In fact,"},{"Start":"00:48.800 ","End":"00:51.155","Text":"you don\u0027t even need C here."},{"Start":"00:51.155 ","End":"00:55.029","Text":"We\u0027ll just take the second axiom."},{"Start":"00:55.029 ","End":"00:59.385","Text":"That T of k scalar times A,"},{"Start":"00:59.385 ","End":"01:07.355","Text":"is it equal to k times T of A. I\u0027ve just copied the formula so I don\u0027t lose it."},{"Start":"01:07.355 ","End":"01:12.050","Text":"As we already said, T of A is a times A transpose."},{"Start":"01:12.050 ","End":"01:16.490","Text":"If I multiply that by k to get what I have on the right-hand side,"},{"Start":"01:16.490 ","End":"01:20.580","Text":"I\u0027ll get k times A, A Transpose."},{"Start":"01:20.580 ","End":"01:25.040","Text":"On the other hand, if I want to compute T of kA,"},{"Start":"01:25.040 ","End":"01:33.415","Text":"it\u0027s kA, that\u0027s like the unit as a whole times its transpose."},{"Start":"01:33.415 ","End":"01:37.630","Text":"Now here I\u0027m doing 2 steps in 1."},{"Start":"01:38.120 ","End":"01:44.045","Text":"I\u0027m using the fact that if I have a scalar times a matrix and I take the transpose,"},{"Start":"01:44.045 ","End":"01:46.880","Text":"I can just take the scalar outside,"},{"Start":"01:46.880 ","End":"01:50.330","Text":"so I take k out and it combines with this k,"},{"Start":"01:50.330 ","End":"01:54.490","Text":"so we get k squared A, A transpose."},{"Start":"01:54.490 ","End":"02:00.020","Text":"Now in general, this will not equal to this not be equal to,"},{"Start":"02:00.020 ","End":"02:03.200","Text":"because the main difference is here I have a k,"},{"Start":"02:03.200 ","End":"02:05.770","Text":"here I have a k squared."},{"Start":"02:05.770 ","End":"02:07.670","Text":"In general they won\u0027t be,"},{"Start":"02:07.670 ","End":"02:12.250","Text":"but it\u0027s customary to give a counter-example."},{"Start":"02:12.250 ","End":"02:16.820","Text":"I\u0027ll make life easy by taking a to be the identity matrix."},{"Start":"02:16.820 ","End":"02:22.700","Text":"For k, I could take almost anything as long as k is not equal to k squared."},{"Start":"02:22.700 ","End":"02:24.740","Text":"For example, k equals 3,"},{"Start":"02:24.740 ","End":"02:27.280","Text":"3 is not equal to 3 squared."},{"Start":"02:27.280 ","End":"02:31.360","Text":"If I do that, let\u0027s see."},{"Start":"02:31.910 ","End":"02:36.300","Text":"T of A is A times A transpose."},{"Start":"02:36.300 ","End":"02:39.060","Text":"But A is transpose is A itself,"},{"Start":"02:39.060 ","End":"02:41.610","Text":"it\u0027s a symmetric matrix."},{"Start":"02:41.610 ","End":"02:43.580","Text":"It\u0027s just identity times identity,"},{"Start":"02:43.580 ","End":"02:45.670","Text":"which is the identity."},{"Start":"02:45.670 ","End":"02:51.615","Text":"If I apply T to 3A its this matrix."},{"Start":"02:51.615 ","End":"02:54.540","Text":"3A is 3 times1,0,0,1,"},{"Start":"02:54.540 ","End":"02:56.055","Text":"which is 3,0,0,3,"},{"Start":"02:56.055 ","End":"02:58.410","Text":"and it is its own transpose."},{"Start":"02:58.410 ","End":"03:00.570","Text":"You can almost throw out the T here."},{"Start":"03:00.570 ","End":"03:03.800","Text":"This times this is this."},{"Start":"03:03.800 ","End":"03:05.180","Text":"If you do the computation,"},{"Start":"03:05.180 ","End":"03:08.120","Text":"actually when you have just numbers along the diagonal,"},{"Start":"03:08.120 ","End":"03:13.670","Text":"you can multiply term-wise for diagonal matrices,"},{"Start":"03:13.670 ","End":"03:17.250","Text":"3 times 3 is 9 and 3 times 3 is 9."},{"Start":"03:17.510 ","End":"03:22.630","Text":"Certainly 3 times T of A is not equal to T of 3A,"},{"Start":"03:22.630 ","End":"03:25.175","Text":"because if I multiply this by 3,"},{"Start":"03:25.175 ","End":"03:29.465","Text":"I would get 3,0,0,3,"},{"Start":"03:29.465 ","End":"03:34.190","Text":"which is certainly not the same as 9,0,0,9."},{"Start":"03:34.190 ","End":"03:39.335","Text":"We conclude that T is not linear because it violates axiom 2."},{"Start":"03:39.335 ","End":"03:41.900","Text":"It also violates axiom 1."},{"Start":"03:41.900 ","End":"03:45.890","Text":"It was just for a change and it actually is easier"},{"Start":"03:45.890 ","End":"03:52.410","Text":"to use ruler axiom number 2 here. We\u0027re done."}],"ID":10232},{"Watched":false,"Name":"Exercise 10","Duration":"4m 10s","ChapterTopicVideoID":9683,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.160","Text":"In this exercise, we have a transformation from this space to itself."},{"Start":"00:05.160 ","End":"00:07.185","Text":"What do I mean by \u0027this space\u0027?"},{"Start":"00:07.185 ","End":"00:15.855","Text":"This means the n by n square matrices over the real numbers."},{"Start":"00:15.855 ","End":"00:20.220","Text":"T, the transformation is given by T of A is"},{"Start":"00:20.220 ","End":"00:24.900","Text":"A cubed with matrices or just means A times A times A."},{"Start":"00:24.900 ","End":"00:31.360","Text":"The question is, is this a linear transformation?"},{"Start":"00:32.000 ","End":"00:35.220","Text":"Gut feeling says no."},{"Start":"00:35.220 ","End":"00:41.520","Text":"Because let\u0027s say to double something and then cube it or to cube it and then double it,"},{"Start":"00:41.520 ","End":"00:43.785","Text":"probably doesn\u0027t come out the same."},{"Start":"00:43.785 ","End":"00:46.040","Text":"Usually, if you suspect that it\u0027s not,"},{"Start":"00:46.040 ","End":"00:49.385","Text":"axiom 2 is the easier one."},{"Start":"00:49.385 ","End":"00:53.405","Text":"Any event, let\u0027s see if axiom 2 holds."},{"Start":"00:53.405 ","End":"00:57.050","Text":"I\u0027ll just say that our axioms were phrased in terms"},{"Start":"00:57.050 ","End":"00:59.900","Text":"of vectors u and v. But for matrices,"},{"Start":"00:59.900 ","End":"01:04.045","Text":"I\u0027m going to use A and C as needed."},{"Start":"01:04.045 ","End":"01:09.895","Text":"The second axiom translates to T of kA equals kT of A."},{"Start":"01:09.895 ","End":"01:11.415","Text":"Because we don\u0027t know this yet,"},{"Start":"01:11.415 ","End":"01:12.490","Text":"hence, the question mark."},{"Start":"01:12.490 ","End":"01:15.715","Text":"We\u0027re going to try and see if this is true or not."},{"Start":"01:15.715 ","End":"01:20.955","Text":"I just copied the T so that when I scroll, it doesn\u0027t vanish."},{"Start":"01:20.955 ","End":"01:24.645","Text":"We\u0027ll need to apply T to A and to kA."},{"Start":"01:24.645 ","End":"01:26.070","Text":"Well, T of A,"},{"Start":"01:26.070 ","End":"01:28.455","Text":"we already have, it\u0027s just this."},{"Start":"01:28.455 ","End":"01:30.290","Text":"To get the right-hand side,"},{"Start":"01:30.290 ","End":"01:32.465","Text":"if I multiply both sides by k,"},{"Start":"01:32.465 ","End":"01:35.605","Text":"kT of A equals kA cubed."},{"Start":"01:35.605 ","End":"01:39.245","Text":"For the left-hand side, I\u0027m going to apply T to kA,"},{"Start":"01:39.245 ","End":"01:43.565","Text":"T of anything is that something cubed."},{"Start":"01:43.565 ","End":"01:47.310","Text":"So T of kA is kA cubed."},{"Start":"01:47.960 ","End":"01:51.665","Text":"This will equal k cubed A cubed."},{"Start":"01:51.665 ","End":"01:53.690","Text":"If you\u0027re not quite sure of this,"},{"Start":"01:53.690 ","End":"01:56.480","Text":"you could just see it by writing it as kA,"},{"Start":"01:56.480 ","End":"02:05.270","Text":"kA, kA, and in each case, you can pull constants in front until you get k, k,"},{"Start":"02:05.270 ","End":"02:10.675","Text":"k, A, A, A, which is k cubed A cubed."},{"Start":"02:10.675 ","End":"02:16.520","Text":"Notice that this and this are different expressions,"},{"Start":"02:16.520 ","End":"02:18.080","Text":"they\u0027re not the same."},{"Start":"02:18.080 ","End":"02:20.839","Text":"Well, they could be the same."},{"Start":"02:20.839 ","End":"02:24.140","Text":"For example, if I chose k equals 1,"},{"Start":"02:24.140 ","End":"02:27.455","Text":"then they would, but they\u0027re not always equal."},{"Start":"02:27.455 ","End":"02:30.500","Text":"Now, when we want to show that something\u0027s not always equal,"},{"Start":"02:30.500 ","End":"02:33.590","Text":"we usually provide a counterexample."},{"Start":"02:33.590 ","End":"02:37.430","Text":"Yeah, just wanted to jot down that this is not equal to this."},{"Start":"02:37.430 ","End":"02:38.810","Text":"This is not equal to this,"},{"Start":"02:38.810 ","End":"02:43.070","Text":"but we\u0027re still not completely sure because we want to find a counterexample."},{"Start":"02:43.070 ","End":"02:46.120","Text":"Doesn\u0027t look the same, but you never know."},{"Start":"02:46.120 ","End":"02:48.965","Text":"I\u0027ll take whatever\u0027s easy to work with."},{"Start":"02:48.965 ","End":"02:53.245","Text":"A take the identity 2 by 2 matrix."},{"Start":"02:53.245 ","End":"02:55.155","Text":"For k, well,"},{"Start":"02:55.155 ","End":"02:57.680","Text":"I\u0027m not going to take k equals 1, that won\u0027t do it."},{"Start":"02:57.680 ","End":"03:00.030","Text":"Take k equals 2."},{"Start":"03:00.910 ","End":"03:04.970","Text":"Let\u0027s compute both sides, see what we get."},{"Start":"03:04.970 ","End":"03:09.750","Text":"T of A is just A cubed,"},{"Start":"03:09.750 ","End":"03:15.170","Text":"which is this cubed and the identity matrix cubed is just itself."},{"Start":"03:15.170 ","End":"03:17.120","Text":"That\u0027s T of A."},{"Start":"03:17.120 ","End":"03:19.980","Text":"Now, I need kA, which is,"},{"Start":"03:19.980 ","End":"03:21.735","Text":"in this case, 2A,"},{"Start":"03:21.735 ","End":"03:27.585","Text":"just comes out to be 2 and 2 here so that the 1 on the 1."},{"Start":"03:27.585 ","End":"03:31.035","Text":"Now, I want to apply T to that."},{"Start":"03:31.035 ","End":"03:34.440","Text":"T of something is that something cubed."},{"Start":"03:34.440 ","End":"03:37.540","Text":"This gives us this."},{"Start":"03:38.510 ","End":"03:41.990","Text":"Note that twice this is not equal to this."},{"Start":"03:41.990 ","End":"03:49.255","Text":"Perhaps I also should have written that twice T of A is twice this,"},{"Start":"03:49.255 ","End":"03:53.025","Text":"which is just 2, 0, 0, 2."},{"Start":"03:53.025 ","End":"03:55.650","Text":"Now, certainly, these 2 are not equal,"},{"Start":"03:55.650 ","End":"04:00.375","Text":"which means that T of 2A is not equal to 2T of A."},{"Start":"04:00.375 ","End":"04:03.875","Text":"This implies that T is not linear."},{"Start":"04:03.875 ","End":"04:05.015","Text":"Because if it were linear,"},{"Start":"04:05.015 ","End":"04:09.750","Text":"we would get equality here and we\u0027re done."}],"ID":10233},{"Watched":false,"Name":"Exercise 11","Duration":"4m 12s","ChapterTopicVideoID":9684,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.900","Text":"In this exercise, we have a transformation from this space to this space."},{"Start":"00:06.900 ","End":"00:13.005","Text":"Now this is the space of polynomials of degree up to 3,"},{"Start":"00:13.005 ","End":"00:16.605","Text":"and this is polynomials up to degree 2."},{"Start":"00:16.605 ","End":"00:21.725","Text":"What it does is it takes a degree 3 polynomial,"},{"Start":"00:21.725 ","End":"00:26.320","Text":"3 or less, because maybe d or c is 0."},{"Start":"00:26.320 ","End":"00:29.420","Text":"It basically just truncates it,"},{"Start":"00:29.420 ","End":"00:36.125","Text":"chops off the x cubed term and gives us just the first 3 terms."},{"Start":"00:36.125 ","End":"00:38.210","Text":"It\u0027s not part of the exercise."},{"Start":"00:38.210 ","End":"00:39.830","Text":"Let me just give you a feel for it."},{"Start":"00:39.830 ","End":"00:49.675","Text":"Suppose I say, what is T of 1 plus 2x plus 3x squared plus 4x cubed?"},{"Start":"00:49.675 ","End":"00:53.280","Text":"The answer would be just throw out the 4x cubed,"},{"Start":"00:53.280 ","End":"00:58.980","Text":"and the answer is 1 plus 2x plus 3x squared."},{"Start":"00:58.980 ","End":"01:04.360","Text":"Let\u0027s see if it\u0027s linear by checking the 2 axioms."},{"Start":"01:05.440 ","End":"01:11.315","Text":"The one with the u plus v and the one with k times u."},{"Start":"01:11.315 ","End":"01:16.945","Text":"Often do I make a preparation before we start."},{"Start":"01:16.945 ","End":"01:19.215","Text":"Just take a typical u,"},{"Start":"01:19.215 ","End":"01:20.625","Text":"this might be it,"},{"Start":"01:20.625 ","End":"01:24.120","Text":"a typical v. Let\u0027s use uppercase A, B, C,"},{"Start":"01:24.120 ","End":"01:28.140","Text":"D. Then I want u plus v and ku."},{"Start":"01:28.140 ","End":"01:30.240","Text":"Here they are."},{"Start":"01:30.240 ","End":"01:32.625","Text":"We start with the first axiom."},{"Start":"01:32.625 ","End":"01:38.945","Text":"I want to know, is T of u plus v equals T of u plus T of v, always?"},{"Start":"01:38.945 ","End":"01:41.740","Text":"There\u0027s a question mark, we\u0027re going to check this."},{"Start":"01:41.740 ","End":"01:47.515","Text":"I just copied the exercise because I\u0027m going to scroll, so we don\u0027t lose it."},{"Start":"01:47.515 ","End":"01:53.460","Text":"If I apply T to u and u is written here,"},{"Start":"01:53.460 ","End":"01:55.290","Text":"and here\u0027s the formula,"},{"Start":"01:55.290 ","End":"01:58.080","Text":"it\u0027s just as written here,"},{"Start":"01:58.080 ","End":"02:01.035","Text":"a plus bx plus cx squared."},{"Start":"02:01.035 ","End":"02:03.950","Text":"T of v is very similar."},{"Start":"02:03.950 ","End":"02:08.585","Text":"Again, take this polynomial and chop off the cube term,"},{"Start":"02:08.585 ","End":"02:11.600","Text":"A plus Bx plus Cx squared with uppercase A, B,"},{"Start":"02:11.600 ","End":"02:18.470","Text":"C. Next we\u0027re going to apply T to u plus v. U plus v is written here,"},{"Start":"02:18.470 ","End":"02:20.825","Text":"so I copied that here."},{"Start":"02:20.825 ","End":"02:24.380","Text":"We just chop off the x cubed term."},{"Start":"02:24.380 ","End":"02:30.860","Text":"We\u0027re left with the first 3 terms."},{"Start":"02:30.860 ","End":"02:33.060","Text":"We have this."},{"Start":"02:33.500 ","End":"02:42.780","Text":"I think it\u0027s fairly clear that this plus this is equal to this,"},{"Start":"02:42.780 ","End":"02:44.955","Text":"a plus A,"},{"Start":"02:44.955 ","End":"02:47.665","Text":"and the x term b plus B,"},{"Start":"02:47.665 ","End":"02:49.790","Text":"and collect the x squareds,"},{"Start":"02:49.790 ","End":"02:52.985","Text":"little c plus big C. We\u0027re okay with"},{"Start":"02:52.985 ","End":"02:59.745","Text":"the u plus v. We\u0027ve answered this question with a yes."},{"Start":"02:59.745 ","End":"03:05.330","Text":"Now we want to move on to the second axiom or condition."},{"Start":"03:05.330 ","End":"03:08.135","Text":"Let me just go to it."},{"Start":"03:08.135 ","End":"03:17.525","Text":"This time we have to check if T of k times u is equal to k times T of u always."},{"Start":"03:17.525 ","End":"03:24.095","Text":"Again, this definition of T. I\u0027ve applied T to both u,"},{"Start":"03:24.095 ","End":"03:27.630","Text":"which we had earlier on also."},{"Start":"03:27.940 ","End":"03:33.090","Text":"K times this would be k times this."},{"Start":"03:33.110 ","End":"03:35.340","Text":"T of ku, well,"},{"Start":"03:35.340 ","End":"03:38.820","Text":"we computed ku just above."},{"Start":"03:38.820 ","End":"03:40.785","Text":"This is what it is."},{"Start":"03:40.785 ","End":"03:42.615","Text":"T of that means,"},{"Start":"03:42.615 ","End":"03:44.550","Text":"chop off the x cubed term,"},{"Start":"03:44.550 ","End":"03:47.130","Text":"so we\u0027re left with this."},{"Start":"03:47.130 ","End":"03:49.250","Text":"Now the question is,"},{"Start":"03:49.250 ","End":"03:51.970","Text":"does this equal this?"},{"Start":"03:51.970 ","End":"03:57.425","Text":"I think it\u0027s fairly clear that it does just apply the distributive law."},{"Start":"03:57.425 ","End":"04:02.120","Text":"We can say that T of ku is kT of u,"},{"Start":"04:02.120 ","End":"04:06.205","Text":"and that\u0027s the second axiom satisfied what we asked here."},{"Start":"04:06.205 ","End":"04:12.700","Text":"Therefore, we may conclude that T indeed is linear. We\u0027re done."}],"ID":10234},{"Watched":false,"Name":"Exercise 12","Duration":"4m 20s","ChapterTopicVideoID":9685,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.400","Text":"In this exercise, we are given a transformation from this space to itself."},{"Start":"00:05.400 ","End":"00:13.105","Text":"This space is the space of polynomials of degree n or less over the reals."},{"Start":"00:13.105 ","End":"00:19.410","Text":"What T does to a polynomial p of x is it sends it"},{"Start":"00:19.410 ","End":"00:27.330","Text":"to that same polynomial with x plus 1 replacing x. I want to give you an example,"},{"Start":"00:27.330 ","End":"00:29.460","Text":"so you\u0027ll see what it really means."},{"Start":"00:29.460 ","End":"00:35.120","Text":"Let\u0027s take the example where p of x is,"},{"Start":"00:35.120 ","End":"00:42.740","Text":"let\u0027s say x squared plus 4x plus 1."},{"Start":"00:42.740 ","End":"00:48.020","Text":"Then T of p of x would be p of x plus"},{"Start":"00:48.020 ","End":"00:54.720","Text":"1 and this is equal to just put x plus 1 instead of x everywhere,"},{"Start":"00:54.720 ","End":"01:00.690","Text":"so 4x plus 1 plus 1 and then this would equal what?"},{"Start":"01:00.690 ","End":"01:03.555","Text":"X squared plus 2x plus 1."},{"Start":"01:03.555 ","End":"01:09.940","Text":"This bit is 4x plus 4 and then plus 1."},{"Start":"01:10.310 ","End":"01:14.905","Text":"It would equal, if we collect everything together, x squared,"},{"Start":"01:14.905 ","End":"01:17.140","Text":"2x plus 4x is 6x,"},{"Start":"01:17.140 ","End":"01:19.675","Text":"1 and 4 and 1 is 6."},{"Start":"01:19.675 ","End":"01:23.535","Text":"It sends this to this."},{"Start":"01:23.535 ","End":"01:26.705","Text":"We have to show that T is linear,"},{"Start":"01:26.705 ","End":"01:30.680","Text":"which means that it satisfies the 2 axioms."},{"Start":"01:30.680 ","End":"01:34.280","Text":"The axioms are phrased in terms of u and v and I"},{"Start":"01:34.280 ","End":"01:39.125","Text":"prefer to use the letters p and q when we\u0027re dealing with polynomials."},{"Start":"01:39.125 ","End":"01:47.045","Text":"The first axiom for a linearity of T says that T of a sum is the sum of the Ts,"},{"Start":"01:47.045 ","End":"01:51.480","Text":"T of p plus q is T of p plus t of q."},{"Start":"01:51.560 ","End":"01:54.210","Text":"It just copied the definition,"},{"Start":"01:54.210 ","End":"01:56.870","Text":"so if it scrolls off."},{"Start":"01:56.870 ","End":"02:01.795","Text":"Now we apply the definition to p and to q,"},{"Start":"02:01.795 ","End":"02:05.830","Text":"and we get this if p of x goes to p of x plus 1,"},{"Start":"02:05.830 ","End":"02:07.615","Text":"q of x is the qx plus 1."},{"Start":"02:07.615 ","End":"02:10.990","Text":"We also need to apply it to p plus q,"},{"Start":"02:10.990 ","End":"02:13.795","Text":"which would give us this."},{"Start":"02:13.795 ","End":"02:19.480","Text":"Now, I claim that this plus this is equal to"},{"Start":"02:19.480 ","End":"02:25.810","Text":"this and the reason is that the definition of p plus q,"},{"Start":"02:25.810 ","End":"02:27.205","Text":"the sum of polynomials,"},{"Start":"02:27.205 ","End":"02:29.305","Text":"says that if we apply it to anything,"},{"Start":"02:29.305 ","End":"02:31.270","Text":"I don\u0027t want to use x, I want to use a different letter,"},{"Start":"02:31.270 ","End":"02:33.220","Text":"let\u0027s say p plus q of a,"},{"Start":"02:33.220 ","End":"02:39.685","Text":"is defined to be just 1 polynomial applied to a plus the other applied to a,"},{"Start":"02:39.685 ","End":"02:44.515","Text":"and instead of a, I replace it by x plus 1,"},{"Start":"02:44.515 ","End":"02:47.220","Text":"then that should also work."},{"Start":"02:47.220 ","End":"02:53.600","Text":"This plus this is equal to this and that is what we wanted to show for the first axiom."},{"Start":"02:53.600 ","End":"02:56.975","Text":"The second axiom says that if we"},{"Start":"02:56.975 ","End":"03:01.190","Text":"first multiply by a constant and then apply the transformation,"},{"Start":"03:01.190 ","End":"03:05.000","Text":"or first apply the transformation and then multiply by a constant,"},{"Start":"03:05.000 ","End":"03:06.500","Text":"that we get the same thing."},{"Start":"03:06.500 ","End":"03:08.135","Text":"Well, let\u0027s check that."},{"Start":"03:08.135 ","End":"03:11.665","Text":"As a reminder, this is how T is defined."},{"Start":"03:11.665 ","End":"03:14.730","Text":"T of p of x is p of x plus 1."},{"Start":"03:14.730 ","End":"03:16.880","Text":"If I multiply that by k,"},{"Start":"03:16.880 ","End":"03:18.935","Text":"which would be like the right-hand side here,"},{"Start":"03:18.935 ","End":"03:22.205","Text":"I just multiply this by k also."},{"Start":"03:22.205 ","End":"03:28.630","Text":"If I apply T to the left-hand side,"},{"Start":"03:28.630 ","End":"03:31.245","Text":"to k p of x,"},{"Start":"03:31.245 ","End":"03:35.640","Text":"what this does is kp is just a polynomial just as much as"},{"Start":"03:35.640 ","End":"03:40.795","Text":"p. So I take that polynomial and apply it to x plus 1 instead of x."},{"Start":"03:40.795 ","End":"03:48.230","Text":"Now the definition of kp in general is that the polynomial kp,"},{"Start":"03:48.230 ","End":"03:49.790","Text":"if I apply it to something,"},{"Start":"03:49.790 ","End":"03:52.475","Text":"I want to use another letter, say a,"},{"Start":"03:52.475 ","End":"03:56.650","Text":"is equal to k times just applying the polynomial to a,"},{"Start":"03:56.650 ","End":"04:00.740","Text":"it\u0027s definition of a constant scalar times the polynomial."},{"Start":"04:00.740 ","End":"04:04.910","Text":"Here, if I replace a by x plus 1,"},{"Start":"04:04.910 ","End":"04:09.450","Text":"it just tells me that this is equal to this,"},{"Start":"04:09.460 ","End":"04:13.780","Text":"which means that this is equal to this."},{"Start":"04:13.780 ","End":"04:21.360","Text":"We verified both axioms and so T is indeed linear and we are done."}],"ID":10235},{"Watched":false,"Name":"Exercise 13","Duration":"5m 32s","ChapterTopicVideoID":9686,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.490","Text":"In this exercise, I\u0027m going to assume you\u0027ve done some calculus,"},{"Start":"00:05.490 ","End":"00:09.330","Text":"that you know what a derivative is and a second derivative."},{"Start":"00:09.330 ","End":"00:16.230","Text":"Now, we have here a transformation from this space to itself."},{"Start":"00:16.230 ","End":"00:17.295","Text":"What is this space?"},{"Start":"00:17.295 ","End":"00:25.320","Text":"It\u0027s the polynomials of degree n or less over the real numbers."},{"Start":"00:25.320 ","End":"00:28.680","Text":"What the transformation does is it takes a"},{"Start":"00:28.680 ","End":"00:35.825","Text":"polynomial and sends it to its derivative plus its second derivative."},{"Start":"00:35.825 ","End":"00:38.210","Text":"Just as an example,"},{"Start":"00:38.210 ","End":"00:42.695","Text":"if I wanted to know what is T of x cubed,"},{"Start":"00:42.695 ","End":"00:44.720","Text":"then first of all,"},{"Start":"00:44.720 ","End":"00:47.510","Text":"I take the derivative 3x squared,"},{"Start":"00:47.510 ","End":"00:49.220","Text":"and then I add the second derivative,"},{"Start":"00:49.220 ","End":"00:51.560","Text":"which is the derivative of the first derivative,"},{"Start":"00:51.560 ","End":"00:53.945","Text":"and that would be 6x."},{"Start":"00:53.945 ","End":"00:57.565","Text":"X cubed gets sent to 3x squared plus 6x."},{"Start":"00:57.565 ","End":"01:02.280","Text":"The question is, is it linear? Let\u0027s check."},{"Start":"01:02.280 ","End":"01:04.980","Text":"There are 2 axioms that have to apply."},{"Start":"01:04.980 ","End":"01:12.350","Text":"The first axiom says that T of p plus q is T of p plus T of q."},{"Start":"01:12.350 ","End":"01:15.920","Text":"Usually we state it with u and v,"},{"Start":"01:15.920 ","End":"01:18.650","Text":"but with polynomials I\u0027d rather use p and q."},{"Start":"01:18.650 ","End":"01:21.185","Text":"The letters don\u0027t matter of course."},{"Start":"01:21.185 ","End":"01:25.630","Text":"I just wrote T again because we\u0027re going to do some scrolling here."},{"Start":"01:25.630 ","End":"01:30.030","Text":"Then I\u0027m going to apply T 3 times once to p, once to q,"},{"Start":"01:30.030 ","End":"01:34.215","Text":"and once to p plus q. T of p of x is,"},{"Start":"01:34.215 ","End":"01:35.580","Text":"follows from the definition,"},{"Start":"01:35.580 ","End":"01:39.300","Text":"is p prime plus p double prime of x,"},{"Start":"01:39.300 ","End":"01:45.545","Text":"and similarly for q. Q gets sent to q prime plus q double prime,"},{"Start":"01:45.545 ","End":"01:48.430","Text":"the first derivative plus a second derivative."},{"Start":"01:48.430 ","End":"01:53.045","Text":"Then if I want to see what p plus q goes to,"},{"Start":"01:53.045 ","End":"01:56.540","Text":"it\u0027s the derivative of p plus"},{"Start":"01:56.540 ","End":"02:04.530","Text":"q at x plus the second derivative of p plus q at x."},{"Start":"02:04.750 ","End":"02:08.420","Text":"There is a theorem in calculus, maybe I should write it,"},{"Start":"02:08.420 ","End":"02:11.010","Text":"that if I take the sum of 2 functions,"},{"Start":"02:11.010 ","End":"02:12.245","Text":"it could be polynomial,"},{"Start":"02:12.245 ","End":"02:15.604","Text":"but it works in general for differentiable functions."},{"Start":"02:15.604 ","End":"02:18.860","Text":"If I take the sum and I take the derivative,"},{"Start":"02:18.860 ","End":"02:24.645","Text":"then it\u0027s the same as taking each one separately the derivative."},{"Start":"02:24.645 ","End":"02:27.780","Text":"Similarly, I could just apply the rule again."},{"Start":"02:27.780 ","End":"02:29.870","Text":"If I had a second derivative here,"},{"Start":"02:29.870 ","End":"02:34.320","Text":"that would be a second derivative here and a second derivative here."},{"Start":"02:34.340 ","End":"02:37.995","Text":"P and q are like f and g here."},{"Start":"02:37.995 ","End":"02:43.850","Text":"The derivative of p plus q is the derivative of p plus the derivative of q."},{"Start":"02:43.850 ","End":"02:46.940","Text":"Similarly for second derivative of p plus q,"},{"Start":"02:46.940 ","End":"02:49.225","Text":"we can apply it separately."},{"Start":"02:49.225 ","End":"02:53.630","Text":"Then from the definition of sum of functions or sum of polynomials,"},{"Start":"02:53.630 ","End":"02:58.265","Text":"the sum applied to x means just apply each one to x and add."},{"Start":"02:58.265 ","End":"02:59.630","Text":"This gives me 3,"},{"Start":"02:59.630 ","End":"03:02.770","Text":"these 2 terms give me these 2 terms."},{"Start":"03:02.770 ","End":"03:08.225","Text":"I think that we can see that if I take this and add it to this,"},{"Start":"03:08.225 ","End":"03:10.715","Text":"that we will get exactly this,"},{"Start":"03:10.715 ","End":"03:12.470","Text":"possibly in a different order,"},{"Start":"03:12.470 ","End":"03:15.630","Text":"but that doesn\u0027t matter because it\u0027s the sum."},{"Start":"03:15.950 ","End":"03:21.135","Text":"We\u0027ve proved axiom 1 for linearity."},{"Start":"03:21.135 ","End":"03:25.500","Text":"Now, let\u0027s move on to the second axiom."},{"Start":"03:25.500 ","End":"03:31.640","Text":"Here we have to show that if I multiply the polynomial by k,"},{"Start":"03:31.640 ","End":"03:34.295","Text":"a scalar, and then take T of that,"},{"Start":"03:34.295 ","End":"03:40.130","Text":"it\u0027s the same as first applying T and then multiplying by k. Once again,"},{"Start":"03:40.130 ","End":"03:45.185","Text":"I\u0027m going to need a result from calculus that if I take a function f,"},{"Start":"03:45.185 ","End":"03:48.110","Text":"and multiply it by a constant a,"},{"Start":"03:48.110 ","End":"03:50.495","Text":"and then take the derivative of that,"},{"Start":"03:50.495 ","End":"03:53.810","Text":"I can pull a outside which is like saying it\u0027s"},{"Start":"03:53.810 ","End":"03:59.119","Text":"a times the derivative of f. If it works for first derivative,"},{"Start":"03:59.119 ","End":"04:01.790","Text":"it will work the second derivative also."},{"Start":"04:01.790 ","End":"04:04.615","Text":"Of course we just have to differentiate again."},{"Start":"04:04.615 ","End":"04:07.935","Text":"Just to remind you of what T was."},{"Start":"04:07.935 ","End":"04:10.590","Text":"Let\u0027s apply it to p,"},{"Start":"04:10.590 ","End":"04:12.870","Text":"to q, and to p plus q."},{"Start":"04:12.870 ","End":"04:14.970","Text":"Applying T to p, well,"},{"Start":"04:14.970 ","End":"04:17.985","Text":"you just have to copy this here."},{"Start":"04:17.985 ","End":"04:23.070","Text":"I\u0027m working on the right-hand side now so let\u0027s see what is k times this."},{"Start":"04:23.070 ","End":"04:25.500","Text":"It\u0027s just k times this sum,"},{"Start":"04:25.500 ","End":"04:29.325","Text":"so I\u0027ll just put k times the sum in brackets."},{"Start":"04:29.325 ","End":"04:31.880","Text":"Now, working on the left-hand side,"},{"Start":"04:31.880 ","End":"04:37.925","Text":"if I apply T to the polynomial k p of function k p,"},{"Start":"04:37.925 ","End":"04:43.990","Text":"then it\u0027s this function derivative plus its second derivative."},{"Start":"04:43.990 ","End":"04:50.180","Text":"This is where we use this result from calculus with a being k and"},{"Start":"04:50.180 ","End":"04:56.060","Text":"f being p. When we take a derivative or a second derivative,"},{"Start":"04:56.060 ","End":"04:58.160","Text":"the constant comes out front."},{"Start":"04:58.160 ","End":"05:00.600","Text":"This is what we get."},{"Start":"05:00.730 ","End":"05:08.540","Text":"I think we can see that this is equal to this or the distributive law,"},{"Start":"05:08.540 ","End":"05:14.125","Text":"just take the constant out or multiply out by the constants whichever way,"},{"Start":"05:14.125 ","End":"05:17.360","Text":"and that means that this equals this,"},{"Start":"05:17.360 ","End":"05:19.910","Text":"which is what we were trying to show."},{"Start":"05:19.910 ","End":"05:23.315","Text":"We\u0027ve proved the equality so axiom 2 holds."},{"Start":"05:23.315 ","End":"05:26.555","Text":"We\u0027ve already done the first axiom."},{"Start":"05:26.555 ","End":"05:30.050","Text":"That just means that T is linear."},{"Start":"05:30.050 ","End":"05:33.090","Text":"We\u0027ve proved it, and we are done."}],"ID":10236},{"Watched":false,"Name":"Exercise 14","Duration":"2m 35s","ChapterTopicVideoID":9687,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:09.210","Text":"In this exercise, we have a transformation which takes polynomials and just squares them."},{"Start":"00:09.210 ","End":"00:10.845","Text":"If I put in p of x,"},{"Start":"00:10.845 ","End":"00:13.125","Text":"I get out p squared of x."},{"Start":"00:13.125 ","End":"00:19.665","Text":"It goes from the space of polynomials up to degree n. But here,"},{"Start":"00:19.665 ","End":"00:20.940","Text":"well after I\u0027ve squared it,"},{"Start":"00:20.940 ","End":"00:24.705","Text":"the degree could go up as high as 2n."},{"Start":"00:24.705 ","End":"00:26.970","Text":"That\u0027s just a technical point."},{"Start":"00:26.970 ","End":"00:33.340","Text":"Main thing is that it\u0027s squares polynomials and is this linear or not?"},{"Start":"00:34.310 ","End":"00:39.739","Text":"The squaring doesn\u0027t really give the impression of being linear."},{"Start":"00:39.739 ","End":"00:44.180","Text":"Let\u0027s see if we can disprove that it\u0027s linear."},{"Start":"00:44.180 ","End":"00:48.200","Text":"Usually it\u0027s easier to attack axiom 2."},{"Start":"00:48.200 ","End":"00:52.620","Text":"Just for convenience, I copied it again."},{"Start":"00:52.780 ","End":"00:55.670","Text":"If axiom 2 were true,"},{"Start":"00:55.670 ","End":"00:58.790","Text":"then it wouldn\u0027t make any difference if I first"},{"Start":"00:58.790 ","End":"01:02.900","Text":"applied the transformation and then multiplied by k,"},{"Start":"01:02.900 ","End":"01:08.105","Text":"or if I first multiply the polynomial by k and then apply the transformation,"},{"Start":"01:08.105 ","End":"01:11.050","Text":"this doesn\u0027t seem very likely."},{"Start":"01:11.050 ","End":"01:15.460","Text":"I think we could look for a counterexample."},{"Start":"01:15.710 ","End":"01:19.870","Text":"For example, if we take k equals 2,"},{"Start":"01:19.870 ","End":"01:22.570","Text":"if you multiply something by 2 and then square it."},{"Start":"01:22.570 ","End":"01:26.420","Text":"It\u0027s not the same as squaring and multiplying by 2."},{"Start":"01:26.570 ","End":"01:32.620","Text":"What I\u0027m claiming is if we take k equals 2 and I\u0027ll take,"},{"Start":"01:32.620 ","End":"01:38.925","Text":"for example, p of x equals x squared."},{"Start":"01:38.925 ","End":"01:40.755","Text":"Or maybe I\u0027ll write it."},{"Start":"01:40.755 ","End":"01:42.465","Text":"K equals 2,"},{"Start":"01:42.465 ","End":"01:46.020","Text":"p of x is x squared so a counterexample."},{"Start":"01:46.020 ","End":"01:49.560","Text":"I\u0027m going to show this is not equal to this."},{"Start":"01:49.560 ","End":"01:51.685","Text":"Now, in general,"},{"Start":"01:51.685 ","End":"01:55.795","Text":"T of a polynomial is that polynomial squared."},{"Start":"01:55.795 ","End":"01:57.490","Text":"If I apply it,"},{"Start":"01:57.490 ","End":"02:01.850","Text":"the 2x squared, I need to square 2x squared."},{"Start":"02:01.850 ","End":"02:05.740","Text":"It will come out to be 4x to the 4th."},{"Start":"02:05.740 ","End":"02:09.620","Text":"On the other hand, if I wanted to evaluate this,"},{"Start":"02:09.620 ","End":"02:14.510","Text":"T of x squared is x squared squared is x to the 4th,"},{"Start":"02:14.510 ","End":"02:17.330","Text":"and then I multiply it by 2."},{"Start":"02:17.330 ","End":"02:19.905","Text":"I\u0027ve got 2x to the 4th."},{"Start":"02:19.905 ","End":"02:24.770","Text":"Now, clearly this and this are not the same polynomial."},{"Start":"02:24.770 ","End":"02:28.500","Text":"So axiom 2 is violated."},{"Start":"02:28.540 ","End":"02:35.220","Text":"T is not linear. We are done."}],"ID":10237},{"Watched":false,"Name":"Exercise 15","Duration":"8m 29s","ChapterTopicVideoID":9688,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.155","Text":"This exercise has 2 parts."},{"Start":"00:04.155 ","End":"00:07.485","Text":"Now, in each case,"},{"Start":"00:07.485 ","End":"00:14.610","Text":"the complex numbers are going to be a vector space."},{"Start":"00:15.080 ","End":"00:19.860","Text":"This vector space could be considered as"},{"Start":"00:19.860 ","End":"00:25.845","Text":"a vector space over the real numbers or over the complex numbers."},{"Start":"00:25.845 ","End":"00:29.250","Text":"I mean, I can take 2 complex numbers,"},{"Start":"00:29.250 ","End":"00:30.960","Text":"I can add them."},{"Start":"00:30.960 ","End":"00:34.230","Text":"I can multiply them by a scalar,"},{"Start":"00:34.230 ","End":"00:36.270","Text":"which could be real or complex,"},{"Start":"00:36.270 ","End":"00:39.270","Text":"and all the axioms would be satisfied."},{"Start":"00:39.270 ","End":"00:41.930","Text":"There are 2 variations of"},{"Start":"00:41.930 ","End":"00:47.250","Text":"the complex numbers over the reals and over the complex numbers themselves."},{"Start":"00:47.250 ","End":"00:49.730","Text":"On the 1 hand, the elements are vectors,"},{"Start":"00:49.730 ","End":"00:52.340","Text":"on the other hand, they\u0027re scalars in the field."},{"Start":"00:52.340 ","End":"00:57.855","Text":"Bit confusing, but it\u0027s not too difficult to follow."},{"Start":"00:57.855 ","End":"01:00.675","Text":"The questions are,"},{"Start":"01:00.675 ","End":"01:06.260","Text":"transformation which sends a complex number to its complex conjugate."},{"Start":"01:06.260 ","End":"01:08.479","Text":"The bar means conjugate."},{"Start":"01:08.479 ","End":"01:15.260","Text":"Is it a linear transformation when I consider the field of real numbers?"},{"Start":"01:15.260 ","End":"01:21.995","Text":"Is it a linear transformation when I take it over the field of complex numbers?"},{"Start":"01:21.995 ","End":"01:27.710","Text":"It turns out that I get different answers for a and for b. Let\u0027s see."},{"Start":"01:27.710 ","End":"01:29.990","Text":"Let\u0027s start with part a,"},{"Start":"01:29.990 ","End":"01:35.285","Text":"where complex numbers are vectors and the scalars are real numbers."},{"Start":"01:35.285 ","End":"01:39.800","Text":"Before I start working on axioms 1 and 2 is to a bit of preparation."},{"Start":"01:39.800 ","End":"01:46.490","Text":"We always need a u and a v and u plus v. For the second axiom at ku."},{"Start":"01:46.490 ","End":"01:51.600","Text":"Let\u0027s take u with a typical complex number, a plus bi."},{"Start":"01:51.600 ","End":"01:52.800","Text":"The a and b are reals,"},{"Start":"01:52.800 ","End":"01:56.390","Text":"of course, vbc plus di,"},{"Start":"01:56.390 ","End":"02:02.865","Text":"and then u plus v would be a plus c plus b plus di."},{"Start":"02:02.865 ","End":"02:05.700","Text":"Multiplying by k, again,"},{"Start":"02:05.700 ","End":"02:07.470","Text":"k is a real number,"},{"Start":"02:07.470 ","End":"02:13.600","Text":"we get k times a plus bi is ka plus kbi."},{"Start":"02:13.960 ","End":"02:16.655","Text":"Because k is a real number,"},{"Start":"02:16.655 ","End":"02:17.975","Text":"and so are a and b,"},{"Start":"02:17.975 ","End":"02:20.495","Text":"then ka and kb are real numbers."},{"Start":"02:20.495 ","End":"02:22.820","Text":"This really is a complex number,"},{"Start":"02:22.820 ","End":"02:28.080","Text":"the proper representation, a real plus a real times i."},{"Start":"02:28.220 ","End":"02:34.005","Text":"The first axiom says that T of the sum,"},{"Start":"02:34.005 ","End":"02:35.840","Text":"is the sum of the ts."},{"Start":"02:35.840 ","End":"02:39.440","Text":"In other words, if I take the sum of 2 vectors and then apply"},{"Start":"02:39.440 ","End":"02:43.780","Text":"T is the same as applying T separately and then adding."},{"Start":"02:43.780 ","End":"02:46.860","Text":"Remember T of z,"},{"Start":"02:46.860 ","End":"02:48.925","Text":"z is a complex number,"},{"Start":"02:48.925 ","End":"02:52.735","Text":"is its complex conjugate z bar."},{"Start":"02:52.735 ","End":"02:58.540","Text":"Now the complex conjugate of u,"},{"Start":"02:58.540 ","End":"03:00.405","Text":"which is a plus bi,"},{"Start":"03:00.405 ","End":"03:03.030","Text":"is simply a minus bi."},{"Start":"03:03.030 ","End":"03:06.120","Text":"The imaginary part is negated."},{"Start":"03:06.120 ","End":"03:09.045","Text":"Similarly, if we have c plus di,"},{"Start":"03:09.045 ","End":"03:12.090","Text":"the conjugate is c minus di."},{"Start":"03:12.090 ","End":"03:15.820","Text":"This is z, this is z bar, same here."},{"Start":"03:16.220 ","End":"03:18.700","Text":"T of u plus v,"},{"Start":"03:18.700 ","End":"03:20.410","Text":"u plus v is here."},{"Start":"03:20.410 ","End":"03:24.525","Text":"We need to take the the complex conjugate of that."},{"Start":"03:24.525 ","End":"03:27.240","Text":"We just make a minus."},{"Start":"03:27.240 ","End":"03:28.760","Text":"We\u0027re here there was a plus,"},{"Start":"03:28.760 ","End":"03:30.380","Text":"we put a minus."},{"Start":"03:30.380 ","End":"03:32.680","Text":"This is what we get."},{"Start":"03:32.680 ","End":"03:39.950","Text":"The question is, is this plus this equals to this?"},{"Start":"03:39.950 ","End":"03:41.420","Text":"If you look at it for a moment,"},{"Start":"03:41.420 ","End":"03:43.595","Text":"you\u0027ll see that, yes, that is true."},{"Start":"03:43.595 ","End":"03:46.770","Text":"The real part, a plus c, in both cases,"},{"Start":"03:46.770 ","End":"03:49.010","Text":"this is a minus I have a bi and a di,"},{"Start":"03:49.010 ","End":"03:50.450","Text":"so it\u0027s b plus di."},{"Start":"03:50.450 ","End":"03:54.685","Text":"That\u0027s fine. Axiom 1 holds."},{"Start":"03:54.685 ","End":"03:58.290","Text":"This plus this is equal to this,"},{"Start":"03:58.290 ","End":"04:00.090","Text":"and that\u0027s what we tried to show."},{"Start":"04:00.090 ","End":"04:03.475","Text":"Let\u0027s get onto axiom 2 now."},{"Start":"04:03.475 ","End":"04:10.235","Text":"For axiom 2, we have to show that if we multiply a vector which is a complex number,"},{"Start":"04:10.235 ","End":"04:12.860","Text":"by k and then apply T,"},{"Start":"04:12.860 ","End":"04:14.690","Text":"it\u0027s the same as first applying T,"},{"Start":"04:14.690 ","End":"04:22.815","Text":"then multiplying by k. But this is important k has to be from our field of real numbers."},{"Start":"04:22.815 ","End":"04:29.839","Text":"This is a reminder what T does to z is it sends it to its complex conjugate."},{"Start":"04:29.839 ","End":"04:32.435","Text":"Now, T of u,"},{"Start":"04:32.435 ","End":"04:34.415","Text":"we already saw this up here,"},{"Start":"04:34.415 ","End":"04:37.585","Text":"is a minus bi."},{"Start":"04:37.585 ","End":"04:40.190","Text":"If I multiply that by k,"},{"Start":"04:40.190 ","End":"04:42.440","Text":"where k is real,"},{"Start":"04:42.440 ","End":"04:48.090","Text":"I get k times a minus bi."},{"Start":"04:48.110 ","End":"04:58.880","Text":"On the other hand, if I first of all multiply u by k. Ku we computed earlier is ka plus,"},{"Start":"05:00.550 ","End":"05:05.075","Text":"I\u0027m sorry, there\u0027s a typo where the k somehow got lost here."},{"Start":"05:05.075 ","End":"05:12.750","Text":"K times u is ka plus kbi and T of that just negates the imaginary parts."},{"Start":"05:12.750 ","End":"05:15.790","Text":"It\u0027s ka minus kbi."},{"Start":"05:15.790 ","End":"05:21.490","Text":"The question is, does this equal this?"},{"Start":"05:21.490 ","End":"05:23.230","Text":"The answer is certainly yes."},{"Start":"05:23.230 ","End":"05:25.090","Text":"Just open the brackets."},{"Start":"05:25.090 ","End":"05:28.495","Text":"Of course, ka and kb are real numbers."},{"Start":"05:28.495 ","End":"05:37.500","Text":"This is the correct expression of the complex number because this has to be real."},{"Start":"05:37.500 ","End":"05:41.930","Text":"We see that in this case where we\u0027re working over the reals,"},{"Start":"05:41.930 ","End":"05:45.905","Text":"then k times T of u is T of ku."},{"Start":"05:45.905 ","End":"05:50.600","Text":"It doesn\u0027t matter if you first multiply by the real number and then"},{"Start":"05:50.600 ","End":"05:57.020","Text":"apply the conjugate or first apply the conjugate and then multiply by a real number,"},{"Start":"05:57.020 ","End":"05:59.090","Text":"comes out the same."},{"Start":"05:59.090 ","End":"06:03.840","Text":"We\u0027ve shown axioms 1 and 2."},{"Start":"06:03.840 ","End":"06:08.430","Text":"Perhaps I should have just written the T is linear,"},{"Start":"06:08.430 ","End":"06:10.670","Text":"at least in part a."},{"Start":"06:10.670 ","End":"06:13.900","Text":"Now, remember there was a part b."},{"Start":"06:13.900 ","End":"06:21.440","Text":"In part b, we had the field as complex numbers, not real numbers."},{"Start":"06:21.440 ","End":"06:26.240","Text":"I\u0027m going to show you that T is not linear in this case."},{"Start":"06:26.240 ","End":"06:32.715","Text":"Over the reals it is an over the complex field, it isn\u0027t."},{"Start":"06:32.715 ","End":"06:34.820","Text":"That is usually the case,"},{"Start":"06:34.820 ","End":"06:39.260","Text":"it\u0027s easier to show that axiom 2 is not satisfied."},{"Start":"06:39.260 ","End":"06:41.390","Text":"This is what axiom 2 says,"},{"Start":"06:41.390 ","End":"06:46.055","Text":"that if you multiply by k and then apply T,"},{"Start":"06:46.055 ","End":"06:47.840","Text":"or first apply T,"},{"Start":"06:47.840 ","End":"06:50.780","Text":"then multiply by k, we should get the same thing."},{"Start":"06:50.780 ","End":"06:56.095","Text":"The difference being that this time k could be a complex number."},{"Start":"06:56.095 ","End":"07:00.125","Text":"Let\u0027s take k as a complex number."},{"Start":"07:00.125 ","End":"07:04.375","Text":"I\u0027ll take k equals i."},{"Start":"07:04.375 ","End":"07:11.470","Text":"As for u, I\u0027ll take u equals just a real number 5."},{"Start":"07:11.470 ","End":"07:16.300","Text":"If we want to apply this particular instance,"},{"Start":"07:16.300 ","End":"07:24.190","Text":"we should get that T of i times 5 is equal to i times T of 5."},{"Start":"07:24.190 ","End":"07:27.859","Text":"I\u0027m claiming that they are not equal."},{"Start":"07:27.859 ","End":"07:32.470","Text":"That is because i times 5 is like 5i."},{"Start":"07:32.480 ","End":"07:36.105","Text":"If you like it\u0027s 0 plus 5i,"},{"Start":"07:36.105 ","End":"07:42.030","Text":"if I take the conjugate of that it\u0027s 0 minus 5i."},{"Start":"07:42.030 ","End":"07:46.575","Text":"T of i times 5 is minus 5i."},{"Start":"07:46.575 ","End":"07:49.275","Text":"On the other hand,"},{"Start":"07:49.275 ","End":"07:51.750","Text":"if do it the other way."},{"Start":"07:51.750 ","End":"08:01.660","Text":"First apply T to 5 and then multiply by i, I get plus 5i."},{"Start":"08:02.010 ","End":"08:07.675","Text":"Because the conjugate of a real number is just itself."},{"Start":"08:07.675 ","End":"08:12.445","Text":"5 plus 0i, 5 minus 0i."},{"Start":"08:12.445 ","End":"08:16.325","Text":"That\u0027s 5 and 5 times i is 5i."},{"Start":"08:16.325 ","End":"08:19.755","Text":"This is not equal to this,"},{"Start":"08:19.755 ","End":"08:22.545","Text":"and so we found 1 counter example,"},{"Start":"08:22.545 ","End":"08:26.775","Text":"and so it is not linear."},{"Start":"08:26.775 ","End":"08:30.040","Text":"We are done."}],"ID":10238},{"Watched":false,"Name":"Exercise 16","Duration":"7m 28s","ChapterTopicVideoID":9689,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.805","Text":"In this exercise, we\u0027re given a transformation from R^2 to R^2,"},{"Start":"00:05.805 ","End":"00:08.700","Text":"given by this formula."},{"Start":"00:08.700 ","End":"00:14.940","Text":"The difference this time is that it has a parameter m in it."},{"Start":"00:14.940 ","End":"00:21.690","Text":"Our task is to find out what value or values of m,"},{"Start":"00:21.690 ","End":"00:25.840","Text":"if any, will make this transformation linear."},{"Start":"00:25.940 ","End":"00:28.635","Text":"Now, how do we go about this?"},{"Start":"00:28.635 ","End":"00:39.000","Text":"1 idea is to use the 2nd axiom for linearity and just try it out on particular numbers."},{"Start":"00:39.000 ","End":"00:42.175","Text":"I wrote something here, let me explain."},{"Start":"00:42.175 ","End":"00:52.510","Text":"In general, we have that k times t of u equals t of k u,"},{"Start":"00:52.510 ","End":"00:54.810","Text":"if t is linear."},{"Start":"00:54.810 ","End":"00:57.620","Text":"I said, okay, let\u0027s just take some values."},{"Start":"00:57.620 ","End":"01:01.365","Text":"Suppose we took k equals 2,"},{"Start":"01:01.365 ","End":"01:06.080","Text":"and I\u0027ll let u be something that\u0027s easy to substitute,"},{"Start":"01:06.080 ","End":"01:08.200","Text":"say 1, 1."},{"Start":"01:08.200 ","End":"01:10.250","Text":"Now, if we do that,"},{"Start":"01:10.250 ","End":"01:12.790","Text":"this interprets to this,"},{"Start":"01:12.790 ","End":"01:15.520","Text":"twice t of 1, 1,"},{"Start":"01:15.520 ","End":"01:19.100","Text":"but it\u0027s got to be t if twice 1, 1."},{"Start":"01:19.100 ","End":"01:22.170","Text":"The plan is after we\u0027ve substituted all this,"},{"Start":"01:22.170 ","End":"01:27.860","Text":"we\u0027ll have an equation in m and that will help us to find m,"},{"Start":"01:27.860 ","End":"01:30.295","Text":"and afterwards, we\u0027ll see what we\u0027ll do."},{"Start":"01:30.295 ","End":"01:34.110","Text":"I\u0027ll start with computing t of 1, 1."},{"Start":"01:34.110 ","End":"01:38.145","Text":"Just put x equals 1 and y equals 1 here,"},{"Start":"01:38.145 ","End":"01:48.510","Text":"and what we\u0027ll get is that m squared and then 1 to the power of anything is 1,"},{"Start":"01:48.510 ","End":"01:51.375","Text":"that just gives us m squared."},{"Start":"01:51.375 ","End":"01:53.640","Text":"Here, again, y is 1,"},{"Start":"01:53.640 ","End":"01:55.200","Text":"1 to the anything is 1,"},{"Start":"01:55.200 ","End":"01:58.750","Text":"and x is 1, 1 plus 1 is 2."},{"Start":"01:58.750 ","End":"02:03.980","Text":"Next, multiply both sides by 2 because we want twice t of 1,"},{"Start":"02:03.980 ","End":"02:10.015","Text":"1, and this will come out to be twice m squared and twice 2 is 4."},{"Start":"02:10.015 ","End":"02:12.030","Text":"That was the left-hand side."},{"Start":"02:12.030 ","End":"02:13.610","Text":"Now, let\u0027s go for the right-hand side."},{"Start":"02:13.610 ","End":"02:14.900","Text":"Of course, 2 times 1,"},{"Start":"02:14.900 ","End":"02:17.940","Text":"1 is 2, 2."},{"Start":"02:17.940 ","End":"02:20.160","Text":"So on the right side,"},{"Start":"02:20.160 ","End":"02:22.040","Text":"we have t of 2, 2."},{"Start":"02:22.040 ","End":"02:27.195","Text":"That means plug in x and y both equal to 2 here,"},{"Start":"02:27.195 ","End":"02:33.685","Text":"and we\u0027ll get this expression in m. Now,"},{"Start":"02:33.685 ","End":"02:35.959","Text":"these will have to be equal,"},{"Start":"02:35.959 ","End":"02:39.840","Text":"but they are 2 dimensional,"},{"Start":"02:39.840 ","End":"02:41.430","Text":"they have 2 components."},{"Start":"02:41.430 ","End":"02:47.150","Text":"The first component will equal the first component here,"},{"Start":"02:47.150 ","End":"02:52.730","Text":"and the second component will equal the second component because when vectors are equal,"},{"Start":"02:52.730 ","End":"02:57.300","Text":"the respective components are also equal."},{"Start":"02:57.380 ","End":"03:02.145","Text":"This equals this is the first equation,"},{"Start":"03:02.145 ","End":"03:06.050","Text":"and this equals this is the second equation."},{"Start":"03:06.050 ","End":"03:08.510","Text":"You\u0027ve got 2 equations and 1 unknown,"},{"Start":"03:08.510 ","End":"03:10.790","Text":"let\u0027s see if it has a solution."},{"Start":"03:10.790 ","End":"03:14.270","Text":"I\u0027m going to need some space here."},{"Start":"03:14.270 ","End":"03:17.315","Text":"From the second equation,"},{"Start":"03:17.315 ","End":"03:20.990","Text":"I just bring the 2 over to the left and we\u0027ll"},{"Start":"03:20.990 ","End":"03:25.205","Text":"get 2 equals 2^2m and just then switch sides."},{"Start":"03:25.205 ","End":"03:27.710","Text":"We get 2^2m equals 2,"},{"Start":"03:27.710 ","End":"03:32.000","Text":"but 2 is actually 2^1."},{"Start":"03:32.000 ","End":"03:35.455","Text":"So 2m must equal 1,"},{"Start":"03:35.455 ","End":"03:39.120","Text":"and that gives us that m equals 1/2."},{"Start":"03:39.120 ","End":"03:43.450","Text":"Now, we have to check that it also satisfies the first equation,"},{"Start":"03:43.450 ","End":"03:45.210","Text":"and I\u0027ll leave you to plug it in,"},{"Start":"03:45.210 ","End":"03:48.125","Text":"and indeed, it does satisfy the first equation."},{"Start":"03:48.125 ","End":"03:50.840","Text":"m equals 1/2 satisfies both,"},{"Start":"03:50.840 ","End":"03:52.970","Text":"so it\u0027s the only possibility for"},{"Start":"03:52.970 ","End":"04:00.770","Text":"m. If you go back to the original definition and plug in m equals 1/2,"},{"Start":"04:00.770 ","End":"04:03.350","Text":"then we get that t of x,"},{"Start":"04:03.350 ","End":"04:08.525","Text":"y equals 1/4x, y plus x."},{"Start":"04:08.525 ","End":"04:17.370","Text":"Now, we still have to check that this is a linear transformation."},{"Start":"04:17.450 ","End":"04:21.349","Text":"See, m equals 1/2 is a necessary condition."},{"Start":"04:21.349 ","End":"04:22.760","Text":"If it\u0027s going to be linear,"},{"Start":"04:22.760 ","End":"04:23.840","Text":"m has to be 1/2,"},{"Start":"04:23.840 ","End":"04:25.520","Text":"but it doesn\u0027t mean that if m is 1/2,"},{"Start":"04:25.520 ","End":"04:27.470","Text":"that this really will be linear."},{"Start":"04:27.470 ","End":"04:35.545","Text":"Now, we\u0027ll check that this transformation from R^2 to R^2 is indeed linear."},{"Start":"04:35.545 ","End":"04:37.400","Text":"Started a new page."},{"Start":"04:37.400 ","End":"04:39.815","Text":"This is what we have to check."},{"Start":"04:39.815 ","End":"04:43.610","Text":"Start with the preparation phase before I check the 2 axioms,"},{"Start":"04:43.610 ","End":"04:47.210","Text":"I\u0027ll take u and v. Since we\u0027re in R^2,"},{"Start":"04:47.210 ","End":"04:49.175","Text":"this will be some a, b."},{"Start":"04:49.175 ","End":"04:50.900","Text":"Here, I\u0027ll use capital letters,"},{"Start":"04:50.900 ","End":"04:52.970","Text":"we\u0027ll need u plus v,"},{"Start":"04:52.970 ","End":"04:54.235","Text":"which is this,"},{"Start":"04:54.235 ","End":"04:57.845","Text":"and we will need k times u, which is this."},{"Start":"04:57.845 ","End":"05:00.530","Text":"Now, let\u0027s get to the axiom,"},{"Start":"05:00.530 ","End":"05:02.615","Text":"starting with the first."},{"Start":"05:02.615 ","End":"05:07.570","Text":"We want to check if this equality always holds."},{"Start":"05:07.570 ","End":"05:11.700","Text":"Let\u0027s see then. T of u and u is a,"},{"Start":"05:11.700 ","End":"05:15.210","Text":"b, gives us this expression,"},{"Start":"05:15.210 ","End":"05:17.400","Text":"if we plug it into t of x."},{"Start":"05:17.400 ","End":"05:20.715","Text":"Why? T of v,"},{"Start":"05:20.715 ","End":"05:24.705","Text":"very similar, just with capital A and B."},{"Start":"05:24.705 ","End":"05:29.745","Text":"Then we want t of u plus v. u plus v is here."},{"Start":"05:29.745 ","End":"05:31.425","Text":"We want t of that."},{"Start":"05:31.425 ","End":"05:34.310","Text":"If you plug it into the formula,"},{"Start":"05:34.310 ","End":"05:37.120","Text":"we get this expression."},{"Start":"05:37.120 ","End":"05:39.995","Text":"Now, the question is,"},{"Start":"05:39.995 ","End":"05:46.890","Text":"does this plus this equal this?"},{"Start":"05:46.890 ","End":"05:50.070","Text":"I think you can see that it does."},{"Start":"05:50.070 ","End":"05:53.100","Text":"You have 1/4 a plus 1/4 of big A,"},{"Start":"05:53.100 ","End":"05:55.625","Text":"so it\u0027s 1/4 of a plus a, and similarly,"},{"Start":"05:55.625 ","End":"06:00.025","Text":"the second component, this plus this will give us this."},{"Start":"06:00.025 ","End":"06:04.710","Text":"This means that if this is this plus this,"},{"Start":"06:04.710 ","End":"06:06.730","Text":"t of u plus v is t of u plus t of v,"},{"Start":"06:06.730 ","End":"06:08.280","Text":"which is what we wanted to show,"},{"Start":"06:08.280 ","End":"06:12.425","Text":"and that takes care of the first axiom."},{"Start":"06:12.425 ","End":"06:17.395","Text":"Now, let\u0027s move on to the second axiom."},{"Start":"06:17.395 ","End":"06:23.090","Text":"Does t of ku always equal k times t of u?"},{"Start":"06:23.150 ","End":"06:27.270","Text":"t of u, we have this already from up here,"},{"Start":"06:27.270 ","End":"06:30.234","Text":"do it again, gives us this."},{"Start":"06:30.234 ","End":"06:38.675","Text":"Then we want to multiply this by k. We get k times t of u, k times this."},{"Start":"06:38.675 ","End":"06:42.705","Text":"k times each of the 2 components is this."},{"Start":"06:42.705 ","End":"06:45.970","Text":"On the other hand, we want the left-hand side,"},{"Start":"06:45.970 ","End":"06:49.450","Text":"t of ku is t of ka,"},{"Start":"06:49.450 ","End":"06:52.765","Text":"kb, because if I take k times u,"},{"Start":"06:52.765 ","End":"06:56.515","Text":"u is ab, so ku is ka, kv."},{"Start":"06:56.515 ","End":"06:58.330","Text":"From the formula,"},{"Start":"06:58.330 ","End":"07:00.575","Text":"that will give us this."},{"Start":"07:00.575 ","End":"07:07.840","Text":"Indeed, this is equal to this."},{"Start":"07:07.840 ","End":"07:11.040","Text":"Indeed, this is equal to this,"},{"Start":"07:11.040 ","End":"07:13.755","Text":"we have that t of ku is kt of u."},{"Start":"07:13.755 ","End":"07:17.880","Text":"That\u0027s the second axiom verified."},{"Start":"07:17.880 ","End":"07:22.925","Text":"The answer to the question is that it\u0027s a linear transformation"},{"Start":"07:22.925 ","End":"07:29.010","Text":"for the value m equals 2. We\u0027re done."}],"ID":10239},{"Watched":false,"Name":"Exercise 17","Duration":"13m 6s","ChapterTopicVideoID":9690,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.295","Text":"In this exercise,"},{"Start":"00:02.295 ","End":"00:07.785","Text":"we are asked to find out if there is a linear transformation, T,"},{"Start":"00:07.785 ","End":"00:10.590","Text":"that goes from R^4 to R^3,"},{"Start":"00:10.590 ","End":"00:16.860","Text":"such that T of this vector is this vector and so on."},{"Start":"00:16.860 ","End":"00:21.615","Text":"Notice that these all have 4 components and that these have 3 components,"},{"Start":"00:21.615 ","End":"00:24.180","Text":"so it is R^4 to R^3."},{"Start":"00:24.180 ","End":"00:27.945","Text":"To find out whether such a T exists,"},{"Start":"00:27.945 ","End":"00:30.405","Text":"and if not explain why,"},{"Start":"00:30.405 ","End":"00:32.655","Text":"and if there is,"},{"Start":"00:32.655 ","End":"00:39.280","Text":"we have to find such a T and to say whether or not it\u0027s unique."},{"Start":"00:39.320 ","End":"00:41.610","Text":"Let\u0027s give them names,"},{"Start":"00:41.610 ","End":"00:43.320","Text":"these 3 will be v_1,"},{"Start":"00:43.320 ","End":"00:45.495","Text":"v_2, and v_3,"},{"Start":"00:45.495 ","End":"00:48.660","Text":"as here, and these will be u_1, u_2,"},{"Start":"00:48.660 ","End":"00:53.130","Text":"and u_3, so we have to have a T of v_1 is u_1 and so on."},{"Start":"00:53.130 ","End":"00:57.330","Text":"The next thing we have to do is check if v_1, v_2,"},{"Start":"00:57.330 ","End":"01:02.820","Text":"and v_3 are linearly dependent because if they are linearly dependent,"},{"Start":"01:02.820 ","End":"01:04.980","Text":"that will put some restrictions on u_1,"},{"Start":"01:04.980 ","End":"01:06.495","Text":"u_2, u_3,"},{"Start":"01:06.495 ","End":"01:12.230","Text":"and maybe it will turn out that it\u0027s impossible to have such a T. You\u0027ll see."},{"Start":"01:12.230 ","End":"01:16.615","Text":"We\u0027ll do that check for v_1, v_2, v_3."},{"Start":"01:16.615 ","End":"01:19.620","Text":"What we do, is we build an augmented matrix,"},{"Start":"01:19.620 ","End":"01:21.120","Text":"this is the vector v_1,"},{"Start":"01:21.120 ","End":"01:23.475","Text":"this is v_2, and this is v_3."},{"Start":"01:23.475 ","End":"01:27.120","Text":"Now if we just wanted to know if they are linearly dependent or not,"},{"Start":"01:27.120 ","End":"01:31.545","Text":"we wouldn\u0027t bother with this extra column on the right,"},{"Start":"01:31.545 ","End":"01:34.740","Text":"but we want to also know if they are dependent,"},{"Start":"01:34.740 ","End":"01:36.525","Text":"how they are dependent."},{"Start":"01:36.525 ","End":"01:37.920","Text":"We need to take v_1,"},{"Start":"01:37.920 ","End":"01:40.695","Text":"v_2, v_3 also."},{"Start":"01:40.695 ","End":"01:47.805","Text":"Now, we need to perform row operations to see if we can get a 0 on the last row."},{"Start":"01:47.805 ","End":"01:50.175","Text":"We bring it to echelon form,"},{"Start":"01:50.175 ","End":"01:54.765","Text":"add the first row to the second row to get a 0 here,"},{"Start":"01:54.765 ","End":"01:56.160","Text":"and this is what we get."},{"Start":"01:56.160 ","End":"01:58.860","Text":"Notice also we have to do it with the right column,"},{"Start":"01:58.860 ","End":"02:05.895","Text":"to add this v_1 to v_2 to get this v_2 plus v_1."},{"Start":"02:05.895 ","End":"02:12.000","Text":"Next, we will subtract twice this row from this row to get a 0 here."},{"Start":"02:12.000 ","End":"02:15.210","Text":"After we do that we do get a 0 here,"},{"Start":"02:15.210 ","End":"02:18.600","Text":"but we also get a 0 here because twice this minus this,"},{"Start":"02:18.600 ","End":"02:23.770","Text":"and also notice that v_3 minus twice this is this."},{"Start":"02:24.950 ","End":"02:29.670","Text":"This last row is important because we got all 0s here,"},{"Start":"02:29.670 ","End":"02:37.935","Text":"it means that they are linearly dependent and that this combination is 0."},{"Start":"02:37.935 ","End":"02:40.545","Text":"I wrote that here."},{"Start":"02:40.545 ","End":"02:47.955","Text":"I prefer to put it in the form that v_3 is twice v_2 plus twice v_1."},{"Start":"02:47.955 ","End":"02:51.240","Text":"Now, why is this important?"},{"Start":"02:51.240 ","End":"02:54.315","Text":"Because I have this dependency,"},{"Start":"02:54.315 ","End":"02:59.670","Text":"we have to make sure that our conditions are consistent by applying T to both sides,"},{"Start":"02:59.670 ","End":"03:01.965","Text":"I should also get an equality."},{"Start":"03:01.965 ","End":"03:06.285","Text":"Let\u0027s check that. T is linear."},{"Start":"03:06.285 ","End":"03:08.805","Text":"This is one of the rules of linearity,"},{"Start":"03:08.805 ","End":"03:12.420","Text":"so I can break the right-hand side into 2 bits,"},{"Start":"03:12.420 ","End":"03:17.490","Text":"and then I can apply the other rule of linearity to take constants out,"},{"Start":"03:17.490 ","End":"03:21.075","Text":"and I bring this 2 upfront and this 2 upfront,"},{"Start":"03:21.075 ","End":"03:24.765","Text":"and now we have to settle whether this is equal,"},{"Start":"03:24.765 ","End":"03:27.480","Text":"but we have T of v_3,"},{"Start":"03:27.480 ","End":"03:30.240","Text":"T of v_2, and T of v_1."},{"Start":"03:30.240 ","End":"03:32.820","Text":"They were part of the condition of the problem."},{"Start":"03:32.820 ","End":"03:33.930","Text":"They were u_3,"},{"Start":"03:33.930 ","End":"03:36.420","Text":"u_2, u_1 like so."},{"Start":"03:36.420 ","End":"03:39.225","Text":"We have to check if this is equal,"},{"Start":"03:39.225 ","End":"03:41.430","Text":"and the answer is, yes."},{"Start":"03:41.430 ","End":"03:44.310","Text":"Because look, multiply the 2 here, I get 2, 0,"},{"Start":"03:44.310 ","End":"03:45.780","Text":"0, here I get 0,"},{"Start":"03:45.780 ","End":"03:47.940","Text":"2 minus 2, and then I add the component-wise,"},{"Start":"03:47.940 ","End":"03:49.650","Text":"2 and 0 is 2,"},{"Start":"03:49.650 ","End":"03:51.495","Text":"0 and 2 is 2,"},{"Start":"03:51.495 ","End":"03:53.100","Text":"0 and minus 2, is minus 2,"},{"Start":"03:53.100 ","End":"03:55.155","Text":"so our answer is yes."},{"Start":"03:55.155 ","End":"03:57.045","Text":"This is important."},{"Start":"03:57.045 ","End":"04:00.090","Text":"If we got an answer, no,"},{"Start":"04:00.090 ","End":"04:01.950","Text":"that this was not equal,"},{"Start":"04:01.950 ","End":"04:07.875","Text":"we would immediately stop and say that such a T does not exist."},{"Start":"04:07.875 ","End":"04:09.510","Text":"Important to remember this."},{"Start":"04:09.510 ","End":"04:12.855","Text":"I don\u0027t have an exercise where this comes out, no,"},{"Start":"04:12.855 ","End":"04:14.850","Text":"but remember if you get a no,"},{"Start":"04:14.850 ","End":"04:20.835","Text":"then you say T doesn\u0027t exist because it violates the condition."},{"Start":"04:20.835 ","End":"04:23.190","Text":"If this is true,"},{"Start":"04:23.190 ","End":"04:25.560","Text":"we have a dependency then we apply T,"},{"Start":"04:25.560 ","End":"04:28.180","Text":"the dependency still has to hold."},{"Start":"04:28.730 ","End":"04:32.380","Text":"That\u0027s okay, let\u0027s continue."},{"Start":"04:33.170 ","End":"04:38.280","Text":"Now we\u0027re back to looking for a T which satisfies these."},{"Start":"04:38.280 ","End":"04:42.855","Text":"I can tell you that when we had the case of the linear dependence,"},{"Start":"04:42.855 ","End":"04:47.190","Text":"T will exist once we\u0027ve passed that check,"},{"Start":"04:47.190 ","End":"04:48.975","Text":"but it won\u0027t be unique."},{"Start":"04:48.975 ","End":"04:55.560","Text":"Let\u0027s see that. I\u0027m going to recall the theory I\u0027ve used it several times before,"},{"Start":"04:55.560 ","End":"04:57.675","Text":"but here it is again."},{"Start":"04:57.675 ","End":"05:01.065","Text":"If we have 2 vector spaces V and U,"},{"Start":"05:01.065 ","End":"05:06.090","Text":"and then we have v_1 through v_n as the basis of V,"},{"Start":"05:06.090 ","End":"05:07.410","Text":"where n, of course,"},{"Start":"05:07.410 ","End":"05:10.455","Text":"is the dimension of the space V,"},{"Start":"05:10.455 ","End":"05:12.270","Text":"and we also have n vectors,"},{"Start":"05:12.270 ","End":"05:13.650","Text":"u_1 through u_n,"},{"Start":"05:13.650 ","End":"05:17.370","Text":"any vectors in the space U,"},{"Start":"05:17.370 ","End":"05:23.490","Text":"then there exists a unique linear transformation T from V to U,"},{"Start":"05:23.490 ","End":"05:26.295","Text":"which takes each v to each corresponding u,"},{"Start":"05:26.295 ","End":"05:27.420","Text":"v_1 to u_1,"},{"Start":"05:27.420 ","End":"05:29.745","Text":"v_2 to u_2 and so on."},{"Start":"05:29.745 ","End":"05:32.130","Text":"How does this relate to our case,"},{"Start":"05:32.130 ","End":"05:34.860","Text":"because here these three are not a basis,"},{"Start":"05:34.860 ","End":"05:39.195","Text":"and they\u0027re not even linearly dependent? I just wrote that."},{"Start":"05:39.195 ","End":"05:41.730","Text":"R set which were these three,"},{"Start":"05:41.730 ","End":"05:47.055","Text":"and not basis like I said they\u0027re not even linearly independent,"},{"Start":"05:47.055 ","End":"05:53.865","Text":"so I can actually throw this one out and then these 2 will have the same span as these 3;"},{"Start":"05:53.865 ","End":"05:59.595","Text":"but what we can do is that because these two are linearly independent,"},{"Start":"05:59.595 ","End":"06:01.290","Text":"without the third one,"},{"Start":"06:01.290 ","End":"06:05.910","Text":"I can complete to a basis."},{"Start":"06:05.910 ","End":"06:09.120","Text":"I have only 2,"},{"Start":"06:09.120 ","End":"06:11.775","Text":"it\u0027s too few, I need 4 in a basis"},{"Start":"06:11.775 ","End":"06:16.350","Text":"but when they\u0027re linearly independent you can complete it to a basis."},{"Start":"06:16.350 ","End":"06:19.320","Text":"There are probably many ways of doing this,"},{"Start":"06:19.320 ","End":"06:24.570","Text":"probably infinitely many but here\u0027s one way I found of completing to a basis."},{"Start":"06:24.570 ","End":"06:26.820","Text":"I won\u0027t go into the technical details,"},{"Start":"06:26.820 ","End":"06:31.605","Text":"but you just have to make sure that if you get a 4 by 4 matrix with these other rows"},{"Start":"06:31.605 ","End":"06:38.220","Text":"that in echelon form it doesn\u0027t have any 0 rows,"},{"Start":"06:38.220 ","End":"06:39.810","Text":"and if you check these 4,"},{"Start":"06:39.810 ","End":"06:44.890","Text":"I won\u0027t check it, these 4 are linearly independent and are a basis."},{"Start":"06:45.080 ","End":"06:48.300","Text":"Because these 4 are a basis,"},{"Start":"06:48.300 ","End":"06:53.595","Text":"we can use the theorem to decide where to send each of these 4."},{"Start":"06:53.595 ","End":"06:59.205","Text":"We can choose any 4 vectors in our 3 and for example,"},{"Start":"06:59.205 ","End":"07:03.030","Text":"we want to choose that this one goes here as part of the conditions,"},{"Start":"07:03.030 ","End":"07:05.400","Text":"and we want to choose that this one goes to this,"},{"Start":"07:05.400 ","End":"07:06.960","Text":"that was also a condition."},{"Start":"07:06.960 ","End":"07:09.810","Text":"For these two we can choose what we like,"},{"Start":"07:09.810 ","End":"07:14.740","Text":"so this is just to send them both to 0."},{"Start":"07:15.140 ","End":"07:22.545","Text":"Now the theorem guarantees that there is a unique T which satisfies these conditions,"},{"Start":"07:22.545 ","End":"07:29.025","Text":"but it\u0027s not unique in terms of the original problem because we made choices."},{"Start":"07:29.025 ","End":"07:33.870","Text":"For one thing, we made a choice of how to complete these two to a basis,"},{"Start":"07:33.870 ","End":"07:36.135","Text":"and we could have done this differently,"},{"Start":"07:36.135 ","End":"07:39.150","Text":"the other choice we made was where to send these two."},{"Start":"07:39.150 ","End":"07:42.240","Text":"We sent them to 0 and 0 but we could have chosen"},{"Start":"07:42.240 ","End":"07:46.980","Text":"2 other vectors so the answer that we find will not be unique,"},{"Start":"07:46.980 ","End":"07:50.955","Text":"but once you\u0027ve written this then that specifies just one T,"},{"Start":"07:50.955 ","End":"07:53.070","Text":"but overall it\u0027s not unique."},{"Start":"07:53.070 ","End":"07:56.710","Text":"I\u0027ve put what I\u0027ve said in a few words here,"},{"Start":"07:57.110 ","End":"08:00.630","Text":"just summarizing where we are."},{"Start":"08:00.630 ","End":"08:06.120","Text":"We want to find what is T in general of x, y, z,"},{"Start":"08:06.120 ","End":"08:11.400","Text":"t. We want it to be something in R^3 something, something,"},{"Start":"08:11.400 ","End":"08:19.200","Text":"something given that these 4 vectors go to these 4 vectors here."},{"Start":"08:19.200 ","End":"08:24.540","Text":"The technique we use is to find the coordinate vector of x, y, z,"},{"Start":"08:24.540 ","End":"08:27.870","Text":"t relative to this basis of these 4 vectors,"},{"Start":"08:27.870 ","End":"08:34.170","Text":"and what we do is we write this one here, this one here."},{"Start":"08:34.170 ","End":"08:38.850","Text":"The rows become columns for the 4 members"},{"Start":"08:38.850 ","End":"08:43.620","Text":"of the basis and the extra column is x, y, z,"},{"Start":"08:43.620 ","End":"08:48.035","Text":"t, and the idea is to do row operations on"},{"Start":"08:48.035 ","End":"08:54.100","Text":"this until we get the identity matrix to the left of the separator."},{"Start":"08:54.100 ","End":"09:01.905","Text":"Subtract twice the first row from the second row and add the first row to the third row,"},{"Start":"09:01.905 ","End":"09:04.215","Text":"and we get this."},{"Start":"09:04.215 ","End":"09:06.930","Text":"Don\u0027t forget to do also it to the x, y, z,"},{"Start":"09:06.930 ","End":"09:10.590","Text":"t. We subtract twice x from y,"},{"Start":"09:10.590 ","End":"09:16.350","Text":"and we add the first row to the third row so here\u0027s z plus x."},{"Start":"09:16.350 ","End":"09:23.010","Text":"Next thing get a 0 here by taking twice the fourth row minus the second row,"},{"Start":"09:23.010 ","End":"09:26.145","Text":"and that will give us this."},{"Start":"09:26.145 ","End":"09:33.580","Text":"Notice that here I get twice t minus this which is minus y plus 2x."},{"Start":"09:34.010 ","End":"09:36.750","Text":"Here the remaining steps,"},{"Start":"09:36.750 ","End":"09:40.680","Text":"I divided this by 2,"},{"Start":"09:40.680 ","End":"09:42.585","Text":"and I divided this by 2,"},{"Start":"09:42.585 ","End":"09:44.520","Text":"that gives us 1\u0027s here."},{"Start":"09:44.520 ","End":"09:48.600","Text":"Then all I have to do is get rid of this minus 1 by adding"},{"Start":"09:48.600 ","End":"09:54.390","Text":"the second row to the first row and this is what we get."},{"Start":"09:54.390 ","End":"09:56.520","Text":"Here\u0027s the identity matrix and these are"},{"Start":"09:56.520 ","End":"10:03.130","Text":"the 4 coordinate vectors relative to the new basis."},{"Start":"10:03.380 ","End":"10:06.030","Text":"Then these coordinates,"},{"Start":"10:06.030 ","End":"10:11.400","Text":"what I do with them is I take each one with its respective basis member."},{"Start":"10:11.400 ","End":"10:14.400","Text":"I take the 1/2y with the first one,"},{"Start":"10:14.400 ","End":"10:15.570","Text":"I can still see it here,"},{"Start":"10:15.570 ","End":"10:20.085","Text":"with this one, then this one with this one."},{"Start":"10:20.085 ","End":"10:21.675","Text":"Maybe I\u0027ll do some coloring."},{"Start":"10:21.675 ","End":"10:31.120","Text":"This one with this one gives me this and then this one with this one gives me this."},{"Start":"10:31.120 ","End":"10:36.500","Text":"This one with this one gives this."},{"Start":"10:36.500 ","End":"10:43.835","Text":"Finally, this coordinate with this vector gives this."},{"Start":"10:43.835 ","End":"10:47.300","Text":"What we have now is x, y, z,"},{"Start":"10:47.300 ","End":"10:52.650","Text":"t as a combination of our basis which was these 2 vectors,"},{"Start":"10:52.650 ","End":"10:56.250","Text":"and the two other ones that we completed to a basis,"},{"Start":"10:56.250 ","End":"10:58.965","Text":"but we want t of this."},{"Start":"10:58.965 ","End":"11:04.935","Text":"We just apply the transformation T to the left-hand side and to the right-hand side,"},{"Start":"11:04.935 ","End":"11:08.370","Text":"and there we\u0027ll use the properties of linearity to break"},{"Start":"11:08.370 ","End":"11:12.015","Text":"up sums and scalars times the vector."},{"Start":"11:12.015 ","End":"11:16.500","Text":"We can do 2 steps in one from the linearity of T. First of all,"},{"Start":"11:16.500 ","End":"11:20.460","Text":"I could apply T to each of the 4 terms separately,"},{"Start":"11:20.460 ","End":"11:23.985","Text":"and then I would take the scalars upfront,"},{"Start":"11:23.985 ","End":"11:26.295","Text":"and we can do it all in one step."},{"Start":"11:26.295 ","End":"11:29.865","Text":"Basically, you just throw T in in front of each vector."},{"Start":"11:29.865 ","End":"11:31.650","Text":"I put a T in here,"},{"Start":"11:31.650 ","End":"11:33.015","Text":"in front of here,"},{"Start":"11:33.015 ","End":"11:36.510","Text":"in front of here, and in front of here and this gives us this,"},{"Start":"11:36.510 ","End":"11:38.860","Text":"that\u0027s 2 steps in one."},{"Start":"11:39.200 ","End":"11:42.330","Text":"Now if you look back we have the conditions."},{"Start":"11:42.330 ","End":"11:45.135","Text":"We know that T of this is this,"},{"Start":"11:45.135 ","End":"11:50.685","Text":"and we know that this one goes to this one and that this goes"},{"Start":"11:50.685 ","End":"11:57.720","Text":"to this and this goes to this."},{"Start":"11:57.720 ","End":"12:01.380","Text":"Now, we chose the last 2 vectors to be 0,"},{"Start":"12:01.380 ","End":"12:03.000","Text":"so they don\u0027t count,"},{"Start":"12:03.000 ","End":"12:06.450","Text":"so we just have the first 2 and if I multiply out,"},{"Start":"12:06.450 ","End":"12:07.800","Text":"this one becomes this,"},{"Start":"12:07.800 ","End":"12:09.945","Text":"this one becomes this,"},{"Start":"12:09.945 ","End":"12:16.485","Text":"and then we do the addition of this 2 component-wise and this gives us our answer."},{"Start":"12:16.485 ","End":"12:18.510","Text":"I could say we\u0027re done,"},{"Start":"12:18.510 ","End":"12:19.980","Text":"but you probably know me,"},{"Start":"12:19.980 ","End":"12:21.975","Text":"I also like to also do a verification."},{"Start":"12:21.975 ","End":"12:23.820","Text":"If you have time on an exam,"},{"Start":"12:23.820 ","End":"12:25.530","Text":"then I would do that."},{"Start":"12:25.530 ","End":"12:29.955","Text":"What you might want to do to see if you have made some silly mistake,"},{"Start":"12:29.955 ","End":"12:34.380","Text":"is to substitute those 3 vectors from the very beginning in"},{"Start":"12:34.380 ","End":"12:40.080","Text":"the original part of the question and see if you apply T to them using this formula,"},{"Start":"12:40.080 ","End":"12:44.385","Text":"that you really get those 3 vectors that we were asked for."},{"Start":"12:44.385 ","End":"12:47.670","Text":"For example, in the first one, x is 1,"},{"Start":"12:47.670 ","End":"12:49.440","Text":"y is 2, z is minus 1,"},{"Start":"12:49.440 ","End":"12:50.760","Text":"and t is 0,"},{"Start":"12:50.760 ","End":"12:59.685","Text":"so the first components are 1/2 y minus x is 1/2 2 minus 1 is 0, and so on."},{"Start":"12:59.685 ","End":"13:03.375","Text":"Just technical and this does work out,"},{"Start":"13:03.375 ","End":"13:06.610","Text":"and we are finally done."}],"ID":10240},{"Watched":false,"Name":"Exercise 18","Duration":"11m 29s","ChapterTopicVideoID":9691,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"In this exercise, we have to determine if there\u0027s"},{"Start":"00:03.390 ","End":"00:09.270","Text":"a linear transformation T from P2 over R to P2 over R,"},{"Start":"00:09.270 ","End":"00:13.010","Text":"such that the following 3 conditions hold."},{"Start":"00:13.010 ","End":"00:17.155","Text":"T of the polynomial 1 is the polynomial 4,"},{"Start":"00:17.155 ","End":"00:20.310","Text":"T of 4x plus x squared is x,"},{"Start":"00:20.310 ","End":"00:23.820","Text":"and T of 1 minus x is x squared plus 1."},{"Start":"00:23.820 ","End":"00:26.040","Text":"P2 of R, just to remind you,"},{"Start":"00:26.040 ","End":"00:31.060","Text":"is the space of polynomials up to degree 2."},{"Start":"00:31.430 ","End":"00:34.125","Text":"Anyway, the question is,"},{"Start":"00:34.125 ","End":"00:35.850","Text":"is there and if there isn\u0027t,"},{"Start":"00:35.850 ","End":"00:37.890","Text":"you explain why there isn\u0027t."},{"Start":"00:37.890 ","End":"00:39.645","Text":"If there is,"},{"Start":"00:39.645 ","End":"00:42.049","Text":"there\u0027s a further question, is it unique?"},{"Start":"00:42.049 ","End":"00:45.635","Text":"If it\u0027s unique, then find its formula."},{"Start":"00:45.635 ","End":"00:53.350","Text":"Now, a typical member of the space is a plus bx plus cx squared."},{"Start":"00:53.900 ","End":"00:57.510","Text":"If we take the standard basis,"},{"Start":"00:57.510 ","End":"01:02.195","Text":"1, x, and x squared for this space,"},{"Start":"01:02.195 ","End":"01:05.565","Text":"then if I take a polynomial,"},{"Start":"01:05.565 ","End":"01:07.550","Text":"a plus bx plus cx squared,"},{"Start":"01:07.550 ","End":"01:12.290","Text":"its coordinate vector relative to this basis is just a, b,"},{"Start":"01:12.290 ","End":"01:18.610","Text":"and c because it\u0027s a times 1 plus b times x plus c times x squared."},{"Start":"01:18.610 ","End":"01:21.885","Text":"Why am I doing all this?"},{"Start":"01:21.885 ","End":"01:27.170","Text":"Well, it\u0027s easier to work in R^3 than in P2 of R. In general,"},{"Start":"01:27.170 ","End":"01:29.090","Text":"we like the spaces RN."},{"Start":"01:29.090 ","End":"01:36.420","Text":"What we can do is to switch from P2 over R"},{"Start":"01:36.420 ","End":"01:44.520","Text":"to R^3 and translate these 3 conditions to conditions on vectors in R^3."},{"Start":"01:44.520 ","End":"01:46.965","Text":"1 corresponds to 1, 0, 0."},{"Start":"01:46.965 ","End":"01:52.470","Text":"4x plus x squared is 0 plus 4x plus 1x squared is 0,"},{"Start":"01:52.470 ","End":"01:54.000","Text":"4, 1 and so on."},{"Start":"01:54.000 ","End":"01:56.940","Text":"X squared plus 1 is 1, 0, 1."},{"Start":"01:56.940 ","End":"01:59.370","Text":"Converting everything to R^3,"},{"Start":"01:59.370 ","End":"02:01.755","Text":"we\u0027re in more familiar territory."},{"Start":"02:01.755 ","End":"02:03.540","Text":"Once we\u0027ve solved it in R^3,"},{"Start":"02:03.540 ","End":"02:11.555","Text":"we\u0027ll get our way back to P2 over R. Now, what I\u0027ve done,"},{"Start":"02:11.555 ","End":"02:16.985","Text":"just to be pedantic is that not really supposed to use the same letter T again,"},{"Start":"02:16.985 ","End":"02:19.805","Text":"so I\u0027ve called this one T bar,"},{"Start":"02:19.805 ","End":"02:22.235","Text":"the one that goes from R^3 to R^3,"},{"Start":"02:22.235 ","End":"02:28.650","Text":"but many would just leave the letter T as is and no harm would be done."},{"Start":"02:28.650 ","End":"02:33.559","Text":"I\u0027m just being extra pedantic because it\u0027s in theory a different function,"},{"Start":"02:33.559 ","End":"02:35.845","Text":"a different transformation,"},{"Start":"02:35.845 ","End":"02:39.070","Text":"but if you don\u0027t like it, ignore the bar."},{"Start":"02:39.580 ","End":"02:43.030","Text":"How do we solve this?"},{"Start":"02:43.030 ","End":"02:48.860","Text":"There is that theorem I could use and I could use it if I knew that these 3,"},{"Start":"02:48.860 ","End":"02:50.240","Text":"I wrote them here,"},{"Start":"02:50.240 ","End":"02:52.820","Text":"is a basis of R^3."},{"Start":"02:52.820 ","End":"02:56.135","Text":"How do we check if something is a basis?"},{"Start":"02:56.135 ","End":"03:00.680","Text":"We take these coordinates and we place them as"},{"Start":"03:00.680 ","End":"03:05.885","Text":"rows in a matrix and then we reduce this to row echelon form."},{"Start":"03:05.885 ","End":"03:09.560","Text":"First subtract the top row from the bottom row."},{"Start":"03:09.560 ","End":"03:11.405","Text":"That brings us here."},{"Start":"03:11.405 ","End":"03:18.290","Text":"Next, what I did was I multiplied the last row by 4 and added the second row,"},{"Start":"03:18.290 ","End":"03:20.430","Text":"so 0 minus 4,"},{"Start":"03:20.430 ","End":"03:21.840","Text":"0, plus 0, 4,"},{"Start":"03:21.840 ","End":"03:23.720","Text":"1 gives us 0, 0, 1."},{"Start":"03:23.720 ","End":"03:28.170","Text":"It\u0027s now in row echelon form and we see that there are no 0\u0027s,"},{"Start":"03:28.170 ","End":"03:30.255","Text":"no rows of 0\u0027s."},{"Start":"03:30.255 ","End":"03:35.490","Text":"They\u0027re linearly independent and therefore they are a basis."},{"Start":"03:35.860 ","End":"03:39.930","Text":"Now, I can use that theorem."},{"Start":"03:40.010 ","End":"03:42.340","Text":"I\u0027m sure you\u0027ve seen this before,"},{"Start":"03:42.340 ","End":"03:44.120","Text":"I\u0027ve used it several times."},{"Start":"03:44.120 ","End":"03:48.950","Text":"V and U are spaces and we have a transformation from V to"},{"Start":"03:48.950 ","End":"03:54.840","Text":"U and we also have a basis of v,"},{"Start":"03:55.010 ","End":"03:57.390","Text":"v1 through vn,"},{"Start":"03:57.390 ","End":"03:59.475","Text":"where n is the dimension."},{"Start":"03:59.475 ","End":"04:03.240","Text":"If we take any n vectors in U,"},{"Start":"04:03.240 ","End":"04:06.320","Text":"there\u0027s a unique transformation which sends each V to"},{"Start":"04:06.320 ","End":"04:10.300","Text":"the corresponding U as is written here."},{"Start":"04:10.300 ","End":"04:14.340","Text":"Now we can apply this to our case."},{"Start":"04:14.340 ","End":"04:17.280","Text":"Our v1, v2, v3,"},{"Start":"04:17.280 ","End":"04:23.040","Text":"the basis of V is this and these"},{"Start":"04:23.040 ","End":"04:29.920","Text":"are any 3 vectors from U. V and U are both R^3."},{"Start":"04:29.920 ","End":"04:37.340","Text":"By the theorem, we\u0027re guaranteed to have such a T that satisfies the above,"},{"Start":"04:37.340 ","End":"04:38.600","Text":"that this goes to this,"},{"Start":"04:38.600 ","End":"04:41.030","Text":"this goes to this, this goes to this."},{"Start":"04:41.030 ","End":"04:45.130","Text":"That\u0027s the theoretical part that exists and it\u0027s unique,"},{"Start":"04:45.130 ","End":"04:47.810","Text":"but now we\u0027re going to actually find it,"},{"Start":"04:47.810 ","End":"04:51.585","Text":"find the formula for T bar."},{"Start":"04:51.585 ","End":"04:54.610","Text":"I copied what goes to what,"},{"Start":"04:54.610 ","End":"04:58.280","Text":"I just changed the order of these 2 rows."},{"Start":"04:58.280 ","End":"04:59.420","Text":"If you look above,"},{"Start":"04:59.420 ","End":"05:00.590","Text":"this came before this."},{"Start":"05:00.590 ","End":"05:02.015","Text":"It suits me better."},{"Start":"05:02.015 ","End":"05:05.435","Text":"You\u0027ll see in a moment why I swapped them around."},{"Start":"05:05.435 ","End":"05:09.905","Text":"Our first step is to find the general coordinate vector,"},{"Start":"05:09.905 ","End":"05:11.480","Text":"the general vector x, y,"},{"Start":"05:11.480 ","End":"05:16.075","Text":"z relative to the basis that we have, which is this."},{"Start":"05:16.075 ","End":"05:18.600","Text":"I said x, y, z. I changed my mind."},{"Start":"05:18.600 ","End":"05:21.120","Text":"I\u0027ll use a, b, c this time."},{"Start":"05:21.120 ","End":"05:26.085","Text":"Here, we put the vectors from the basis."},{"Start":"05:26.085 ","End":"05:29.465","Text":"Let\u0027s say this one goes here,"},{"Start":"05:29.465 ","End":"05:31.190","Text":"they become column vectors."},{"Start":"05:31.190 ","End":"05:36.620","Text":"This one here and this one here and there\u0027s the a,"},{"Start":"05:36.620 ","End":"05:37.670","Text":"b, c or x, y,"},{"Start":"05:37.670 ","End":"05:39.985","Text":"z or whatever we chose."},{"Start":"05:39.985 ","End":"05:44.810","Text":"The plan is to do row operations on"},{"Start":"05:44.810 ","End":"05:54.410","Text":"this augmented matrix until we get the identity matrix in the first 3 columns,"},{"Start":"05:54.410 ","End":"05:56.285","Text":"the first 3 rows."},{"Start":"05:56.285 ","End":"06:01.415","Text":"Identity matrix, meaning 1\u0027s here and 0\u0027s elsewhere."},{"Start":"06:01.415 ","End":"06:07.160","Text":"What I did here was to subtract 4 times the third row from"},{"Start":"06:07.160 ","End":"06:13.810","Text":"the second row to make a 0 here and here I get b minus 4c."},{"Start":"06:13.810 ","End":"06:20.345","Text":"Then I added this row to this row to get a 0 here."},{"Start":"06:20.345 ","End":"06:24.730","Text":"We get here a plus b minus 4c."},{"Start":"06:24.730 ","End":"06:27.370","Text":"We\u0027re almost there."},{"Start":"06:27.370 ","End":"06:34.760","Text":"Just had to multiply this middle row by minus 1 to make this a plus 1"},{"Start":"06:34.760 ","End":"06:42.710","Text":"so the b minus 4c swaps order and now we have the identity matrix here."},{"Start":"06:42.710 ","End":"06:51.180","Text":"These are the coordinates for the coordinate vector of a,"},{"Start":"06:51.180 ","End":"06:54.155","Text":"b, c relative to this basis."},{"Start":"06:54.155 ","End":"07:01.790","Text":"What it means is that we can write our general a,"},{"Start":"07:01.790 ","End":"07:09.450","Text":"b, c in terms of the 3 basis vectors and these 3 are the coordinates."},{"Start":"07:09.450 ","End":"07:13.140","Text":"It\u0027s this one here times the first basis vector,"},{"Start":"07:13.140 ","End":"07:17.160","Text":"this here times the second basis vector,"},{"Start":"07:17.160 ","End":"07:21.420","Text":"and c, from here times the third basis vector."},{"Start":"07:21.420 ","End":"07:25.110","Text":"This is just the important step."},{"Start":"07:25.110 ","End":"07:27.270","Text":"It\u0027s just the first step."},{"Start":"07:27.270 ","End":"07:34.380","Text":"Next, I want to apply T to both sides because what I need is T of a, b,"},{"Start":"07:34.380 ","End":"07:40.710","Text":"c. I apply T or T-bar,"},{"Start":"07:40.710 ","End":"07:43.440","Text":"whatever, to each side."},{"Start":"07:43.440 ","End":"07:46.905","Text":"Here it is here and on the right also,"},{"Start":"07:46.905 ","End":"07:49.425","Text":"T of this whole expression,"},{"Start":"07:49.425 ","End":"07:54.100","Text":"but we can use the linearity of T to break this up."},{"Start":"07:54.100 ","End":"07:57.065","Text":"There\u0027s more than one way of tackling this."},{"Start":"07:57.065 ","End":"08:01.130","Text":"We could use the 2 axioms for linearity,"},{"Start":"08:01.130 ","End":"08:02.780","Text":"which is this and this,"},{"Start":"08:02.780 ","End":"08:06.950","Text":"but if you recall, there\u0027s also a single axiom that we can use."},{"Start":"08:06.950 ","End":"08:11.285","Text":"That whenever we have T of ku plus lv,"},{"Start":"08:11.285 ","End":"08:13.580","Text":"we can take k times T of u,"},{"Start":"08:13.580 ","End":"08:18.855","Text":"l times T of v and we can even extend this from 2 terms to 3 terms."},{"Start":"08:18.855 ","End":"08:23.360","Text":"Suppose I put also ku plus lv plus mw or whatever."},{"Start":"08:23.360 ","End":"08:27.230","Text":"Basically, what it means is I can throw T in before the vector in"},{"Start":"08:27.230 ","End":"08:33.670","Text":"each place and if I do that to this row here,"},{"Start":"08:34.030 ","End":"08:37.040","Text":"then this is what we get."},{"Start":"08:37.040 ","End":"08:38.930","Text":"You can also do it in 2 steps."},{"Start":"08:38.930 ","End":"08:43.670","Text":"If you want to use these 2, you can first of all say it\u0027s T of the whole first term,"},{"Start":"08:43.670 ","End":"08:44.975","Text":"T of this, T of this,"},{"Start":"08:44.975 ","End":"08:46.880","Text":"and then bring the constants in front."},{"Start":"08:46.880 ","End":"08:53.180","Text":"Either way, this is where we end up with. This is good."},{"Start":"08:53.180 ","End":"08:58.800","Text":"The reason we\u0027re in good shape now is because we know what T of 1,"},{"Start":"08:58.800 ","End":"09:01.580","Text":"0, 0 should be and 1 minus 1,"},{"Start":"09:01.580 ","End":"09:03.110","Text":"0 and 0, 4, 1."},{"Start":"09:03.110 ","End":"09:06.934","Text":"This was the initial conditions for the problem."},{"Start":"09:06.934 ","End":"09:09.410","Text":"This, if you look at the beginning,"},{"Start":"09:09.410 ","End":"09:17.595","Text":"was required to be this and T of this has got to be this and T of this is got to be this."},{"Start":"09:17.595 ","End":"09:20.885","Text":"Now we\u0027re almost at a formula for T bar."},{"Start":"09:20.885 ","End":"09:24.715","Text":"This is the formula. We just have to tidy it up a bit."},{"Start":"09:24.715 ","End":"09:28.865","Text":"Multiply this scalar by each of the components here."},{"Start":"09:28.865 ","End":"09:33.319","Text":"Do the same thing here and c times this vector."},{"Start":"09:33.319 ","End":"09:36.320","Text":"Now let\u0027s add component-wise."},{"Start":"09:36.320 ","End":"09:40.655","Text":"After doing that, we get the formula for T,"},{"Start":"09:40.655 ","End":"09:44.060","Text":"only it\u0027s not quite T. I distinguished,"},{"Start":"09:44.060 ","End":"09:45.440","Text":"I called it T bar."},{"Start":"09:45.440 ","End":"09:52.700","Text":"We went over from the world of polynomials to the world of vectors from P2 of R to R^3."},{"Start":"09:52.700 ","End":"09:54.455","Text":"Now we have to go back."},{"Start":"09:54.455 ","End":"10:02.495","Text":"Remember that a, b, c was our way of representing a plus bx plus cx squared."},{"Start":"10:02.495 ","End":"10:12.890","Text":"Likewise, this is this times 1 and c times x and 4c minus b times x squared."},{"Start":"10:12.890 ","End":"10:17.135","Text":"This really is the answer, not this."},{"Start":"10:17.135 ","End":"10:20.550","Text":"We just used R^3 because it was easier to work with,"},{"Start":"10:20.550 ","End":"10:26.240","Text":"but at the end we convert back to polynomials of degree 2 and less."},{"Start":"10:26.240 ","End":"10:28.520","Text":"I would say that we\u0027re done,"},{"Start":"10:28.520 ","End":"10:29.810","Text":"but you know me,"},{"Start":"10:29.810 ","End":"10:32.285","Text":"I like to add an extra step,"},{"Start":"10:32.285 ","End":"10:35.300","Text":"step 3, which is the verification."},{"Start":"10:35.300 ","End":"10:38.750","Text":"This is optional, but if you have the time on the exam or whatever,"},{"Start":"10:38.750 ","End":"10:42.905","Text":"you should do it and see if the initial conditions are met,"},{"Start":"10:42.905 ","End":"10:46.255","Text":"if T of 1 is 4."},{"Start":"10:46.255 ","End":"10:51.600","Text":"You would just do this by substituting here."},{"Start":"10:53.180 ","End":"10:55.350","Text":"Well, maybe I\u0027ll do one of them."},{"Start":"10:55.350 ","End":"10:56.895","Text":"The polynomial 1,"},{"Start":"10:56.895 ","End":"10:58.485","Text":"if I plug it in,"},{"Start":"10:58.485 ","End":"11:00.330","Text":"then a is 1, b is 0,"},{"Start":"11:00.330 ","End":"11:01.680","Text":"c is 0,"},{"Start":"11:01.680 ","End":"11:05.780","Text":"so we get 4 here and this is 0,"},{"Start":"11:05.780 ","End":"11:08.450","Text":"this is 0, this also is 0,"},{"Start":"11:08.450 ","End":"11:11.735","Text":"and c and b are 0, so we just get 4."},{"Start":"11:11.735 ","End":"11:15.830","Text":"Likewise, check if a is 0, b is 1,"},{"Start":"11:15.830 ","End":"11:17.270","Text":"c is 1,"},{"Start":"11:17.270 ","End":"11:22.050","Text":"whether we really get this."},{"Start":"11:22.050 ","End":"11:24.390","Text":"I\u0027ll leave that to you."},{"Start":"11:24.390 ","End":"11:30.210","Text":"Anyway, this here is the answer and we are done."}],"ID":10241},{"Watched":false,"Name":"Exercise 19","Duration":"11m 40s","ChapterTopicVideoID":9692,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.680","Text":"In this exercise, we\u0027re asked to find out if there is"},{"Start":"00:04.680 ","End":"00:09.990","Text":"a linear transformation from R^3 to R^3, call it T,"},{"Start":"00:09.990 ","End":"00:18.285","Text":"which satisfies the following 3 conditions that T of this vector is this vector,"},{"Start":"00:18.285 ","End":"00:22.030","Text":"and so on for the other 2 as written here."},{"Start":"00:23.000 ","End":"00:26.310","Text":"If there isn\u0027t such a transformation,"},{"Start":"00:26.310 ","End":"00:28.095","Text":"we have to explain why."},{"Start":"00:28.095 ","End":"00:31.134","Text":"If there is and it happens to be unique,"},{"Start":"00:31.134 ","End":"00:36.080","Text":"then we want to find its formula."},{"Start":"00:36.080 ","End":"00:38.360","Text":"Now to solve this exercise,"},{"Start":"00:38.360 ","End":"00:40.780","Text":"we need a theorem."},{"Start":"00:40.780 ","End":"00:43.715","Text":"Without this, I don\u0027t see how we would do it"},{"Start":"00:43.715 ","End":"00:46.010","Text":"but here\u0027s the theorem I\u0027ll give you in"},{"Start":"00:46.010 ","End":"00:48.890","Text":"general and then we\u0027ll see how it applies in our case."},{"Start":"00:48.890 ","End":"00:53.590","Text":"In general, V and U are 2 vector spaces."},{"Start":"00:53.590 ","End":"00:57.600","Text":"Suppose that we have n vectors,"},{"Start":"00:57.600 ","End":"00:59.200","Text":"v_1 through v_n,"},{"Start":"00:59.200 ","End":"01:05.160","Text":"and they\u0027re a basis of V. N of course is the dimension."},{"Start":"01:05.380 ","End":"01:11.030","Text":"U_1 through u_n are any n vectors in U,"},{"Start":"01:11.030 ","End":"01:13.290","Text":"in the other one."},{"Start":"01:13.720 ","End":"01:19.640","Text":"In those conditions, there exists a unique linear transformation"},{"Start":"01:19.640 ","End":"01:25.125","Text":"T from V to U such that T of v_1 is u_1,"},{"Start":"01:25.125 ","End":"01:28.355","Text":"T of v_2 is u_2, and so on."},{"Start":"01:28.355 ","End":"01:32.490","Text":"T of the last one is the last one here."},{"Start":"01:33.010 ","End":"01:36.305","Text":"How does this help us?"},{"Start":"01:36.305 ","End":"01:38.735","Text":"First the general idea,"},{"Start":"01:38.735 ","End":"01:42.935","Text":"these will be the vectors v_1, v_2, v_3."},{"Start":"01:42.935 ","End":"01:47.845","Text":"These 3 are a basis of R^3."},{"Start":"01:47.845 ","End":"01:50.415","Text":"V and U are both going to be R^3."},{"Start":"01:50.415 ","End":"01:56.340","Text":"These 3 are going to be R any 3 vectors from here."},{"Start":"01:56.340 ","End":"02:04.150","Text":"The theorem will guarantee that there is a transformation which takes v_1 to u_1,"},{"Start":"02:04.150 ","End":"02:06.790","Text":"v_2, to u_2, v_3 to u_3."},{"Start":"02:06.790 ","End":"02:08.485","Text":"Let me write this,"},{"Start":"02:08.485 ","End":"02:11.450","Text":"but I\u0027m going to have to scroll."},{"Start":"02:11.680 ","End":"02:17.770","Text":"These were the 3 vectors that I\u0027ve just lost off-screen."},{"Start":"02:17.770 ","End":"02:20.110","Text":"These are a basis."},{"Start":"02:20.110 ","End":"02:23.720","Text":"I\u0027m not going to spend time showing that this is a basis."},{"Start":"02:23.720 ","End":"02:26.590","Text":"Well, I\u0027ll just give you a hint of how we would go about it."},{"Start":"02:26.590 ","End":"02:32.905","Text":"You would write each vector as a row in a matrix,"},{"Start":"02:32.905 ","End":"02:36.905","Text":"0, 1, 1 0, 0, 1."},{"Start":"02:36.905 ","End":"02:40.930","Text":"Then we would bring this to row echelon form."},{"Start":"02:40.930 ","End":"02:43.130","Text":"It already is in row echelon form."},{"Start":"02:43.130 ","End":"02:45.290","Text":"We see there are no rows of 0s."},{"Start":"02:45.290 ","End":"02:53.630","Text":"These 3 are 3 linearly independent vectors in R^3 and so they\u0027re a basis."},{"Start":"02:53.630 ","End":"02:57.730","Text":"These are our v_1, v_2, v_3,"},{"Start":"02:57.730 ","End":"03:00.500","Text":"and the other 3, the ones that were on the right,"},{"Start":"03:00.500 ","End":"03:01.880","Text":"I remember those numbers, there were 0,"},{"Start":"03:01.880 ","End":"03:04.040","Text":"1, 2, 3, 4, 5, 6, 7, 8, 9."},{"Start":"03:04.040 ","End":"03:06.575","Text":"Those are the u_1, u_2, u_3."},{"Start":"03:06.575 ","End":"03:12.030","Text":"They are the any 3 vectors in U because V and U are both R^3 here."},{"Start":"03:13.850 ","End":"03:21.410","Text":"By the theorem, we know that there is a linear transformation T from V to U,"},{"Start":"03:21.410 ","End":"03:24.800","Text":"which is in our case from R^3 to R^3"},{"Start":"03:24.800 ","End":"03:29.180","Text":"which satisfies our conditions that T of this is this,"},{"Start":"03:29.180 ","End":"03:33.860","Text":"T of this is this and T of this is this as we required."},{"Start":"03:33.860 ","End":"03:40.280","Text":"Now, the difference between knowing that there\u0027s a unique T and actually finding it."},{"Start":"03:40.280 ","End":"03:44.690","Text":"The rest of the exercise will be actually finding the formula,"},{"Start":"03:44.690 ","End":"03:48.630","Text":"which takes these 3 to these 3."},{"Start":"03:48.950 ","End":"03:53.800","Text":"The formula for T will be of the form T of x, y,"},{"Start":"03:53.800 ","End":"03:56.650","Text":"z equals something, something,"},{"Start":"03:56.650 ","End":"04:00.790","Text":"something, because it takes R^3 and it also goes to R^3."},{"Start":"04:00.790 ","End":"04:05.830","Text":"What are these 3 things are going to be?"},{"Start":"04:05.830 ","End":"04:07.300","Text":"That\u0027s what we have to find out."},{"Start":"04:07.300 ","End":"04:10.210","Text":"There\u0027ll be expressions with x, y, and z."},{"Start":"04:10.210 ","End":"04:14.230","Text":"It turns out not to be so simple to find it."},{"Start":"04:14.230 ","End":"04:15.955","Text":"It\u0027s more than 1 step."},{"Start":"04:15.955 ","End":"04:22.090","Text":"In the first step we find the general coordinate vector relative to this basis B,"},{"Start":"04:22.090 ","End":"04:24.160","Text":"which means writing x, y,"},{"Start":"04:24.160 ","End":"04:28.110","Text":"z as a linear combination of these 3."},{"Start":"04:28.110 ","End":"04:30.055","Text":"The way to do this,"},{"Start":"04:30.055 ","End":"04:31.915","Text":"I\u0027m going to remind you,"},{"Start":"04:31.915 ","End":"04:36.430","Text":"we take the first vector and put it as the first column,"},{"Start":"04:36.430 ","End":"04:39.415","Text":"the second vector, the second column,"},{"Start":"04:39.415 ","End":"04:41.940","Text":"third vector as the third column."},{"Start":"04:41.940 ","End":"04:46.315","Text":"We put an extra column for x, y, z."},{"Start":"04:46.315 ","End":"04:53.155","Text":"The task is now to do row operations on this augmented matrix"},{"Start":"04:53.155 ","End":"04:59.950","Text":"until we get the identity matrix in the first 3 by 3,"},{"Start":"04:59.950 ","End":"05:02.275","Text":"not including the x y z"},{"Start":"05:02.275 ","End":"05:07.015","Text":"but they\u0027re all operations we do on everything including the x, y, z."},{"Start":"05:07.015 ","End":"05:12.610","Text":"The first row operation would be to subtract the first row from the second row,"},{"Start":"05:12.610 ","End":"05:15.175","Text":"and that would give me a 0 here."},{"Start":"05:15.175 ","End":"05:18.305","Text":"That brings us to this."},{"Start":"05:18.305 ","End":"05:23.135","Text":"Notice that they also subtract on the last column,"},{"Start":"05:23.135 ","End":"05:25.130","Text":"the first row from the second row."},{"Start":"05:25.130 ","End":"05:27.080","Text":"Here we have y minus x."},{"Start":"05:27.080 ","End":"05:29.765","Text":"The next thing to do is you want a 0 here."},{"Start":"05:29.765 ","End":"05:34.225","Text":"We\u0027ll subtract the second row from the third row."},{"Start":"05:34.225 ","End":"05:37.910","Text":"Now we can stop row operations because what we have"},{"Start":"05:37.910 ","End":"05:43.415","Text":"here is the identity matrix for 3 by 3."},{"Start":"05:43.415 ","End":"05:47.395","Text":"These will be our coordinates."},{"Start":"05:47.395 ","End":"05:54.950","Text":"I can use these 3 coordinates relative to this basis to write x, y, z."},{"Start":"05:54.950 ","End":"05:56.825","Text":"Well, I\u0027ll show you."},{"Start":"05:56.825 ","End":"06:02.900","Text":"I can write x, y, z as a linear combination of these 3,"},{"Start":"06:02.900 ","End":"06:06.170","Text":"or if you like these 3 columns as follows,"},{"Start":"06:06.170 ","End":"06:12.545","Text":"I take the x with the first one,"},{"Start":"06:12.545 ","End":"06:17.525","Text":"say this, and I write that here."},{"Start":"06:17.525 ","End":"06:24.665","Text":"Next, I take this y minus x with this vector here."},{"Start":"06:24.665 ","End":"06:28.050","Text":"That gives me this."},{"Start":"06:28.490 ","End":"06:36.900","Text":"Finally, this coordinate with this factor over here."},{"Start":"06:36.900 ","End":"06:39.050","Text":"Lets us write x, y,"},{"Start":"06:39.050 ","End":"06:43.340","Text":"z as a linear combination of the 3 basis vectors."},{"Start":"06:43.340 ","End":"06:48.775","Text":"The coordinates are the 3 numbers here which I\u0027ve now put here, here, and here."},{"Start":"06:48.775 ","End":"06:51.870","Text":"Now all this was just step 1."},{"Start":"06:51.870 ","End":"06:55.175","Text":"Now we\u0027ll get on to step 2."},{"Start":"06:55.175 ","End":"06:57.760","Text":"What we need is not x, y,"},{"Start":"06:57.760 ","End":"07:01.285","Text":"z but T of x, y, z"},{"Start":"07:01.285 ","End":"07:04.345","Text":"but we use this to find this."},{"Start":"07:04.345 ","End":"07:10.615","Text":"What I\u0027m going to do is apply the transformation T to both sides of the equation."},{"Start":"07:10.615 ","End":"07:14.270","Text":"Both sides of the equality, I should say."},{"Start":"07:15.440 ","End":"07:20.640","Text":"On the left we\u0027ve just put T with x, y, z,"},{"Start":"07:20.640 ","End":"07:22.925","Text":"and on the right T,"},{"Start":"07:22.925 ","End":"07:27.445","Text":"with all this inside the brackets here."},{"Start":"07:27.445 ","End":"07:29.800","Text":"Now, T has to be linear."},{"Start":"07:29.800 ","End":"07:34.650","Text":"I can take the right hand side and break it up."},{"Start":"07:34.650 ","End":"07:38.255","Text":"I have T of something plus something plus something."},{"Start":"07:38.255 ","End":"07:44.855","Text":"I can put T and apply it to each of these 3 terms."},{"Start":"07:44.855 ","End":"07:47.450","Text":"It\u0027s T of the first plus T of the second term,"},{"Start":"07:47.450 ","End":"07:49.595","Text":"plus T of the third term."},{"Start":"07:49.595 ","End":"07:55.505","Text":"I mean, the rule says T of u plus v equals T of u"},{"Start":"07:55.505 ","End":"08:01.190","Text":"plus T of v. But obviously we can extend it to 3 vectors."},{"Start":"08:01.190 ","End":"08:04.370","Text":"If I had u plus v plus w here I\u0027ll have T of u"},{"Start":"08:04.370 ","End":"08:07.700","Text":"plus T of v plus T of w. There could be more than 2 of them,"},{"Start":"08:07.700 ","End":"08:11.080","Text":"the 3 of them, we get this."},{"Start":"08:11.080 ","End":"08:19.745","Text":"Then there\u0027s the other rule that T of ku equals k times T of u."},{"Start":"08:19.745 ","End":"08:23.360","Text":"Here, x is like my K and I bring it out front."},{"Start":"08:23.360 ","End":"08:26.840","Text":"It basically means that you bring the scalar to the front here,"},{"Start":"08:26.840 ","End":"08:32.320","Text":"y minus x goes to the front and here has z minus y plus x goes to the front."},{"Start":"08:32.320 ","End":"08:39.830","Text":"Now we are in good shape because we were given initially that T of 1,"},{"Start":"08:39.830 ","End":"08:42.740","Text":"1, 0 is going to be 1, 2,"},{"Start":"08:42.740 ","End":"08:44.180","Text":"3, and this 1 is 4,"},{"Start":"08:44.180 ","End":"08:46.580","Text":"5, 6, and this one is 7, 8, 9."},{"Start":"08:46.580 ","End":"08:49.475","Text":"If you go back and look at the original conditions,"},{"Start":"08:49.475 ","End":"08:53.600","Text":"it said we are looking for T linear such that T of this is this,"},{"Start":"08:53.600 ","End":"08:54.875","Text":"T of this is this,"},{"Start":"08:54.875 ","End":"08:57.275","Text":"and T of this is this."},{"Start":"08:57.275 ","End":"09:00.980","Text":"Now we\u0027ve gotten rid of the T on"},{"Start":"09:00.980 ","End":"09:06.455","Text":"the right-hand side and all we have to do is a bit of expansion here, a bit of algebra."},{"Start":"09:06.455 ","End":"09:09.920","Text":"Yeah it\u0027s a little bit messy x times 1,"},{"Start":"09:09.920 ","End":"09:11.765","Text":"x times 2, x times 3."},{"Start":"09:11.765 ","End":"09:15.185","Text":"Then y minus x times 4 is 4y minus 4x,"},{"Start":"09:15.185 ","End":"09:18.080","Text":"5y minus 5x, and so on."},{"Start":"09:18.080 ","End":"09:20.610","Text":"These are bit longer."},{"Start":"09:20.610 ","End":"09:23.085","Text":"The last one would be 9 times this,"},{"Start":"09:23.085 ","End":"09:27.790","Text":"which is 9z minus 9y plus 9x. Here it is."},{"Start":"09:27.790 ","End":"09:30.800","Text":"Next we add the components,"},{"Start":"09:30.800 ","End":"09:37.140","Text":"like the first component here is x and here it\u0027s 4y minus 4x."},{"Start":"09:37.140 ","End":"09:38.795","Text":"Here it\u0027s this."},{"Start":"09:38.795 ","End":"09:40.235","Text":"I\u0027ll highlight it a bit."},{"Start":"09:40.235 ","End":"09:41.975","Text":"This is the first component here."},{"Start":"09:41.975 ","End":"09:44.495","Text":"This is the first component here up to the comma."},{"Start":"09:44.495 ","End":"09:46.730","Text":"This is the first component here."},{"Start":"09:46.730 ","End":"09:55.060","Text":"I put all these 3 together and I get the first component up to there."},{"Start":"09:55.730 ","End":"09:58.980","Text":"The second and third components similarly,"},{"Start":"09:58.980 ","End":"10:02.295","Text":"I won\u0027t bother highlighting those."},{"Start":"10:02.295 ","End":"10:04.970","Text":"Then we just simplify each components so that"},{"Start":"10:04.970 ","End":"10:07.910","Text":"this one would come out to be this because look,"},{"Start":"10:07.910 ","End":"10:10.775","Text":"x minus 4x plus 7x."},{"Start":"10:10.775 ","End":"10:13.010","Text":"The next is minus 4x is 4x,"},{"Start":"10:13.010 ","End":"10:22.290","Text":"4y minus 7y is minus 3y and z we just have 7z and that\u0027s here."},{"Start":"10:22.290 ","End":"10:25.145","Text":"Similarly the second component and the third component."},{"Start":"10:25.145 ","End":"10:28.045","Text":"This is simplified."},{"Start":"10:28.045 ","End":"10:31.365","Text":"This is our answer."},{"Start":"10:31.365 ","End":"10:33.740","Text":"We could say that we\u0027re done"},{"Start":"10:33.740 ","End":"10:39.670","Text":"but I would recommend the third step a verification."},{"Start":"10:39.670 ","End":"10:44.750","Text":"In those so many computations we could have made a mistake. It\u0027s best to check."},{"Start":"10:44.750 ","End":"10:50.630","Text":"Let\u0027s see if really the initial conditions are true by just substituting 1,"},{"Start":"10:50.630 ","End":"10:53.765","Text":"1, 0 in this formula,"},{"Start":"10:53.765 ","End":"10:55.730","Text":"then 0, 1, 1 in this formula."},{"Start":"10:55.730 ","End":"10:57.320","Text":"Let\u0027s see if we get these."},{"Start":"10:57.320 ","End":"11:02.185","Text":"I\u0027ll just check one of them, I will check the last one because it has lot of 0s in it."},{"Start":"11:02.185 ","End":"11:04.380","Text":"Let\u0027s see what is T of 0,"},{"Start":"11:04.380 ","End":"11:06.090","Text":"0, 1 from this formula."},{"Start":"11:06.090 ","End":"11:07.500","Text":"This means x is 0,"},{"Start":"11:07.500 ","End":"11:09.510","Text":"y is 0, z is 1."},{"Start":"11:09.510 ","End":"11:11.420","Text":"In the first component, 0,"},{"Start":"11:11.420 ","End":"11:14.030","Text":"0 and z is 1 is 7."},{"Start":"11:14.030 ","End":"11:16.070","Text":"7, that\u0027s good."},{"Start":"11:16.070 ","End":"11:21.230","Text":"Next component, check this off."},{"Start":"11:21.230 ","End":"11:24.080","Text":"Next component, x and y is 0,"},{"Start":"11:24.080 ","End":"11:26.570","Text":"z is 1, so that gives us 8."},{"Start":"11:26.570 ","End":"11:30.050","Text":"Good. Last component, x is 0,"},{"Start":"11:30.050 ","End":"11:31.550","Text":"y is 0, z is 1."},{"Start":"11:31.550 ","End":"11:32.975","Text":"That gives us the 9."},{"Start":"11:32.975 ","End":"11:34.730","Text":"Good. The other 2,"},{"Start":"11:34.730 ","End":"11:37.205","Text":"I\u0027ll leave you to check if you want."},{"Start":"11:37.205 ","End":"11:41.160","Text":"Then we are finally done."}],"ID":10242},{"Watched":false,"Name":"Exercise 20","Duration":"8m 39s","ChapterTopicVideoID":9693,"CourseChapterTopicPlaylistID":7314,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.815","Text":"In this exercise, we\u0027re looking for a linear transformation T from R^3 to R^3,"},{"Start":"00:07.815 ","End":"00:10.800","Text":"which satisfies 3 conditions."},{"Start":"00:10.800 ","End":"00:13.410","Text":"That it takes the vector 1,"},{"Start":"00:13.410 ","End":"00:15.780","Text":"0, 1 to the vector 1,"},{"Start":"00:15.780 ","End":"00:18.210","Text":"1, 0 and T of 0, 1, 1 is 1,"},{"Start":"00:18.210 ","End":"00:19.530","Text":"2 1 and 0,"},{"Start":"00:19.530 ","End":"00:21.225","Text":"0, 1 goes to 0, 1, 1."},{"Start":"00:21.225 ","End":"00:24.600","Text":"Well, it\u0027s a question. Is there such a linear transformation?"},{"Start":"00:24.600 ","End":"00:27.705","Text":"That\u0027s the first part, and if there isn\u0027t,"},{"Start":"00:27.705 ","End":"00:29.475","Text":"explain why not,"},{"Start":"00:29.475 ","End":"00:31.785","Text":"and if it does exist,"},{"Start":"00:31.785 ","End":"00:33.120","Text":"is it unique,"},{"Start":"00:33.120 ","End":"00:35.105","Text":"and if it happens to be unique,"},{"Start":"00:35.105 ","End":"00:37.985","Text":"then we have to find its formula."},{"Start":"00:37.985 ","End":"00:40.610","Text":"Let\u0027s get started."},{"Start":"00:40.610 ","End":"00:43.280","Text":"For this exercise we really need"},{"Start":"00:43.280 ","End":"00:48.065","Text":"a certain theorem which is going to be very useful for this kind of exercise,"},{"Start":"00:48.065 ","End":"00:52.585","Text":"and what the theorem says is that if we have V and U,"},{"Start":"00:52.585 ","End":"00:57.980","Text":"2 vector spaces and we are given a basis for V,"},{"Start":"00:57.980 ","End":"00:59.690","Text":"v_1 through v_n,"},{"Start":"00:59.690 ","End":"01:01.715","Text":"where N is the dimension of V,"},{"Start":"01:01.715 ","End":"01:06.625","Text":"and we\u0027re given any n-vectors in U,"},{"Start":"01:06.625 ","End":"01:09.750","Text":"that\u0027s the other space,"},{"Start":"01:09.750 ","End":"01:17.629","Text":"then theorem guarantees that there is a unique linear transformation from V to U,"},{"Start":"01:17.629 ","End":"01:21.875","Text":"which takes each V to its corresponding U."},{"Start":"01:21.875 ","End":"01:23.585","Text":"Takes v_1 to u_1,"},{"Start":"01:23.585 ","End":"01:26.700","Text":"takes v_2 to u_2 and so on."},{"Start":"01:26.710 ","End":"01:29.705","Text":"How is this good for us?"},{"Start":"01:29.705 ","End":"01:34.920","Text":"Well, it is because let\u0027s take V and U both to"},{"Start":"01:34.920 ","End":"01:41.460","Text":"be R^3 and then these 3 vectors are like v_1,"},{"Start":"01:41.460 ","End":"01:45.060","Text":"v_2, v_3, they\u0027re a basis."},{"Start":"01:45.060 ","End":"01:48.565","Text":"We have to show this as a basis."},{"Start":"01:48.565 ","End":"01:52.040","Text":"I\u0027ll just quickly give you the idea of how to do it."},{"Start":"01:52.040 ","End":"01:55.190","Text":"What you would do would be to write them in a matrix."},{"Start":"01:55.190 ","End":"01:58.190","Text":"The first vector is in the first row and 0,"},{"Start":"01:58.190 ","End":"02:01.460","Text":"1, 1, and then 0, 0, 1,"},{"Start":"02:01.460 ","End":"02:06.410","Text":"and then we do row operations until we get in row echelon form,"},{"Start":"02:06.410 ","End":"02:08.720","Text":"but here it is in row echelon form,"},{"Start":"02:08.720 ","End":"02:11.315","Text":"and we see that they\u0027re all non-zero,"},{"Start":"02:11.315 ","End":"02:17.300","Text":"and that would tell us that these 3 are linearly independent and so a basis."},{"Start":"02:17.300 ","End":"02:19.405","Text":"That\u0027s that part."},{"Start":"02:19.405 ","End":"02:22.610","Text":"That\u0027s v_1, v_2, v_3."},{"Start":"02:22.610 ","End":"02:24.875","Text":"If you go back and look,"},{"Start":"02:24.875 ","End":"02:30.200","Text":"you\u0027ll see that these 3 vectors where what we have to send these 3 to."},{"Start":"02:30.200 ","End":"02:34.805","Text":"There were these 3 and they are just the u_1, u_2, u_3."},{"Start":"02:34.805 ","End":"02:36.650","Text":"There\u0027s no condition on them."},{"Start":"02:36.650 ","End":"02:40.800","Text":"Any 3 vectors in U here N is 3."},{"Start":"02:41.450 ","End":"02:45.950","Text":"So we fulfill the conditions of the theorem,"},{"Start":"02:45.950 ","End":"02:48.830","Text":"and so there is a T which satisfies what"},{"Start":"02:48.830 ","End":"02:51.830","Text":"we had to show originally that T of this is this,"},{"Start":"02:51.830 ","End":"02:52.970","Text":"T of this is this,"},{"Start":"02:52.970 ","End":"02:55.505","Text":"and T of this is this."},{"Start":"02:55.505 ","End":"03:01.380","Text":"Now, so far we\u0027ve just guaranteed in theory that T exists."},{"Start":"03:02.830 ","End":"03:08.500","Text":"The continuation of this would be to actually find T. Now,"},{"Start":"03:08.500 ","End":"03:13.610","Text":"here\u0027s the practical part of actually finding T. We start off with the basis B and"},{"Start":"03:13.610 ","End":"03:18.935","Text":"we want to figure out the general coordinate vector relative to this basis."},{"Start":"03:18.935 ","End":"03:21.990","Text":"What that means is writing a general x,"},{"Start":"03:21.990 ","End":"03:25.415","Text":"y, z as a linear combination of these."},{"Start":"03:25.415 ","End":"03:30.365","Text":"The way we do this, as we build an augmented matrix where these 3,"},{"Start":"03:30.365 ","End":"03:34.550","Text":"1, 2, 3 are these 3 columns,"},{"Start":"03:34.550 ","End":"03:36.110","Text":"rows become columns,"},{"Start":"03:36.110 ","End":"03:39.355","Text":"and the last column we put x, y, z."},{"Start":"03:39.355 ","End":"03:42.950","Text":"Now, the plan is to keep doing row operations on"},{"Start":"03:42.950 ","End":"03:47.014","Text":"this augmented matrix until we get the unit,"},{"Start":"03:47.014 ","End":"03:53.580","Text":"the identity matrix in this 3 by 3 place."},{"Start":"03:54.110 ","End":"04:01.500","Text":"At the end, what\u0027s on the right column will be the coordinates. Well, we\u0027ll see."},{"Start":"04:01.720 ","End":"04:03.830","Text":"First thing we\u0027ll do,"},{"Start":"04:03.830 ","End":"04:06.500","Text":"we\u0027ll subtract the first row from the third row,"},{"Start":"04:06.500 ","End":"04:10.710","Text":"that will give us a 0 here and then we\u0027re up to this point,"},{"Start":"04:10.710 ","End":"04:12.590","Text":"and also of course, on the right column,"},{"Start":"04:12.590 ","End":"04:16.040","Text":"we subtracted the x from the z."},{"Start":"04:16.040 ","End":"04:23.405","Text":"Next step was to subtract the second row from the third row to get the 0 here,"},{"Start":"04:23.405 ","End":"04:25.025","Text":"and we get this,"},{"Start":"04:25.025 ","End":"04:31.440","Text":"and if we subtract this from this,"},{"Start":"04:31.440 ","End":"04:34.005","Text":"we get z minus x minus y,"},{"Start":"04:34.005 ","End":"04:40.300","Text":"and here we have the identity matrix,"},{"Start":"04:40.300 ","End":"04:42.235","Text":"so this part is done."},{"Start":"04:42.235 ","End":"04:45.210","Text":"Now, how do we put it all together?"},{"Start":"04:45.210 ","End":"04:47.810","Text":"What we do is we take these 3,"},{"Start":"04:47.810 ","End":"04:50.860","Text":"these 3 are coordinates relative to the basis."},{"Start":"04:50.860 ","End":"04:54.970","Text":"We take a linear combination of these times these,"},{"Start":"04:54.970 ","End":"04:57.385","Text":"I\u0027ll show you more specifically,"},{"Start":"04:57.385 ","End":"05:02.455","Text":"we take this x with the first vector,"},{"Start":"05:02.455 ","End":"05:05.260","Text":"and that goes here."},{"Start":"05:05.260 ","End":"05:10.570","Text":"Then we take the second coordinate with the second basis vector,"},{"Start":"05:10.570 ","End":"05:12.845","Text":"and that goes here."},{"Start":"05:12.845 ","End":"05:17.440","Text":"Then the third coordinate with the third vector,"},{"Start":"05:17.440 ","End":"05:19.405","Text":"and that goes here."},{"Start":"05:19.405 ","End":"05:21.610","Text":"This gives us x, y,"},{"Start":"05:21.610 ","End":"05:28.290","Text":"z as the linear combination of those 3 vectors,"},{"Start":"05:28.290 ","End":"05:29.670","Text":"1, 0, 1, 0, 1,"},{"Start":"05:29.670 ","End":"05:32.325","Text":"1 and 0, 0, 1, which were a basis."},{"Start":"05:32.325 ","End":"05:37.610","Text":"The second step will be to use this to find what we really want is T of x, y,"},{"Start":"05:37.610 ","End":"05:43.360","Text":"z, and we\u0027ll just apply T to both sides of this equality."},{"Start":"05:43.360 ","End":"05:48.180","Text":"T of the left-hand side equals T,"},{"Start":"05:48.180 ","End":"05:51.390","Text":"brackets, of the right-hand side."},{"Start":"05:51.390 ","End":"05:53.700","Text":"T is going to be linear,"},{"Start":"05:53.700 ","End":"05:54.870","Text":"so T of something,"},{"Start":"05:54.870 ","End":"05:56.570","Text":"plus something, plus something,"},{"Start":"05:56.570 ","End":"06:00.410","Text":"we can apply T to each P separately,"},{"Start":"06:00.410 ","End":"06:02.090","Text":"T of this, plus T of this,"},{"Start":"06:02.090 ","End":"06:04.180","Text":"plus T of this."},{"Start":"06:04.180 ","End":"06:06.500","Text":"Also because of linearity,"},{"Start":"06:06.500 ","End":"06:09.110","Text":"remember T of k_u is k_t of U."},{"Start":"06:09.110 ","End":"06:13.460","Text":"We can take the scalars in front of the t. This x goes in front,"},{"Start":"06:13.460 ","End":"06:15.230","Text":"this y goes in front,"},{"Start":"06:15.230 ","End":"06:16.790","Text":"z minus x minus y,"},{"Start":"06:16.790 ","End":"06:18.744","Text":"these go in front."},{"Start":"06:18.744 ","End":"06:23.130","Text":"Now, we know what these 3 are."},{"Start":"06:23.130 ","End":"06:24.260","Text":"This T of 1, 1,"},{"Start":"06:24.260 ","End":"06:27.020","Text":"0 that was given originally,"},{"Start":"06:27.020 ","End":"06:29.525","Text":"these 3 were specified,"},{"Start":"06:29.525 ","End":"06:31.340","Text":"I\u0027m not going to scroll all the way back,"},{"Start":"06:31.340 ","End":"06:32.720","Text":"but if you look at the beginning,"},{"Start":"06:32.720 ","End":"06:35.480","Text":"you\u0027ll see that what was required was T of 1, 1,"},{"Start":"06:35.480 ","End":"06:36.950","Text":"0 was 1, 1 0,"},{"Start":"06:36.950 ","End":"06:38.540","Text":"T of this was this,"},{"Start":"06:38.540 ","End":"06:41.270","Text":"and T of this was this,"},{"Start":"06:41.270 ","End":"06:49.490","Text":"and so we\u0027ve got an expression for T but we want it separated by coordinates."},{"Start":"06:49.490 ","End":"06:52.760","Text":"This is not really an answer,"},{"Start":"06:52.760 ","End":"06:58.480","Text":"but we want it as something comma something comma something."},{"Start":"06:58.480 ","End":"07:01.415","Text":"Let\u0027s do some simplifications calculations."},{"Start":"07:01.415 ","End":"07:04.070","Text":"X times this first vector is x,"},{"Start":"07:04.070 ","End":"07:06.835","Text":"x, 0, y times this is this,"},{"Start":"07:06.835 ","End":"07:09.690","Text":"z minus y, plus x, times this,"},{"Start":"07:09.690 ","End":"07:11.580","Text":"times 0, times 1,"},{"Start":"07:11.580 ","End":"07:13.485","Text":"and times 1, anyway, this is what we get."},{"Start":"07:13.485 ","End":"07:16.650","Text":"Now we\u0027ll add each component,"},{"Start":"07:16.650 ","End":"07:20.175","Text":"for example, x plus y,"},{"Start":"07:20.175 ","End":"07:23.520","Text":"plus 0 is x plus y,"},{"Start":"07:23.520 ","End":"07:27.030","Text":"and x, plus 2y,"},{"Start":"07:27.030 ","End":"07:30.570","Text":"plus z, minus x, minus y,"},{"Start":"07:30.570 ","End":"07:33.780","Text":"if you check it is just y plus z,"},{"Start":"07:33.780 ","End":"07:35.970","Text":"and finally, 0,"},{"Start":"07:35.970 ","End":"07:38.940","Text":"plus y, plus z, minus x,"},{"Start":"07:38.940 ","End":"07:43.170","Text":"minus y, gives us z, minus x."},{"Start":"07:43.170 ","End":"07:47.725","Text":"Now, this is actually the final answer,"},{"Start":"07:47.725 ","End":"07:51.715","Text":"but I would recommend a third step."},{"Start":"07:51.715 ","End":"07:54.490","Text":"Optionally, I would do a verification because of"},{"Start":"07:54.490 ","End":"07:56.980","Text":"so many computations might have just"},{"Start":"07:56.980 ","End":"08:00.010","Text":"missed something or might have arithmetical error somewhere,"},{"Start":"08:00.010 ","End":"08:03.310","Text":"so let\u0027s check that the original conditions really do apply."},{"Start":"08:03.310 ","End":"08:04.885","Text":"That T of this is this."},{"Start":"08:04.885 ","End":"08:06.990","Text":"That these 3 apply,"},{"Start":"08:06.990 ","End":"08:09.765","Text":"so I\u0027ll just check 1 of them,"},{"Start":"08:09.765 ","End":"08:11.370","Text":"see which is the easiest,"},{"Start":"08:11.370 ","End":"08:13.635","Text":"last 1 has couple of 0s in it."},{"Start":"08:13.635 ","End":"08:15.810","Text":"Let\u0027s see, T of 0, 0,"},{"Start":"08:15.810 ","End":"08:17.790","Text":"1, x is 0, y is 0,"},{"Start":"08:17.790 ","End":"08:21.990","Text":"z is 1, so x and y are both 0,"},{"Start":"08:21.990 ","End":"08:23.610","Text":"so that\u0027s 0,"},{"Start":"08:23.610 ","End":"08:26.115","Text":"y is 0 and z is 1,"},{"Start":"08:26.115 ","End":"08:27.765","Text":"that gives us the 1,"},{"Start":"08:27.765 ","End":"08:30.555","Text":"z is 1, x is 0,"},{"Start":"08:30.555 ","End":"08:33.210","Text":"that also gives us 1, so it\u0027s okay,"},{"Start":"08:33.210 ","End":"08:37.895","Text":"check the other 2 and see that they work out and that\u0027s it."},{"Start":"08:37.895 ","End":"08:39.930","Text":"We are done."}],"ID":10243}],"Thumbnail":null,"ID":7314},{"Name":"Image and Kernel","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Lesson 1 - Kernel","Duration":"5m 17s","ChapterTopicVideoID":9641,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.435","Text":"In this clip, we\u0027ll learn a new concept,"},{"Start":"00:03.435 ","End":"00:06.870","Text":"the Kernel of a linear transformation."},{"Start":"00:06.870 ","End":"00:09.585","Text":"We start with a transformation,"},{"Start":"00:09.585 ","End":"00:13.380","Text":"a linear transformation T from V-U."},{"Start":"00:13.380 ","End":"00:17.865","Text":"We define the kernel of T as follows."},{"Start":"00:17.865 ","End":"00:26.430","Text":"It\u0027s the set of all those vectors in V that T takes to 0, the 0 in U."},{"Start":"00:26.430 ","End":"00:30.690","Text":"It\u0027s written like this in formal language,"},{"Start":"00:30.690 ","End":"00:35.040","Text":"which basically means everything that goes to 0 by T."},{"Start":"00:35.040 ","End":"00:42.425","Text":"It turns out that the kernel of T is a subspace of V,"},{"Start":"00:42.425 ","End":"00:46.115","Text":"the source or domain space."},{"Start":"00:46.115 ","End":"00:47.660","Text":"It\u0027s very easy to prove,"},{"Start":"00:47.660 ","End":"00:49.660","Text":"but I won\u0027t bother doing it."},{"Start":"00:49.660 ","End":"00:56.610","Text":"Let\u0027s look at an example where V and U are both R^2."},{"Start":"00:56.610 ","End":"00:58.790","Text":"Here\u0027s the definition of T."},{"Start":"00:58.790 ","End":"01:01.250","Text":"Typical vector in R^2 is x, y,"},{"Start":"01:01.250 ","End":"01:07.050","Text":"and it takes it to the vector x plus y, 2x plus 2y."},{"Start":"01:07.160 ","End":"01:18.555","Text":"For example, T of 1, 2 is x plus y is 3 and 2x plus 2, y is 6, so it\u0027s 3, 6."},{"Start":"01:18.555 ","End":"01:22.155","Text":"Since 3, 6 is not the 0 vector,"},{"Start":"01:22.155 ","End":"01:26.179","Text":"then the vector 1, 2 is not in the kernel."},{"Start":"01:26.179 ","End":"01:31.200","Text":"When we say 0, it\u0027s the 0 vector, the 0 of U"},{"Start":"01:31.200 ","End":"01:38.390","Text":"and in our case the 0 is the vector 0, 0."},{"Start":"01:38.390 ","End":"01:43.065","Text":"Maybe obvious but I\u0027ll rewrite it anyway."},{"Start":"01:43.065 ","End":"01:44.730","Text":"On the other hand,"},{"Start":"01:44.730 ","End":"01:49.935","Text":"if we take the vector V to be 1, minus 1,"},{"Start":"01:49.935 ","End":"01:57.225","Text":"then T of this is x plus y is 0 and 2x plus 2y is 2 minus 2 is also 0."},{"Start":"01:57.225 ","End":"01:59.820","Text":"This is the 0 vector."},{"Start":"01:59.820 ","End":"02:03.441","Text":"The vector 1, minus 1 is in the kernel,"},{"Start":"02:03.441 ","End":"02:04.993","Text":"kernel of T."},{"Start":"02:04.993 ","End":"02:10.030","Text":"Okay, just a few more examples."},{"Start":"02:10.910 ","End":"02:16.750","Text":"If you computed T of 2 minus 2 is also 0, 0."},{"Start":"02:16.750 ","End":"02:22.375","Text":"X plus y plus 2y, and 3 minus 3 also goes to 0, 0."},{"Start":"02:22.375 ","End":"02:25.070","Text":"It looks like there\u0027s a pattern here."},{"Start":"02:25.070 ","End":"02:28.300","Text":"What is written here in mathematical language"},{"Start":"02:28.300 ","End":"02:35.920","Text":"just says that the kernel consists of all the pairs t, minus t,"},{"Start":"02:35.920 ","End":"02:38.365","Text":"where t is a real number."},{"Start":"02:38.365 ","End":"02:41.260","Text":"Now in 1 direction, it\u0027s obvious."},{"Start":"02:41.260 ","End":"02:43.900","Text":"If I start with t, minus t,"},{"Start":"02:43.900 ","End":"02:45.900","Text":"like 3 minus 3,"},{"Start":"02:45.900 ","End":"02:47.325","Text":"and I apply T to it."},{"Start":"02:47.325 ","End":"02:52.250","Text":"This plus this is 0 and 2x plus 2y"},{"Start":"02:52.250 ","End":"02:54.990","Text":"is actually twice the first components so if this is 0,"},{"Start":"02:54.990 ","End":"02:57.250","Text":"this is also going to be 0."},{"Start":"02:57.250 ","End":"02:59.390","Text":"Also the other way around."},{"Start":"02:59.390 ","End":"03:14.150","Text":"If x, y is in the kernel, then T of it,"},{"Start":"03:14.150 ","End":"03:18.260","Text":"which is this is 0, 0 and if x plus y is 0,"},{"Start":"03:18.260 ","End":"03:20.550","Text":"you could say that,"},{"Start":"03:22.550 ","End":"03:25.505","Text":"let\u0027s say that y is t,"},{"Start":"03:25.505 ","End":"03:28.610","Text":"then x being minus y is minus t."},{"Start":"03:28.610 ","End":"03:29.855","Text":"It works both ways."},{"Start":"03:29.855 ","End":"03:34.310","Text":"Basically the kernel is all these pairs t, minus t."},{"Start":"03:34.310 ","End":"03:42.890","Text":"Often we want to just find a basis for a subspace."},{"Start":"03:42.890 ","End":"03:48.020","Text":"I want a basis of the subspace of all the t, minus t."},{"Start":"03:48.020 ","End":"03:53.278","Text":"This is just all the multiples of 1, minus 1"},{"Start":"03:53.278 ","End":"03:56.780","Text":"and the multiple of a vector is a linear combination of it."},{"Start":"03:56.780 ","End":"04:03.480","Text":"The kernel is the span of the vector 1, minus 1."},{"Start":"04:03.650 ","End":"04:06.710","Text":"The dimension of the kernel,"},{"Start":"04:06.710 ","End":"04:13.770","Text":"which is important, is 1 because there\u0027s only 1 vector in it."},{"Start":"04:14.210 ","End":"04:18.965","Text":"There\u0027s a name for the dimension of the kernel."},{"Start":"04:18.965 ","End":"04:22.940","Text":"It\u0027s written N-U-L-L."},{"Start":"04:22.940 ","End":"04:28.070","Text":"Optionally, you could put a bracket just to show,"},{"Start":"04:28.070 ","End":"04:31.220","Text":"separated from the transformation T."},{"Start":"04:31.220 ","End":"04:34.640","Text":"And it\u0027s called the nullity of T."},{"Start":"04:34.640 ","End":"04:37.560","Text":"You write it N-U-L-L,"},{"Start":"04:37.560 ","End":"04:39.310","Text":"but it\u0027s called the nullity."},{"Start":"04:39.310 ","End":"04:41.760","Text":"It\u0027s the dimension of the kernel."},{"Start":"04:41.760 ","End":"04:43.460","Text":"That was our example."},{"Start":"04:43.460 ","End":"04:47.375","Text":"Let\u0027s summarize and write it in general."},{"Start":"04:47.375 ","End":"04:52.070","Text":"When we have a linear transformation from V-U,"},{"Start":"04:52.070 ","End":"05:00.095","Text":"then the kernel is the set of all v in V such that T of that v is 0."},{"Start":"05:00.095 ","End":"05:02.960","Text":"As we said, this is a subspace of V."},{"Start":"05:02.960 ","End":"05:06.740","Text":"The nullity of the transformation T"},{"Start":"05:06.740 ","End":"05:12.110","Text":"is simply the dimension of this subspace, the kernel."},{"Start":"05:13.150 ","End":"05:16.860","Text":"That\u0027s it for kernel."}],"ID":10197},{"Watched":false,"Name":"Lesson 2 - Image","Duration":"6m 29s","ChapterTopicVideoID":9642,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.787","Text":"In this clip, we\u0027ll learn a couple of new concepts"},{"Start":"00:05.787 ","End":"00:09.749","Text":"for linear transformation, image and rank."},{"Start":"00:09.749 ","End":"00:12.990","Text":"We start with a linear transformation."},{"Start":"00:12.990 ","End":"00:18.740","Text":"As usual, we like to call the source V and the destination U."},{"Start":"00:18.740 ","End":"00:22.620","Text":"I\u0027ll explain through an example."},{"Start":"00:22.620 ","End":"00:26.295","Text":"After the example, we\u0027ll do it formally."},{"Start":"00:26.295 ","End":"00:32.640","Text":"In our example, V and U are R^3, 3D space."},{"Start":"00:32.640 ","End":"00:42.400","Text":"We define T of x, y, z to be x plus y, y plus z, x plus z."},{"Start":"00:42.520 ","End":"00:46.940","Text":"We need a bit of preparation before we can get to the concept of image."},{"Start":"00:46.940 ","End":"00:52.290","Text":"The first thing we do is get a basis for V,"},{"Start":"00:52.290 ","End":"00:54.375","Text":"which in this case is our 3."},{"Start":"00:54.375 ","End":"00:57.515","Text":"Here we\u0027ll just take the standard basis,"},{"Start":"00:57.515 ","End":"01:05.070","Text":"which is this with just 1s and 1 and 2, 0s like this."},{"Start":"01:05.080 ","End":"01:12.095","Text":"Then we apply T to each of these basis vectors."},{"Start":"01:12.095 ","End":"01:15.770","Text":"For the first 1, it goes to 1, 0, 1,"},{"Start":"01:15.770 ","End":"01:20.780","Text":"and then the other 2 just do the computation like here,"},{"Start":"01:20.780 ","End":"01:26.240","Text":"x plus y is 1, and y plus z is also 1,"},{"Start":"01:26.240 ","End":"01:29.690","Text":"but x plus z is 0, and so on,"},{"Start":"01:29.690 ","End":"01:32.645","Text":"3 of the basis vectors."},{"Start":"01:32.645 ","End":"01:38.440","Text":"Now we have these 3 vectors."},{"Start":"01:40.790 ","End":"01:43.925","Text":"The image of T,"},{"Start":"01:43.925 ","End":"01:46.970","Text":"and it\u0027s written image,"},{"Start":"01:46.970 ","End":"01:52.250","Text":"but we write it as Im of T and sometimes we put brackets,"},{"Start":"01:52.250 ","End":"01:58.840","Text":"but mostly not, is the span of these 3 vectors."},{"Start":"01:58.840 ","End":"02:03.980","Text":"However, we usually do a bit of processing on this because"},{"Start":"02:03.980 ","End":"02:10.145","Text":"we would like the vectors that we take the span of to be linearly independent."},{"Start":"02:10.145 ","End":"02:12.815","Text":"I\u0027m not saying these are, not saying these aren\u0027t,"},{"Start":"02:12.815 ","End":"02:15.240","Text":"but let\u0027s check that."},{"Start":"02:15.290 ","End":"02:17.870","Text":"We take these 3 vectors"},{"Start":"02:17.870 ","End":"02:22.130","Text":"and put them into a matrix as rows as follows,"},{"Start":"02:22.130 ","End":"02:26.075","Text":"and we want to bring this to row echelon form."},{"Start":"02:26.075 ","End":"02:30.940","Text":"Subtract the first row from the second row,"},{"Start":"02:30.940 ","End":"02:32.955","Text":"and that will give us this."},{"Start":"02:32.955 ","End":"02:38.370","Text":"Now we want to subtract the second row from the third row."},{"Start":"02:38.370 ","End":"02:43.010","Text":"Now, this is in row echelon form and there are no 0s."},{"Start":"02:43.010 ","End":"02:45.350","Text":"These 3 are linearly independent,"},{"Start":"02:45.350 ","End":"02:48.080","Text":"and it means that the original 3 were,"},{"Start":"02:48.080 ","End":"02:49.360","Text":"but we didn\u0027t know that."},{"Start":"02:49.360 ","End":"02:55.160","Text":"It\u0027s possible in a different example that you might only get 2 vectors"},{"Start":"02:55.160 ","End":"02:58.760","Text":"and then a 0, or even only just 1."},{"Start":"02:58.760 ","End":"03:05.960","Text":"Usually, instead of defining the image to be the span of the original 3,"},{"Start":"03:05.960 ","End":"03:12.655","Text":"we take the span of the 3 we get after the row echelon process."},{"Start":"03:12.655 ","End":"03:19.080","Text":"We define the image to be the span of this vector."},{"Start":"03:19.790 ","End":"03:24.070","Text":"Then this 1 here, and this 1."},{"Start":"03:24.070 ","End":"03:27.620","Text":"Now, we could have kept these vectors"},{"Start":"03:27.620 ","End":"03:30.740","Text":"because it turned out that they were linearly independent."},{"Start":"03:30.740 ","End":"03:38.040","Text":"But in general, we take any non-zero vectors after the echelon form."},{"Start":"03:38.300 ","End":"03:43.460","Text":"It turns out that the image of a transformation is"},{"Start":"03:43.460 ","End":"03:47.750","Text":"always going to be a subspace of the target,"},{"Start":"03:47.750 ","End":"03:49.565","Text":"not of v but of u."},{"Start":"03:49.565 ","End":"03:51.260","Text":"Because here they both R^3,"},{"Start":"03:51.260 ","End":"03:57.035","Text":"but this will be a subspace in general of u and here subspace of R^3."},{"Start":"03:57.035 ","End":"03:58.670","Text":"In this particular case,"},{"Start":"03:58.670 ","End":"04:03.485","Text":"it\u0027s actually all of R^3 because we have 3 independent vectors."},{"Start":"04:03.485 ","End":"04:06.835","Text":"But in general, it might be a proper subspace."},{"Start":"04:06.835 ","End":"04:09.645","Text":"Now, the concept of rank,"},{"Start":"04:09.645 ","End":"04:11.989","Text":"we just write it as rank"},{"Start":"04:11.989 ","End":"04:15.470","Text":"and then the transformation you could put brackets in"},{"Start":"04:15.470 ","End":"04:17.640","Text":"if you wanted to,"},{"Start":"04:17.800 ","End":"04:20.210","Text":"usually don\u0027t need to."},{"Start":"04:20.210 ","End":"04:24.174","Text":"The rank is the dimension of the image."},{"Start":"04:24.174 ","End":"04:27.260","Text":"Now, the image is the span of these 3,"},{"Start":"04:27.260 ","End":"04:29.825","Text":"and these 3 are linearly independent,"},{"Start":"04:29.825 ","End":"04:33.445","Text":"and so the dimension of the image is 3."},{"Start":"04:33.445 ","End":"04:36.080","Text":"Like I said before, in this particular case,"},{"Start":"04:36.080 ","End":"04:44.995","Text":"because the dimension was the same as the dimension of the whole space U of R^3,"},{"Start":"04:44.995 ","End":"04:48.920","Text":"a subspace of R^3 with dimension 3 is the whole thing."},{"Start":"04:48.920 ","End":"04:52.175","Text":"That\u0027s just something that happened in this particular case."},{"Start":"04:52.175 ","End":"04:56.070","Text":"I said in general we\u0027ll get a proper subspace."},{"Start":"04:56.660 ","End":"05:02.195","Text":"Now I just want to generalize and summarize what we\u0027ve done so far,"},{"Start":"05:02.195 ","End":"05:05.015","Text":"more than our particular example in general,"},{"Start":"05:05.015 ","End":"05:10.775","Text":"we start with a transformation T from V-U, linear of course."},{"Start":"05:10.775 ","End":"05:17.930","Text":"We find a basis for V, V_1 through V_n."},{"Start":"05:17.930 ","End":"05:20.945","Text":"We apply T to each of the Vs,"},{"Start":"05:20.945 ","End":"05:24.465","Text":"Tv_1, Tv_2, up to T of V_n."},{"Start":"05:24.465 ","End":"05:27.275","Text":"We take the span of these vectors."},{"Start":"05:27.275 ","End":"05:29.240","Text":"These vectors are in U."},{"Start":"05:29.240 ","End":"05:32.965","Text":"The span of these vectors in U is called the image of T."},{"Start":"05:32.965 ","End":"05:37.470","Text":"We note that the image is a subspace of U."},{"Start":"05:37.470 ","End":"05:39.540","Text":"There is no much symbol for subspace,"},{"Start":"05:39.540 ","End":"05:41.675","Text":"so I wrote the symbol for subset,"},{"Start":"05:41.675 ","End":"05:44.600","Text":"which is subset of U, but it really is a subspace,"},{"Start":"05:44.600 ","End":"05:47.515","Text":"because the span is always a subspace."},{"Start":"05:47.515 ","End":"05:55.785","Text":"In practice, we replaced these Vs by other Vs when we did a row echelon."},{"Start":"05:55.785 ","End":"05:59.570","Text":"That whole process with row echelon form on the matrix"},{"Start":"05:59.570 ","End":"06:03.920","Text":"can sometimes condense this to less than n"},{"Start":"06:03.920 ","End":"06:06.720","Text":"and make them linearly independent."},{"Start":"06:06.720 ","End":"06:10.530","Text":"But in principle, this is correct."},{"Start":"06:10.530 ","End":"06:15.080","Text":"We also introduced the concept of rank for a transformation"},{"Start":"06:15.080 ","End":"06:18.980","Text":"is simply the dimension of the image."},{"Start":"06:18.980 ","End":"06:21.620","Text":"It\u0027s a subspace, so it has a dimension,"},{"Start":"06:21.620 ","End":"06:25.385","Text":"and that\u0027s called the rank."},{"Start":"06:25.385 ","End":"06:29.460","Text":"That\u0027s it for image and rank."}],"ID":10198},{"Watched":false,"Name":"Lesson 3 - Dimension Theorem","Duration":"2m 41s","ChapterTopicVideoID":9643,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.370","Text":"Now we come to an important theorem called the rank nullity theorem."},{"Start":"00:05.370 ","End":"00:09.554","Text":"Sometimes it\u0027s called the dimension theorem in this context."},{"Start":"00:09.554 ","End":"00:14.069","Text":"The previous 2 clips we learned about nullity,"},{"Start":"00:14.069 ","End":"00:16.440","Text":"which was the dimension of the kernel and rank,"},{"Start":"00:16.440 ","End":"00:18.555","Text":"which was the dimension of the image,"},{"Start":"00:18.555 ","End":"00:22.750","Text":"and they combine together to form a nice theorem."},{"Start":"00:22.750 ","End":"00:24.860","Text":"We start with the usual setup,"},{"Start":"00:24.860 ","End":"00:29.690","Text":"a linear transformation from a space V to a space U."},{"Start":"00:29.690 ","End":"00:33.050","Text":"There are a couple of ways of phrasing this theorem,"},{"Start":"00:33.050 ","End":"00:34.415","Text":"this is 1 of them."},{"Start":"00:34.415 ","End":"00:38.900","Text":"The theorem just says that the dimension of V is equal"},{"Start":"00:38.900 ","End":"00:43.310","Text":"to the dimension of the kernel of T plus the dimension of the image of"},{"Start":"00:43.310 ","End":"00:47.480","Text":"T. We can rewrite this using the words rank and"},{"Start":"00:47.480 ","End":"00:51.540","Text":"nullity and say that the dimension of V"},{"Start":"00:51.540 ","End":"00:55.850","Text":"is the nullity of T plus the rank of T. This is because,"},{"Start":"00:55.850 ","End":"00:59.480","Text":"like I said, the nullity is just the dimension of"},{"Start":"00:59.480 ","End":"01:04.440","Text":"the kernel and the rank is just the dimension of the image."},{"Start":"01:04.790 ","End":"01:09.570","Text":"Really, that\u0027s all there is to this theorem,"},{"Start":"01:09.570 ","End":"01:13.380","Text":"and this will be used"},{"Start":"01:13.380 ","End":"01:21.315","Text":"a few times in the exercises following this tutorial."},{"Start":"01:21.315 ","End":"01:24.035","Text":"I don\u0027t need to say anymore,"},{"Start":"01:24.035 ","End":"01:29.750","Text":"but I\u0027ll try and give you some picture which may or may not help."},{"Start":"01:29.750 ","End":"01:31.310","Text":"If you\u0027re not 1 for diagrams,"},{"Start":"01:31.310 ","End":"01:33.385","Text":"you can just quit now."},{"Start":"01:33.385 ","End":"01:37.775","Text":"Here\u0027s a picture diagram I found on the internet."},{"Start":"01:37.775 ","End":"01:39.935","Text":"This will serve our purposes,"},{"Start":"01:39.935 ","End":"01:45.690","Text":"except that I want this to be U not W. Some people like V to"},{"Start":"01:45.690 ","End":"01:55.460","Text":"W. T goes from V to U. T is the transformation that goes this way,"},{"Start":"01:55.460 ","End":"02:02.065","Text":"and kernel of T is a subspace of V and it goes to 0."},{"Start":"02:02.065 ","End":"02:05.475","Text":"It looks like everything shrinks a bit."},{"Start":"02:05.475 ","End":"02:07.980","Text":"Kernel T shrinks down to 0,"},{"Start":"02:07.980 ","End":"02:13.565","Text":"and V goes to the image of T which will be somewhat lower down."},{"Start":"02:13.565 ","End":"02:16.940","Text":"If you think of it like in geometrical terms as if it was"},{"Start":"02:16.940 ","End":"02:21.825","Text":"a rhombus or parallelogram and this height is equal to this height."},{"Start":"02:21.825 ","End":"02:23.865","Text":"This is the dimension of the image,"},{"Start":"02:23.865 ","End":"02:25.965","Text":"and this is the dimension of the image,"},{"Start":"02:25.965 ","End":"02:28.275","Text":"and this is the dimension of the kernel."},{"Start":"02:28.275 ","End":"02:34.070","Text":"This plus this equals the dimension of all V. Maybe this helps, maybe it doesn\u0027t."},{"Start":"02:34.070 ","End":"02:42.130","Text":"If it doesn\u0027t, then just forget what I said about the diagram. We\u0027re done."}],"ID":10199},{"Watched":false,"Name":"Exercise 1 Part a","Duration":"5m 2s","ChapterTopicVideoID":9650,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.450","Text":"This exercise uses the concept of kernel and the image of a transformation."},{"Start":"00:06.450 ","End":"00:10.245","Text":"We have T from R^4-R^3,"},{"Start":"00:10.245 ","End":"00:14.700","Text":"a linear transformation and it\u0027s defined as follows, T."},{"Start":"00:14.700 ","End":"00:17.280","Text":"Here we have 4 components."},{"Start":"00:17.280 ","End":"00:19.159","Text":"It goes to this expression."},{"Start":"00:19.159 ","End":"00:21.185","Text":"Notice there are 3 components,"},{"Start":"00:21.185 ","End":"00:23.270","Text":"there are 2 commas here."},{"Start":"00:23.270 ","End":"00:28.400","Text":"What we have to do is to find the basis and dimension of the kernel of T"},{"Start":"00:28.400 ","End":"00:35.035","Text":"and likewise for the image of T. We\u0027ll start with the kernel."},{"Start":"00:35.035 ","End":"00:39.620","Text":"The kernel of a transformation is all the vectors that go to 0."},{"Start":"00:39.620 ","End":"00:44.000","Text":"We simply take the expression for T and assign it to 0."},{"Start":"00:44.000 ","End":"00:48.130","Text":"Notice this is the 0 of R^3."},{"Start":"00:48.130 ","End":"00:50.970","Text":"It\u0027s in the target space."},{"Start":"00:50.970 ","End":"00:54.330","Text":"Makes sense because there\u0027s a 3 components here."},{"Start":"00:54.330 ","End":"00:59.010","Text":"Now, if these 2 vectors are equal and each of the 3 components is equal,"},{"Start":"00:59.010 ","End":"01:05.250","Text":"then that gives us a system of linear equations like x plus y is 0"},{"Start":"01:05.250 ","End":"01:10.565","Text":"and then y minus 4z plus t is 0 and so on."},{"Start":"01:10.565 ","End":"01:14.570","Text":"How do we solve the system of linear equations?"},{"Start":"01:14.570 ","End":"01:19.160","Text":"Most straightforward way is to use matrices."},{"Start":"01:19.160 ","End":"01:21.550","Text":"Because it\u0027s a homogeneous system"},{"Start":"01:21.550 ","End":"01:25.730","Text":"we don\u0027t need the augmented matrix with an extra column of 0s."},{"Start":"01:25.730 ","End":"01:30.229","Text":"We just take the coefficients on the left of the equals."},{"Start":"01:30.229 ","End":"01:36.380","Text":"Here we have 1,1 and then there is no y and there is no t."},{"Start":"01:36.380 ","End":"01:39.810","Text":"Here, there\u0027s no x. Y is 1,"},{"Start":"01:39.810 ","End":"01:41.505","Text":"z is minus 4, and so on."},{"Start":"01:41.505 ","End":"01:43.815","Text":"This is the matrix we get."},{"Start":"01:43.815 ","End":"01:48.125","Text":"Then we want to do row operations to get it to echelon form."},{"Start":"01:48.125 ","End":"01:54.480","Text":"We subtract this row or rather 4 times this row."},{"Start":"01:54.480 ","End":"01:58.230","Text":"From this row we\u0027ll get a 0 here."},{"Start":"01:58.230 ","End":"02:01.955","Text":"We do get a 0 here and this is what we end up with."},{"Start":"02:01.955 ","End":"02:10.145","Text":"The next step would be to add 3 times this row to this row to get a 0 here."},{"Start":"02:10.145 ","End":"02:12.155","Text":"This is the row notation."},{"Start":"02:12.155 ","End":"02:18.180","Text":"If we do that, we get the 0 here and this is what we get."},{"Start":"02:19.210 ","End":"02:25.040","Text":"Let us divide the last row by 2 or multiply by a 1/2."},{"Start":"02:25.040 ","End":"02:27.860","Text":"It gets smaller numbers to deal with."},{"Start":"02:27.860 ","End":"02:32.750","Text":"Now we want to go back from matrices to systems of linear equations."},{"Start":"02:32.750 ","End":"02:37.160","Text":"From this, we get this system like the first row,"},{"Start":"02:37.160 ","End":"02:41.300","Text":"x plus y and there\u0027s no z and no t equals 0"},{"Start":"02:41.300 ","End":"02:44.150","Text":"and so on for the other 2 rows."},{"Start":"02:44.150 ","End":"02:48.260","Text":"Perhaps I could have written it so the x\u0027s are all above each other,"},{"Start":"02:48.260 ","End":"02:49.370","Text":"like with the matrix."},{"Start":"02:49.370 ","End":"02:52.400","Text":"Maybe it\u0027s easier from the matrix to see the pivot elements"},{"Start":"02:52.400 ","End":"02:55.980","Text":"would be this 1, this 1, and this 1,"},{"Start":"02:55.980 ","End":"02:58.605","Text":"the first non-zero in each row."},{"Start":"02:58.605 ","End":"03:07.930","Text":"That would correspond here to the x here to this y and to this 4z."},{"Start":"03:08.750 ","End":"03:12.980","Text":"The leftover t would be the free variable."},{"Start":"03:12.980 ","End":"03:16.295","Text":"The other variables x, y, z are all dependent."},{"Start":"03:16.295 ","End":"03:19.295","Text":"They can be computed once we know t,"},{"Start":"03:19.295 ","End":"03:23.760","Text":"we just have boxed each of these x, y, z\u0027s."},{"Start":"03:24.080 ","End":"03:29.180","Text":"Perhaps you recall our technique of the wondering ones I called it."},{"Start":"03:29.180 ","End":"03:31.370","Text":"Any event when you have the free variables"},{"Start":"03:31.370 ","End":"03:34.445","Text":"when that 1 of them equal 1 and the other 0,"},{"Start":"03:34.445 ","End":"03:36.840","Text":"there is only 1 free variable."},{"Start":"03:36.840 ","End":"03:39.175","Text":"We try letting it equal 1."},{"Start":"03:39.175 ","End":"03:43.520","Text":"But I don\u0027t really like the value 1 because if I put it in here,"},{"Start":"03:43.520 ","End":"03:44.680","Text":"I\u0027ll get z equals a 1/4."},{"Start":"03:44.680 ","End":"03:46.730","Text":"We\u0027ll have to work with fractions."},{"Start":"03:46.730 ","End":"03:52.115","Text":"We actually can take it to be any non-zero number."},{"Start":"03:52.115 ","End":"03:54.835","Text":"Let\u0027s make it 4."},{"Start":"03:54.835 ","End":"03:59.240","Text":"If we let t equals 4, everything else follows backward substitution."},{"Start":"03:59.240 ","End":"04:01.820","Text":"You work away from the last equation to the top."},{"Start":"04:01.820 ","End":"04:04.910","Text":"T is 4, so z is 1."},{"Start":"04:04.910 ","End":"04:07.825","Text":"You put t is 4z is 1 here."},{"Start":"04:07.825 ","End":"04:11.900","Text":"Then we get y is 0."},{"Start":"04:11.900 ","End":"04:14.960","Text":"We get here minus 4 and plus 4."},{"Start":"04:14.960 ","End":"04:17.390","Text":"Yeah, y is 0. If y is 0,"},{"Start":"04:17.390 ","End":"04:19.775","Text":"then from this equation x is 0."},{"Start":"04:19.775 ","End":"04:21.980","Text":"We\u0027ve got x, y, z, and t,"},{"Start":"04:21.980 ","End":"04:24.430","Text":"not in the correct order."},{"Start":"04:24.430 ","End":"04:26.570","Text":"If I put them in the right order,"},{"Start":"04:26.570 ","End":"04:31.400","Text":"I get a vector which will be the basis for the solution space,"},{"Start":"04:31.400 ","End":"04:33.320","Text":"which is actually the kernel."},{"Start":"04:33.320 ","End":"04:39.535","Text":"A kernel is the solution space for the transformation equaling 0."},{"Start":"04:39.535 ","End":"04:41.235","Text":"This is what we have."},{"Start":"04:41.235 ","End":"04:45.930","Text":"That\u0027s the basis."},{"Start":"04:45.930 ","End":"04:47.820","Text":"What else were we asked for?"},{"Start":"04:47.820 ","End":"04:49.065","Text":"The dimension."},{"Start":"04:49.065 ","End":"04:52.620","Text":"It\u0027s only 1 vector in the basis so the dimension is 1."},{"Start":"04:52.620 ","End":"04:56.935","Text":"Here, we found the basis for the kernel and it\u0027s dimension."},{"Start":"04:56.935 ","End":"05:02.040","Text":"We\u0027ll continue in the next clip so you can take a break."}],"ID":10200},{"Watched":false,"Name":"Exercise 1 Part b","Duration":"4m 50s","ChapterTopicVideoID":9651,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:02.400","Text":"Back after the break,"},{"Start":"00:02.400 ","End":"00:03.570","Text":"we took care of the kernel."},{"Start":"00:03.570 ","End":"00:10.500","Text":"Now it\u0027s gone to the image and this is our transformation from R^4-R^3,"},{"Start":"00:10.500 ","End":"00:12.030","Text":"this was the formula."},{"Start":"00:12.030 ","End":"00:20.835","Text":"For the image, we first have to find a basis for the source for the R^4."},{"Start":"00:20.835 ","End":"00:23.640","Text":"But we have a basis is the standard basis,"},{"Start":"00:23.640 ","End":"00:28.660","Text":"it\u0027s 1,0,0,0 and then 0,1,0,0 and so on."},{"Start":"00:28.820 ","End":"00:36.545","Text":"We apply the transformation T to each of the 4 basis vectors, these 4."},{"Start":"00:36.545 ","End":"00:39.095","Text":"Do the computations here,"},{"Start":"00:39.095 ","End":"00:44.170","Text":"we get these 4 vectors in R^3."},{"Start":"00:44.170 ","End":"00:50.120","Text":"Now, these 4 vectors certainly span the image,"},{"Start":"00:50.120 ","End":"00:54.160","Text":"but they are not linearly independent."},{"Start":"00:54.160 ","End":"00:58.550","Text":"Yeah, you can\u0027t have 4 linearly independent vectors in R^3."},{"Start":"00:58.550 ","End":"01:05.030","Text":"Remember what we do is we take the transform of"},{"Start":"01:05.030 ","End":"01:06.470","Text":"each of the basis members"},{"Start":"01:06.470 ","End":"01:08.030","Text":"and then we put them in a matrix"},{"Start":"01:08.030 ","End":"01:10.405","Text":"and bring it to echelon form,"},{"Start":"01:10.405 ","End":"01:16.160","Text":"and here they are, 1 ,0,4, 1, 1, 1 as the roads of this."},{"Start":"01:16.160 ","End":"01:18.980","Text":"I want to bring this to echelon form."},{"Start":"01:18.980 ","End":"01:21.410","Text":"Now we\u0027re going to get some rows of zeros"},{"Start":"01:21.410 ","End":"01:24.530","Text":"so we can reduce it from 4 vectors to less"},{"Start":"01:24.530 ","End":"01:27.155","Text":"and then will be linearly independent."},{"Start":"01:27.155 ","End":"01:29.180","Text":"Let\u0027s just get to doing it."},{"Start":"01:29.180 ","End":"01:35.540","Text":"First thing to do would be to subtract the first row from the second row"},{"Start":"01:35.540 ","End":"01:42.060","Text":"and if we do that, we get this."},{"Start":"01:42.060 ","End":"01:50.370","Text":"All 0s here, but now we want also to get 0s here and here."},{"Start":"01:50.370 ","End":"01:51.915","Text":"This 1 is good for us."},{"Start":"01:51.915 ","End":"01:58.670","Text":"Add 4 times this row to this row and subtract this row from this row,"},{"Start":"01:58.670 ","End":"02:00.380","Text":"that\u0027s what I\u0027ve written here."},{"Start":"02:00.380 ","End":"02:04.580","Text":"Now what I\u0027d like to do is just divide this row by minus 8"},{"Start":"02:04.580 ","End":"02:07.270","Text":"and this row by 2 and we\u0027ll have 1 and 1,"},{"Start":"02:07.270 ","End":"02:10.750","Text":"and that will be easier to deal with."},{"Start":"02:11.390 ","End":"02:15.225","Text":"That gives us this the 1 and the 1 here,"},{"Start":"02:15.225 ","End":"02:19.675","Text":"so all we have to do now is subtract the third row from the fourth row,"},{"Start":"02:19.675 ","End":"02:21.510","Text":"and this is what we get."},{"Start":"02:21.510 ","End":"02:23.790","Text":"Notice that we have a row of 0s,"},{"Start":"02:23.790 ","End":"02:25.875","Text":"so we can throw that out."},{"Start":"02:25.875 ","End":"02:32.245","Text":"Now, we can settle for just 3 vectors to span the image."},{"Start":"02:32.245 ","End":"02:36.995","Text":"The basis for the image will be this vector here,"},{"Start":"02:36.995 ","End":"02:41.295","Text":"this vector here, and this vector here."},{"Start":"02:41.295 ","End":"02:46.170","Text":"If we count, there are 3 vectors in this basis"},{"Start":"02:46.170 ","End":"02:51.960","Text":"and so the dimension of the image is 3."},{"Start":"02:53.230 ","End":"02:58.505","Text":"I\u0027d like to show you another way of solving this particular example."},{"Start":"02:58.505 ","End":"03:06.755","Text":"We can find the image of T using the rank nullity theorem, the dimension theorem."},{"Start":"03:06.755 ","End":"03:10.280","Text":"Remember the dimension of the space on the left,"},{"Start":"03:10.280 ","End":"03:15.350","Text":"the source is equal to the dimension of the kernel plus the dimension of the image,"},{"Start":"03:15.350 ","End":"03:18.980","Text":"I the nullity plus the rank."},{"Start":"03:18.980 ","End":"03:22.580","Text":"Now the dimension of our 4 is 4"},{"Start":"03:22.580 ","End":"03:27.620","Text":"and we already computed the dimension of kernel earlier in the exercise"},{"Start":"03:27.620 ","End":"03:29.345","Text":"and we got it to be 1."},{"Start":"03:29.345 ","End":"03:33.245","Text":"That means that the image has dimension 3."},{"Start":"03:33.245 ","End":"03:37.475","Text":"Now the image is also a subspace of R^3."},{"Start":"03:37.475 ","End":"03:43.410","Text":"I use the symbol for subset to indicate subspace in linear algebra."},{"Start":"03:44.170 ","End":"03:50.170","Text":"From this, I can conclude that image of T is all of R^3,"},{"Start":"03:50.170 ","End":"03:52.740","Text":"because the dimension of this is 3"},{"Start":"03:52.740 ","End":"03:57.215","Text":"and if a subspace has the same dimension as the whole space,"},{"Start":"03:57.215 ","End":"03:59.150","Text":"then it has to be the whole space."},{"Start":"03:59.150 ","End":"04:04.310","Text":"In this case, we computed the image to be all of R^3."},{"Start":"04:04.310 ","End":"04:10.160","Text":"Actually, this other method is better than the previous method in this case,"},{"Start":"04:10.160 ","End":"04:12.700","Text":"because in the previous method,"},{"Start":"04:12.700 ","End":"04:16.920","Text":"we just got the basis for the image,"},{"Start":"04:16.920 ","End":"04:20.560","Text":"and we would have said that the image is the span of this,"},{"Start":"04:20.560 ","End":"04:23.390","Text":"but then it would take a bit more explaining to say"},{"Start":"04:23.390 ","End":"04:26.870","Text":"why the span of this is all of R^3 because"},{"Start":"04:26.870 ","End":"04:31.280","Text":"we have 3 linearly independent vectors in R^3"},{"Start":"04:31.280 ","End":"04:34.540","Text":"and so it always spans all of R^3."},{"Start":"04:34.540 ","End":"04:39.645","Text":"Anyway, you\u0027ve seen both methods here,"},{"Start":"04:39.645 ","End":"04:44.115","Text":"so that will sometimes work, the other method."},{"Start":"04:44.115 ","End":"04:46.845","Text":"This method will always work."},{"Start":"04:46.845 ","End":"04:50.590","Text":"I\u0027ve said enough, we are done."}],"ID":10201},{"Watched":false,"Name":"Exercise 2 Part a","Duration":"3m 53s","ChapterTopicVideoID":9652,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.300","Text":"In this exercise, we have a linear transformation T from R^3-R^4,"},{"Start":"00:06.300 ","End":"00:09.465","Text":"and it\u0027s defined using this formula."},{"Start":"00:09.465 ","End":"00:12.750","Text":"Notice that we have 3 components, x, y, z,"},{"Start":"00:12.750 ","End":"00:15.390","Text":"and over here we have 4 of them."},{"Start":"00:15.390 ","End":"00:17.520","Text":"You can tell by looking at the commas,"},{"Start":"00:17.520 ","End":"00:19.575","Text":"there\u0027s 4 components here."},{"Start":"00:19.575 ","End":"00:23.850","Text":"Our task is first of all to find a basis"},{"Start":"00:23.850 ","End":"00:26.910","Text":"and the dimension of the kernel of T"},{"Start":"00:26.910 ","End":"00:30.330","Text":"and then same thing but with the image of T."},{"Start":"00:30.330 ","End":"00:34.890","Text":"We\u0027ll start first with the kernel"},{"Start":"00:34.890 ","End":"00:42.230","Text":"and the kernel is obtained by letting T equals 0."},{"Start":"00:42.230 ","End":"00:45.650","Text":"See what x, y, z goes to 0."},{"Start":"00:45.650 ","End":"00:49.190","Text":"In other words, we want this vector to be the 0 vector,"},{"Start":"00:49.190 ","End":"00:54.570","Text":"but it\u0027s the 0 from R^4, meaning 0, 0, 0, 0."},{"Start":"00:54.770 ","End":"00:57.915","Text":"This is really 4 equality."},{"Start":"00:57.915 ","End":"00:59.880","Text":"Each component we can compare,"},{"Start":"00:59.880 ","End":"01:04.520","Text":"and then we get a system of linear equations,"},{"Start":"01:04.520 ","End":"01:12.285","Text":"4 equations, but only 3 unknowns, x, y, z."},{"Start":"01:12.285 ","End":"01:16.510","Text":"We\u0027ll solve this system with matrices."},{"Start":"01:16.510 ","End":"01:22.250","Text":"Notice that we don\u0027t need the augmented matrix"},{"Start":"01:22.250 ","End":"01:25.490","Text":"because it\u0027s a homogeneous system and all 0 here,"},{"Start":"01:25.490 ","End":"01:29.090","Text":"so we just need the x, y, and z coefficients."},{"Start":"01:29.090 ","End":"01:32.000","Text":"From here we get 1, 4 minus 1,"},{"Start":"01:32.000 ","End":"01:34.190","Text":"from here, 1, 1, and the 0"},{"Start":"01:34.190 ","End":"01:36.665","Text":"because there\u0027s no z and so on."},{"Start":"01:36.665 ","End":"01:41.335","Text":"We start doing row operations to bring it to echelon form."},{"Start":"01:41.335 ","End":"01:46.550","Text":"First I subtract the top row from the second row"},{"Start":"01:46.550 ","End":"01:53.665","Text":"and also subtract it from the bottom row and then we get this."},{"Start":"01:53.665 ","End":"01:56.340","Text":"Then we want 0s here and here,"},{"Start":"01:56.340 ","End":"02:00.125","Text":"so I\u0027ll take 5 times this row and subtract this row."},{"Start":"02:00.125 ","End":"02:04.850","Text":"Here, we\u0027ll take 5 times this row minus 4 times this row,"},{"Start":"02:04.850 ","End":"02:09.150","Text":"we get 20 minus 20, that will be 0 here."},{"Start":"02:09.470 ","End":"02:13.340","Text":"This is what we get and the 0s here."},{"Start":"02:13.340 ","End":"02:15.640","Text":"Now I\u0027ll divide the bottom 2 rows,"},{"Start":"02:15.640 ","End":"02:19.280","Text":"this 1 by minus 6 and this 1 by 21."},{"Start":"02:19.280 ","End":"02:22.790","Text":"That will bring us to here and now obviously"},{"Start":"02:22.790 ","End":"02:26.720","Text":"we\u0027re going to subtract the third row from the fourth row."},{"Start":"02:26.720 ","End":"02:29.210","Text":"That would give us a 0 in the last row,"},{"Start":"02:29.210 ","End":"02:32.210","Text":"and I\u0027ll just cross it out as if we\u0027ve done it."},{"Start":"02:32.210 ","End":"02:35.090","Text":"From here, we go back to a system of linear equations."},{"Start":"02:35.090 ","End":"02:36.440","Text":"There are no free variables."},{"Start":"02:36.440 ","End":"02:40.445","Text":"Everything is determined by back substitution, z is 0,"},{"Start":"02:40.445 ","End":"02:44.795","Text":"plug it in here we get y is 0 plug z and y is 0 here."},{"Start":"02:44.795 ","End":"02:47.164","Text":"We get that x equals 0."},{"Start":"02:47.164 ","End":"02:53.485","Text":"This means that the kernel just contains the 0 vector."},{"Start":"02:53.485 ","End":"02:57.815","Text":"When this happens, when the kernel is the trivial subspace,"},{"Start":"02:57.815 ","End":"03:01.130","Text":"then the basis is considered to be the empty set."},{"Start":"03:01.130 ","End":"03:04.340","Text":"The number of elements in the empty set is 0,"},{"Start":"03:04.340 ","End":"03:08.605","Text":"so the dimension of the kernel is 0."},{"Start":"03:08.605 ","End":"03:10.580","Text":"That\u0027s it for the kernel."},{"Start":"03:10.580 ","End":"03:12.545","Text":"Now on to the image."},{"Start":"03:12.545 ","End":"03:15.050","Text":"Let\u0027s first compute the dimension of the image because"},{"Start":"03:15.050 ","End":"03:18.455","Text":"that\u0027s easy using the nullity-rank theorem."},{"Start":"03:18.455 ","End":"03:23.380","Text":"Remember that our transformation is from R^3-R^4."},{"Start":"03:23.380 ","End":"03:28.370","Text":"The nullity-rank theorem says we need the dimension of this space."},{"Start":"03:28.370 ","End":"03:31.490","Text":"It\u0027s going to equal the dimension of the kernel plus"},{"Start":"03:31.490 ","End":"03:35.530","Text":"the dimension of the image or the nullity plus the rank."},{"Start":"03:35.530 ","End":"03:41.835","Text":"Now, this is 3 and this we just said was 0,"},{"Start":"03:41.835 ","End":"03:44.710","Text":"so this also has to be 3."},{"Start":"03:44.710 ","End":"03:47.720","Text":"The dimension of the image we know already is 3."},{"Start":"03:47.720 ","End":"03:49.490","Text":"Now we actually have to find the image"},{"Start":"03:49.490 ","End":"03:53.339","Text":"and let\u0027s do that after we take a break."}],"ID":10202},{"Watched":false,"Name":"Exercise 2 Part b","Duration":"2m 14s","ChapterTopicVideoID":9653,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"Continuing after the break,"},{"Start":"00:02.070 ","End":"00:05.370","Text":"we just finished the kernel of T basis and dimension."},{"Start":"00:05.370 ","End":"00:07.469","Text":"Now we\u0027re doing the image."},{"Start":"00:07.469 ","End":"00:12.375","Text":"We already computed the dimension of the image to be 3 before the break."},{"Start":"00:12.375 ","End":"00:15.660","Text":"We\u0027ll check that we do really get 3."},{"Start":"00:15.660 ","End":"00:17.880","Text":"This is the formula for T."},{"Start":"00:17.880 ","End":"00:19.605","Text":"T is from R^3 to R^4."},{"Start":"00:19.605 ","End":"00:22.725","Text":"What we have to do first is find the basis for R^3."},{"Start":"00:22.725 ","End":"00:25.395","Text":"We\u0027ll take the standard basis."},{"Start":"00:25.395 ","End":"00:29.970","Text":"The standard basis is 1, 0, 0, 0, 1, 0, 0, 0, 1."},{"Start":"00:29.970 ","End":"00:33.349","Text":"We apply T to each of the members of the basis."},{"Start":"00:33.349 ","End":"00:36.440","Text":"We\u0027ve now got vectors in R^4."},{"Start":"00:36.440 ","End":"00:39.770","Text":"Now, these span the image,"},{"Start":"00:39.770 ","End":"00:44.990","Text":"but we want to have them linearly independent to be a basis."},{"Start":"00:44.990 ","End":"00:49.880","Text":"What we do is we put them into a matrix and start a row echelon process."},{"Start":"00:49.880 ","End":"00:54.270","Text":"Maybe we\u0027ll get some rows of 0s and throw them out."},{"Start":"00:54.320 ","End":"00:58.610","Text":"Here\u0027s the matrix where I take these as rows,"},{"Start":"00:58.610 ","End":"01:00.530","Text":"just copy the numbers in."},{"Start":"01:00.530 ","End":"01:04.320","Text":"The first set of row operations we\u0027ll do is"},{"Start":"01:04.320 ","End":"01:08.310","Text":"to add 4 times this row to this row"},{"Start":"01:08.310 ","End":"01:12.715","Text":"and 1 time this row to this row as specified here."},{"Start":"01:12.715 ","End":"01:14.715","Text":"That gives us this."},{"Start":"01:14.715 ","End":"01:16.380","Text":"We have a 0 here and here."},{"Start":"01:16.380 ","End":"01:19.500","Text":"We\u0027d like to have a 0 also here."},{"Start":"01:19.500 ","End":"01:23.745","Text":"What we can do, and it\u0027s written here,"},{"Start":"01:23.745 ","End":"01:31.250","Text":"we take 5 times the third row and subtract the second row,"},{"Start":"01:31.250 ","End":"01:33.955","Text":"that will give us a 0 here."},{"Start":"01:33.955 ","End":"01:37.220","Text":"Now we are in row echelon form already."},{"Start":"01:37.220 ","End":"01:40.865","Text":"We didn\u0027t encounter, we can get any rows of 0s."},{"Start":"01:40.865 ","End":"01:47.215","Text":"These 3 rows would be a basis for the image."},{"Start":"01:47.215 ","End":"01:51.275","Text":"Basis for the image is this vector here,"},{"Start":"01:51.275 ","End":"01:55.165","Text":"this vector here, this vector, that\u0027s this."},{"Start":"01:55.165 ","End":"01:59.450","Text":"The dimension of the image just have to count 1, 2, 3."},{"Start":"01:59.450 ","End":"02:00.470","Text":"So it\u0027s 3."},{"Start":"02:00.470 ","End":"02:02.780","Text":"It\u0027s also called the rank of T."},{"Start":"02:02.780 ","End":"02:05.270","Text":"That agrees with what we got earlier."},{"Start":"02:05.270 ","End":"02:07.130","Text":"We earlier found that it was 3,"},{"Start":"02:07.130 ","End":"02:08.650","Text":"so all is well."},{"Start":"02:08.650 ","End":"02:11.179","Text":"Here is the basis and here\u0027s the dimension."},{"Start":"02:11.179 ","End":"02:13.910","Text":"We are done."}],"ID":10203},{"Watched":false,"Name":"Exercise 3 Part a","Duration":"3m 44s","ChapterTopicVideoID":9654,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.995","Text":"Here we have a linear transformation from R^4-R^3,"},{"Start":"00:04.995 ","End":"00:08.295","Text":"and it\u0027s defined in matrix form."},{"Start":"00:08.295 ","End":"00:14.805","Text":"T takes a 4-dimensional vector x, y, z, t and sends it to this."},{"Start":"00:14.805 ","End":"00:17.190","Text":"This will be a 3-dimensional vector"},{"Start":"00:17.190 ","End":"00:20.040","Text":"because this has 3 rows and 4 columns,"},{"Start":"00:20.040 ","End":"00:21.765","Text":"that has 4 rows and 1 column."},{"Start":"00:21.765 ","End":"00:24.435","Text":"So we\u0027ll end up getting 3 rows and 1 column."},{"Start":"00:24.435 ","End":"00:29.340","Text":"Anyway, what we have to do is to find the basis"},{"Start":"00:29.340 ","End":"00:31.935","Text":"and the dimension of the kernel of T,"},{"Start":"00:31.935 ","End":"00:34.140","Text":"and likewise for the image of T."},{"Start":"00:34.140 ","End":"00:36.830","Text":"We\u0027ll start with the kernel."},{"Start":"00:36.830 ","End":"00:45.920","Text":"The kernel of T is all the vectors in R^4 that T takes to 0 in R^3."},{"Start":"00:45.920 ","End":"00:53.675","Text":"Essentially, what we have is that this matrix product has to be the 0 vector,"},{"Start":"00:53.675 ","End":"00:57.125","Text":"and in R^3 it\u0027s 0, 0, 0."},{"Start":"00:57.125 ","End":"01:00.950","Text":"This is actually a system of linear equations,"},{"Start":"01:00.950 ","End":"01:06.570","Text":"homogeneous, 3 equations, 4 unknowns."},{"Start":"01:06.570 ","End":"01:11.945","Text":"We want to work on that matrix to bring it in row echelon form."},{"Start":"01:11.945 ","End":"01:15.165","Text":"We\u0027ll start by getting 0s here."},{"Start":"01:15.165 ","End":"01:17.430","Text":"We\u0027ll subtract the first row from the second row,"},{"Start":"01:17.430 ","End":"01:22.095","Text":"and twice the first row from the third row as specified here."},{"Start":"01:22.095 ","End":"01:24.475","Text":"Next, we want a 0 here."},{"Start":"01:24.475 ","End":"01:28.315","Text":"So we subtract twice the second row from the third row."},{"Start":"01:28.315 ","End":"01:29.690","Text":"Look after we did that,"},{"Start":"01:29.690 ","End":"01:31.190","Text":"we have a row of 0s."},{"Start":"01:31.190 ","End":"01:32.900","Text":"Just cross that out."},{"Start":"01:32.900 ","End":"01:37.885","Text":"At this point, you want to go back from matrices to the SLA."},{"Start":"01:37.885 ","End":"01:41.880","Text":"The pivot terms, the x and the y,"},{"Start":"01:41.880 ","End":"01:47.175","Text":"and the remaining 1, z and t are the free variables."},{"Start":"01:47.175 ","End":"01:51.135","Text":"Remember we had this technique called the wondering 1s,"},{"Start":"01:51.135 ","End":"01:57.374","Text":"where each time we let 1 of the free variables be 1 and the remaining 0."},{"Start":"01:57.374 ","End":"02:00.780","Text":"First time, let\u0027s take t is 1 and z is 0,"},{"Start":"02:00.780 ","End":"02:02.675","Text":"and then we\u0027ll do the other way around."},{"Start":"02:02.675 ","End":"02:04.775","Text":"Once we have t and z,"},{"Start":"02:04.775 ","End":"02:07.295","Text":"we can compute y from here."},{"Start":"02:07.295 ","End":"02:09.960","Text":"We plug in t equals 1 and z equals 0,"},{"Start":"02:09.960 ","End":"02:11.900","Text":"we get that y equals 3."},{"Start":"02:11.900 ","End":"02:15.035","Text":"Then we can plug in t, z, and y into here,"},{"Start":"02:15.035 ","End":"02:18.265","Text":"and check you get x equals minus 7."},{"Start":"02:18.265 ","End":"02:22.610","Text":"The next time we take t is 0 and z is 1."},{"Start":"02:22.610 ","End":"02:23.540","Text":"The other way around."},{"Start":"02:23.540 ","End":"02:26.360","Text":"Again, we plug in t and z into here,"},{"Start":"02:26.360 ","End":"02:30.725","Text":"and we get y, and we plug t, z, and y in here, and we get x."},{"Start":"02:30.725 ","End":"02:31.940","Text":"These are the results."},{"Start":"02:31.940 ","End":"02:34.405","Text":"We have to put them in the right order."},{"Start":"02:34.405 ","End":"02:40.185","Text":"This 1, if I take x, y, z, t is minus 7, 3, 0, 1, that\u0027s this."},{"Start":"02:40.185 ","End":"02:43.575","Text":"Here I have 1 minus 2, 1, 0, it\u0027s this."},{"Start":"02:43.575 ","End":"02:48.089","Text":"These 2 form a basis for the kernel."},{"Start":"02:48.089 ","End":"02:49.940","Text":"As for the dimension,"},{"Start":"02:49.940 ","End":"02:52.460","Text":"you just have to count 1, 2."},{"Start":"02:52.460 ","End":"02:54.980","Text":"The dimension of the kernel is 2."},{"Start":"02:54.980 ","End":"02:58.370","Text":"That\u0027s also called the nullity of T."},{"Start":"02:58.370 ","End":"03:03.980","Text":"After the break, we\u0027ll do the same thing for the image,"},{"Start":"03:03.980 ","End":"03:06.319","Text":"we\u0027ll find the basis and the dimension."},{"Start":"03:06.319 ","End":"03:07.700","Text":"But before the break,"},{"Start":"03:07.700 ","End":"03:11.090","Text":"it\u0027s very easy to find just the dimension of the image"},{"Start":"03:11.090 ","End":"03:14.015","Text":"using the rank-nullity theorem."},{"Start":"03:14.015 ","End":"03:15.890","Text":"By the rank-nullity theorem,"},{"Start":"03:15.890 ","End":"03:18.410","Text":"if I have T from this space to this space,"},{"Start":"03:18.410 ","End":"03:21.380","Text":"the dimension of this space, the source space,"},{"Start":"03:21.380 ","End":"03:24.950","Text":"is the dimension of the image plus the dimension of the kernel."},{"Start":"03:24.950 ","End":"03:28.130","Text":"Now, the dimension of R^4 is 4,"},{"Start":"03:28.130 ","End":"03:32.210","Text":"and the dimension of the kernel we have it here is 2."},{"Start":"03:32.210 ","End":"03:35.615","Text":"The dimension of the image must also be 2."},{"Start":"03:35.615 ","End":"03:38.345","Text":"Let\u0027s remember this and after the break,"},{"Start":"03:38.345 ","End":"03:42.275","Text":"we\u0027ll see that we really do get to for the dimension of the image."},{"Start":"03:42.275 ","End":"03:44.850","Text":"So break."}],"ID":10204},{"Watched":false,"Name":"Exercise 3 Part b","Duration":"2m 21s","ChapterTopicVideoID":9655,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.519","Text":"Continuing after the break,"},{"Start":"00:02.519 ","End":"00:05.010","Text":"we just did the basis and dimension of the kernel,"},{"Start":"00:05.010 ","End":"00:07.900","Text":"now we want to do it for the image."},{"Start":"00:08.660 ","End":"00:12.640","Text":"Before the break, we already found that the dimension is 2,"},{"Start":"00:12.640 ","End":"00:15.940","Text":"and let\u0027s see if indeed we do get 2 for that."},{"Start":"00:16.430 ","End":"00:21.345","Text":"The transformation was given in matrix form like this."},{"Start":"00:21.345 ","End":"00:24.960","Text":"We\u0027ll apply T to a basis."},{"Start":"00:24.960 ","End":"00:29.280","Text":"The easiest thing is to take the standard basis of R^4,"},{"Start":"00:29.280 ","End":"00:32.805","Text":"the first basis member is 1, 0, 0, 0."},{"Start":"00:32.805 ","End":"00:36.180","Text":"If we put it here and do a matrix multiplication,"},{"Start":"00:36.180 ","End":"00:37.710","Text":"we get 1, 1, 2."},{"Start":"00:37.710 ","End":"00:39.795","Text":"Now, we have 3 more to go."},{"Start":"00:39.795 ","End":"00:42.015","Text":"Let me just get some space here."},{"Start":"00:42.015 ","End":"00:45.390","Text":"Here they are, quite fitting, there we are."},{"Start":"00:45.390 ","End":"00:46.100","Text":"We had that 1"},{"Start":"00:46.100 ","End":"00:51.950","Text":"and then there\u0027s these 3 for the other 3 basis members of R^4."},{"Start":"00:51.950 ","End":"00:56.570","Text":"Now we have to keep these 4 vectors."},{"Start":"00:56.570 ","End":"00:58.700","Text":"Well, this 1s scrolled off."},{"Start":"00:58.700 ","End":"01:00.140","Text":"Never mind, we know what it is."},{"Start":"01:00.140 ","End":"01:03.090","Text":"It\u0027s 1, 1, 2."},{"Start":"01:03.580 ","End":"01:08.285","Text":"Then we\u0027ll take these 4 vectors, 1, 1, 2, 2, 3, 6,"},{"Start":"01:08.285 ","End":"01:13.710","Text":"and so on, and put them as rows in a matrix."},{"Start":"01:13.710 ","End":"01:15.550","Text":"Now, these 4 rows,"},{"Start":"01:15.550 ","End":"01:18.335","Text":"they do span the image,"},{"Start":"01:18.335 ","End":"01:20.650","Text":"but they might be dependent linearly,"},{"Start":"01:20.650 ","End":"01:23.330","Text":"so we want to do a row echelon process"},{"Start":"01:23.330 ","End":"01:27.450","Text":"and maybe get some rows with 0s and throw them out."},{"Start":"01:27.470 ","End":"01:33.210","Text":"Here we are again, we want to get 0s below the 1,"},{"Start":"01:33.210 ","End":"01:36.450","Text":"so we subtract twice the first row from the second,"},{"Start":"01:36.450 ","End":"01:39.000","Text":"3 times the first from the third,"},{"Start":"01:39.000 ","End":"01:42.825","Text":"and subtract it from the fourth as written here."},{"Start":"01:42.825 ","End":"01:44.925","Text":"This is what we get."},{"Start":"01:44.925 ","End":"01:48.920","Text":"Now we want to make these 2 entries below the 1 here,"},{"Start":"01:48.920 ","End":"01:50.915","Text":"this and this we want 0."},{"Start":"01:50.915 ","End":"01:54.680","Text":"Subtract twice second row from third row"},{"Start":"01:54.680 ","End":"01:59.235","Text":"and add 3 times the second row to the fourth row."},{"Start":"01:59.235 ","End":"02:01.870","Text":"Well, we got 2 rows of 0,"},{"Start":"02:01.870 ","End":"02:05.180","Text":"so we can check this 1 out and check this 1 out."},{"Start":"02:05.180 ","End":"02:10.490","Text":"We just have these 2 vectors as a basis for the image."},{"Start":"02:10.490 ","End":"02:13.130","Text":"This 1 is here, this 1 is here."},{"Start":"02:13.130 ","End":"02:17.120","Text":"If you count them, 1, 2 dimension of the image is 2."},{"Start":"02:17.120 ","End":"02:19.430","Text":"That\u0027s what we had earlier, so it confirmed."},{"Start":"02:19.430 ","End":"02:22.410","Text":"We are done."}],"ID":10205},{"Watched":false,"Name":"Exercise 4 Part a","Duration":"5m 18s","ChapterTopicVideoID":9656,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.810","Text":"In this exercise, we\u0027re given the transformation to and from this space M_2 over R,"},{"Start":"00:06.810 ","End":"00:11.325","Text":"which is the space of 2 by 2 real matrices."},{"Start":"00:11.325 ","End":"00:16.110","Text":"It\u0027s defined as follows that T of a matrix A is"},{"Start":"00:16.110 ","End":"00:21.495","Text":"A times this matrix here minus this matrix times A."},{"Start":"00:21.495 ","End":"00:26.625","Text":"Our task is to find the basis and the dimension of the kernel of T,"},{"Start":"00:26.625 ","End":"00:33.125","Text":"and likewise for the image of T. Actually I never showed that T really is linear."},{"Start":"00:33.125 ","End":"00:37.055","Text":"I\u0027ll leave that as an exercise for you or you can just take my word for it."},{"Start":"00:37.055 ","End":"00:42.215","Text":"I\u0027d like to start with some preliminary calculations that will serve as for both parts,"},{"Start":"00:42.215 ","End":"00:46.010","Text":"I don\u0027t if prolog is the right word, but anyway, preliminary."},{"Start":"00:46.010 ","End":"00:49.800","Text":"Let\u0027s compute if the matrix A is x,"},{"Start":"00:49.800 ","End":"00:54.105","Text":"y, z, t, what is T of A?"},{"Start":"00:54.105 ","End":"00:56.240","Text":"If I plug A in here as x,"},{"Start":"00:56.240 ","End":"00:58.930","Text":"y, z, t, this is what we get."},{"Start":"00:58.930 ","End":"01:01.820","Text":"The product of these 2 is here,"},{"Start":"01:01.820 ","End":"01:03.980","Text":"for example, x y times 1,"},{"Start":"01:03.980 ","End":"01:06.715","Text":"0 is just x and so on,"},{"Start":"01:06.715 ","End":"01:09.390","Text":"and here 1, 2 times x,"},{"Start":"01:09.390 ","End":"01:11.100","Text":"z is x plus 2z."},{"Start":"01:11.100 ","End":"01:13.890","Text":"Complete all the computations,"},{"Start":"01:13.890 ","End":"01:16.695","Text":"and now we need to do a subtraction."},{"Start":"01:16.695 ","End":"01:19.900","Text":"This is what we get."},{"Start":"01:21.620 ","End":"01:24.740","Text":"This is an important intermediate result."},{"Start":"01:24.740 ","End":"01:29.130","Text":"I\u0027ll just put it in a frame."},{"Start":"01:29.240 ","End":"01:31.875","Text":"Next thing we\u0027re going to do is,"},{"Start":"01:31.875 ","End":"01:34.520","Text":"it\u0027s awkward to work with 2-by-2 matrices."},{"Start":"01:34.520 ","End":"01:36.545","Text":"Remember, we used to flatten them out."},{"Start":"01:36.545 ","End":"01:39.530","Text":"We can take it like first x, then y, then z,"},{"Start":"01:39.530 ","End":"01:43.220","Text":"then t. We call it like a snake this, this, this."},{"Start":"01:43.220 ","End":"01:44.930","Text":"If we do that,"},{"Start":"01:44.930 ","End":"01:48.410","Text":"then we could rewrite the transformation T in"},{"Start":"01:48.410 ","End":"01:54.595","Text":"this form with just row vectors rather than matrices."},{"Start":"01:54.595 ","End":"01:59.860","Text":"We\u0027ll start with the kernel and let\u0027s see what the kernel is."},{"Start":"01:59.860 ","End":"02:06.745","Text":"The kernel is the space of vectors that T takes to the 0 vector,"},{"Start":"02:06.745 ","End":"02:09.640","Text":"and since T of x, y, z,"},{"Start":"02:09.640 ","End":"02:13.855","Text":"t is this, we need this to be equal to 0."},{"Start":"02:13.855 ","End":"02:16.855","Text":"That is, each component must be 0."},{"Start":"02:16.855 ","End":"02:22.090","Text":"Notice that we have 3 components the same."},{"Start":"02:22.090 ","End":"02:23.710","Text":"The first, the third,"},{"Start":"02:23.710 ","End":"02:25.965","Text":"and the fourth are all minus 2z,"},{"Start":"02:25.965 ","End":"02:30.630","Text":"so there\u0027s no point in writing this 3 times minus 2z is 0."},{"Start":"02:30.630 ","End":"02:32.950","Text":"Really we only have 2 equations."},{"Start":"02:32.950 ","End":"02:34.450","Text":"Minus 2z is 0,"},{"Start":"02:34.450 ","End":"02:37.490","Text":"and 2x plus 2y minus 2t is 0,"},{"Start":"02:37.490 ","End":"02:39.725","Text":"2 equations, 4 unknowns."},{"Start":"02:39.725 ","End":"02:43.130","Text":"Divide this equation by minus 2 and this equation by 2,"},{"Start":"02:43.130 ","End":"02:44.480","Text":"and this is what we get,"},{"Start":"02:44.480 ","End":"02:47.870","Text":"the leading term in each row."},{"Start":"02:47.870 ","End":"02:50.405","Text":"That would be this one and this one."},{"Start":"02:50.405 ","End":"02:55.579","Text":"These are the pivot terms and they are dependent or constraint,"},{"Start":"02:55.579 ","End":"02:56.735","Text":"but the other 2,"},{"Start":"02:56.735 ","End":"02:59.530","Text":"t and y are free."},{"Start":"02:59.530 ","End":"03:04.160","Text":"Now we\u0027ll apply the technique that I informally called the wandering ones,"},{"Start":"03:04.160 ","End":"03:09.460","Text":"where we let each free variable be 1 in turn and the other 0."},{"Start":"03:09.460 ","End":"03:13.110","Text":"First, we let y be 1 and t is 0."},{"Start":"03:13.110 ","End":"03:14.870","Text":"Then in this equation,"},{"Start":"03:14.870 ","End":"03:19.715","Text":"we get that x is minus 1 and z is just 0."},{"Start":"03:19.715 ","End":"03:23.980","Text":"That gives us this and now this to the other way around."},{"Start":"03:23.980 ","End":"03:26.975","Text":"Meaning t is 1 and y is 0."},{"Start":"03:26.975 ","End":"03:28.580","Text":"If we plug that in,"},{"Start":"03:28.580 ","End":"03:31.190","Text":"then we get from this equation,"},{"Start":"03:31.190 ","End":"03:36.535","Text":"x equals 1 and still z equals 0 from the first equation."},{"Start":"03:36.535 ","End":"03:39.710","Text":"Now, remember that vectors,"},{"Start":"03:39.710 ","End":"03:41.180","Text":"we flatten them out."},{"Start":"03:41.180 ","End":"03:43.100","Text":"But really the x, y,"},{"Start":"03:43.100 ","End":"03:45.500","Text":"z, t go like this, x,"},{"Start":"03:45.500 ","End":"03:50.900","Text":"y, z, t. If I put each of these in this form,"},{"Start":"03:50.900 ","End":"03:54.860","Text":"and you put them in the order x, y, z,"},{"Start":"03:54.860 ","End":"03:59.045","Text":"t. This will give us this matrix,"},{"Start":"03:59.045 ","End":"04:01.850","Text":"and the other one also x,"},{"Start":"04:01.850 ","End":"04:05.790","Text":"y, z, t gives us this matrix."},{"Start":"04:05.790 ","End":"04:10.175","Text":"These 2 are now a basis for the kernel of"},{"Start":"04:10.175 ","End":"04:17.435","Text":"T. If you count how many members there are in the basis is 1,2."},{"Start":"04:17.435 ","End":"04:19.550","Text":"The dimension of the kernel,"},{"Start":"04:19.550 ","End":"04:23.820","Text":"also known as the nullity, is 2."},{"Start":"04:23.860 ","End":"04:28.400","Text":"Now before we move on to finding the image of T,"},{"Start":"04:28.400 ","End":"04:35.690","Text":"we can actually easily find the dimension of the image T using the rank-nullity theorem."},{"Start":"04:35.690 ","End":"04:39.725","Text":"We have a transformation from this space to itself,"},{"Start":"04:39.725 ","End":"04:43.500","Text":"the space and the rank-nullity theorem."},{"Start":"04:43.500 ","End":"04:45.710","Text":"The dimension of the one on the left, well,"},{"Start":"04:45.710 ","End":"04:50.135","Text":"are the same is equal to the dimension of the kernel plus the dimension of the image."},{"Start":"04:50.135 ","End":"04:52.870","Text":"The dimension of this is 4."},{"Start":"04:52.870 ","End":"04:54.885","Text":"That\u0027s the 2-by-2 matrices."},{"Start":"04:54.885 ","End":"04:58.685","Text":"Dimension of kernel we found from here is 2,"},{"Start":"04:58.685 ","End":"05:03.085","Text":"so the dimension of the image must also be 2."},{"Start":"05:03.085 ","End":"05:05.705","Text":"If 4 equals 2 plus something,"},{"Start":"05:05.705 ","End":"05:08.225","Text":"then that something must be 2."},{"Start":"05:08.225 ","End":"05:09.620","Text":"The dimension of the image,"},{"Start":"05:09.620 ","End":"05:12.715","Text":"also known as the rank of T is 2."},{"Start":"05:12.715 ","End":"05:18.210","Text":"Now we\u0027ll take a break and then we\u0027ll deal with the image."}],"ID":10206},{"Watched":false,"Name":"Exercise 4 Part b","Duration":"3m 1s","ChapterTopicVideoID":9657,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.520","Text":"Continuing the exercise we started,"},{"Start":"00:02.520 ","End":"00:04.350","Text":"we finished with the kernel."},{"Start":"00:04.350 ","End":"00:05.895","Text":"Now, we want to look at the image."},{"Start":"00:05.895 ","End":"00:09.960","Text":"We already computed T as this."},{"Start":"00:09.960 ","End":"00:16.095","Text":"But it was more convenient to flatten them out using the snake method."},{"Start":"00:16.095 ","End":"00:21.614","Text":"Say T of x, y, z, t as a row vector equals this."},{"Start":"00:21.614 ","End":"00:25.709","Text":"Basically what we\u0027ve done here is we simplified the problem"},{"Start":"00:25.709 ","End":"00:33.615","Text":"from M_2 over R to R^4 just by flattening out."},{"Start":"00:33.615 ","End":"00:35.460","Text":"We\u0027ll work in this form"},{"Start":"00:35.460 ","End":"00:39.035","Text":"and later we have to remember at the end to convert back to a matrix."},{"Start":"00:39.035 ","End":"00:43.490","Text":"We need a basis for R^4 and we\u0027ll use the standard basis"},{"Start":"00:43.490 ","End":"00:46.610","Text":"and we have to apply T to each member of the basis."},{"Start":"00:46.610 ","End":"00:51.485","Text":"I\u0027ve given you all 4 at once, the 1, 0, 0, 0, then 0, 1, 0, 0,"},{"Start":"00:51.485 ","End":"00:53.870","Text":"each 1 of these you plug into here."},{"Start":"00:53.870 ","End":"00:58.505","Text":"For example, in the first 1, just x is not 0."},{"Start":"00:58.505 ","End":"01:04.290","Text":"This is 0, this is 2, and this is 0, this is 0."},{"Start":"01:04.290 ","End":"01:06.195","Text":"That\u0027s the first 1 and so on,"},{"Start":"01:06.195 ","End":"01:11.850","Text":"all 4 of the basis members."},{"Start":"01:11.850 ","End":"01:18.510","Text":"Next, we take these 4 vectors and put them into a matrix."},{"Start":"01:18.510 ","End":"01:25.810","Text":"Here they are, 0, 2, 0, 0 is here, 0, 2, 0, 0 is here, this 1 is here."},{"Start":"01:25.810 ","End":"01:28.820","Text":"We want to bring this to row echelon form"},{"Start":"01:28.820 ","End":"01:31.974","Text":"because these might not be linearly independent"},{"Start":"01:31.974 ","End":"01:37.570","Text":"and we\u0027ll see that if we get rows of 0 we\u0027ll be able to throw those out."},{"Start":"01:37.790 ","End":"01:41.450","Text":"It\u0027s convenient for me to actually work with the second column,"},{"Start":"01:41.450 ","End":"01:44.135","Text":"not the first and it\u0027ll work out, you\u0027ll see."},{"Start":"01:44.135 ","End":"01:50.090","Text":"Subtract this row from the second and add it to the fourth."},{"Start":"01:50.090 ","End":"01:51.485","Text":"We get this."},{"Start":"01:51.485 ","End":"01:53.960","Text":"Now, everything\u0027s divisible by 2,"},{"Start":"01:53.960 ","End":"01:57.360","Text":"so this is what we get."},{"Start":"01:57.430 ","End":"02:01.055","Text":"Now, look, we have 2 rows of 0s."},{"Start":"02:01.055 ","End":"02:04.025","Text":"It\u0027s not officially in row echelon form."},{"Start":"02:04.025 ","End":"02:08.390","Text":"If you switch the first and second columns,"},{"Start":"02:08.390 ","End":"02:10.834","Text":"it would be, but that\u0027s okay."},{"Start":"02:10.834 ","End":"02:15.990","Text":"We have 2 rows of 0s and these 2 are linearly independent."},{"Start":"02:15.990 ","End":"02:18.900","Text":"I meant to say we can swap these 2 rows around"},{"Start":"02:18.900 ","End":"02:21.570","Text":"and then it would be in echelon form."},{"Start":"02:21.570 ","End":"02:26.519","Text":"These 2 vectors will span the image."},{"Start":"02:26.519 ","End":"02:29.430","Text":"But now I also want to go back to the matrix form."},{"Start":"02:29.430 ","End":"02:38.380","Text":"So 0, 1, 0, 0, we fill it in 0, 1, 0, 0, and then 1, 0, 1 minus 1."},{"Start":"02:38.380 ","End":"02:42.545","Text":"These 2 matrices are the basis for the image."},{"Start":"02:42.545 ","End":"02:44.715","Text":"If I count them, there\u0027s 2 of them."},{"Start":"02:44.715 ","End":"02:48.745","Text":"So the dimension of the image is 2."},{"Start":"02:48.745 ","End":"02:51.669","Text":"That\u0027s what we expected anyway."},{"Start":"02:51.669 ","End":"02:55.925","Text":"Remember we got this result when we did the kernel."},{"Start":"02:55.925 ","End":"02:57.995","Text":"So it\u0027s a confirmation that this is 2."},{"Start":"02:57.995 ","End":"03:01.140","Text":"We are done."}],"ID":10207},{"Watched":false,"Name":"Exercise 5 Part a","Duration":"4m 56s","ChapterTopicVideoID":9658,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:10.320","Text":"In this exercise, we have a linear transformation T from the space of polynomials of"},{"Start":"00:10.320 ","End":"00:16.730","Text":"degree 2 or less over the reals to the same space and it\u0027s defined with"},{"Start":"00:16.730 ","End":"00:25.710","Text":"the formula that T of a polynomial p of x is p of x plus 1 minus p of x plus 4."},{"Start":"00:25.710 ","End":"00:31.260","Text":"In a moment, I\u0027ll give an example of how to interpret this."},{"Start":"00:31.490 ","End":"00:41.030","Text":"Our task is to find a basis of the kernel of T and also the dimension and then"},{"Start":"00:41.030 ","End":"00:49.710","Text":"the same thing for the image of T. Let\u0027s start the solution by,"},{"Start":"00:49.710 ","End":"00:58.860","Text":"I called it prologue somewhere between preliminaries and preparations,"},{"Start":"00:58.860 ","End":"01:01.650","Text":"a few calculations, and I\u0027ll start,"},{"Start":"01:01.650 ","End":"01:04.610","Text":"for example, by showing what this formula means."},{"Start":"01:04.610 ","End":"01:10.595","Text":"If we take p of x to be the polynomial x squared and apply T to it,"},{"Start":"01:10.595 ","End":"01:14.120","Text":"what this formula says is that instead of x,"},{"Start":"01:14.120 ","End":"01:22.670","Text":"we replace it with x plus 1 once then we replace it with x plus 4 and do the subtraction."},{"Start":"01:22.670 ","End":"01:25.085","Text":"First of all, instead of x,"},{"Start":"01:25.085 ","End":"01:27.280","Text":"I put x plus 1,"},{"Start":"01:27.280 ","End":"01:30.940","Text":"then I put x plus 4 instead of the x,"},{"Start":"01:30.940 ","End":"01:33.140","Text":"and then do the subtraction."},{"Start":"01:33.140 ","End":"01:34.340","Text":"Now if you do the calculation,"},{"Start":"01:34.340 ","End":"01:38.965","Text":"it just comes out to be minus 6x minus 15."},{"Start":"01:38.965 ","End":"01:41.760","Text":"This is x squared plus 2x plus 1,"},{"Start":"01:41.760 ","End":"01:46.590","Text":"this one is x squared plus 8x plus 16."},{"Start":"01:46.590 ","End":"01:49.655","Text":"Do the arithmetic. This is what we get."},{"Start":"01:49.655 ","End":"01:55.060","Text":"Another example, if the polynomial is 4x plus 1 and instead of x,"},{"Start":"01:55.060 ","End":"01:57.560","Text":"I once put x plus 1,"},{"Start":"01:57.560 ","End":"02:02.015","Text":"and the next time I put x plus 4 and do the computation,"},{"Start":"02:02.015 ","End":"02:06.340","Text":"you\u0027ll see this boils down to just minus 12."},{"Start":"02:06.340 ","End":"02:08.300","Text":"Now that you\u0027ve got the idea,"},{"Start":"02:08.300 ","End":"02:10.445","Text":"let\u0027s do it in general."},{"Start":"02:10.445 ","End":"02:17.540","Text":"A general polynomial in P_2 of R is ax squared plus bx plus c,"},{"Start":"02:17.540 ","End":"02:19.850","Text":"where a, b and c are real numbers."},{"Start":"02:19.850 ","End":"02:22.220","Text":"Just like before, first of all,"},{"Start":"02:22.220 ","End":"02:27.110","Text":"we replace x by x plus 1 and that will give us this first square bracket."},{"Start":"02:27.110 ","End":"02:31.855","Text":"Then we replace x by x plus 4 and that will give us this."},{"Start":"02:31.855 ","End":"02:33.690","Text":"Just some simple algebra,"},{"Start":"02:33.690 ","End":"02:36.210","Text":"x plus 1 squared is x squared plus 2x plus 1,"},{"Start":"02:36.210 ","End":"02:39.080","Text":"x plus 4 squared is x squared plus 8x plus 16."},{"Start":"02:39.080 ","End":"02:41.890","Text":"Start opening brackets a bit."},{"Start":"02:41.890 ","End":"02:44.745","Text":"This first bracket simplifies to this,"},{"Start":"02:44.745 ","End":"02:48.040","Text":"this simplifies to this, and at the end,"},{"Start":"02:48.040 ","End":"02:49.775","Text":"all we get is this,"},{"Start":"02:49.775 ","End":"02:53.420","Text":"which is actually a linear polynomial"},{"Start":"02:53.420 ","End":"02:57.605","Text":"in x even though we started with a quadratic, but that\u0027s okay."},{"Start":"02:57.605 ","End":"03:03.520","Text":"Now, we can describe T in this form which I\u0027ve framed,"},{"Start":"03:03.520 ","End":"03:09.980","Text":"but we generally prefer not to work with polynomials,"},{"Start":"03:09.980 ","End":"03:14.245","Text":"but we like to work with RN, with regular vectors."},{"Start":"03:14.245 ","End":"03:18.860","Text":"If I use the standard basis for the polynomials of degree 2 or less,"},{"Start":"03:18.860 ","End":"03:22.985","Text":"which is 1, x, and x squared,"},{"Start":"03:22.985 ","End":"03:27.590","Text":"then this polynomial has coordinates a,"},{"Start":"03:27.590 ","End":"03:31.410","Text":"b, c, and this one has coordinates 0,"},{"Start":"03:31.410 ","End":"03:35.970","Text":"x squared, and then minus 6a with x,"},{"Start":"03:35.970 ","End":"03:40.635","Text":"and minus 15a minus 3b with the 1."},{"Start":"03:40.635 ","End":"03:42.960","Text":"I\u0027ve written it backwards."},{"Start":"03:42.960 ","End":"03:45.345","Text":"Here, this is the basis I meant."},{"Start":"03:45.345 ","End":"03:47.870","Text":"It actually is an ordered basis."},{"Start":"03:47.870 ","End":"03:50.045","Text":"The order makes a difference."},{"Start":"03:50.045 ","End":"04:00.485","Text":"Now, I wrote T bar instead of using another letter but actually, there\u0027s no confusion."},{"Start":"04:00.485 ","End":"04:07.380","Text":"We could actually just use the same letter T and there won\u0027t be any confusion."},{"Start":"04:07.380 ","End":"04:10.710","Text":"There\u0027s one thing I was going to do."},{"Start":"04:10.710 ","End":"04:13.590","Text":"I wanted to just verify this formula."},{"Start":"04:13.590 ","End":"04:14.940","Text":"Earlier, if you recall,"},{"Start":"04:14.940 ","End":"04:23.255","Text":"we checked what is T of x squared and got minus 6x minus 15 from direct computation."},{"Start":"04:23.255 ","End":"04:27.335","Text":"Let\u0027s see if we can go back and look at that. Yeah, here it is."},{"Start":"04:27.335 ","End":"04:31.430","Text":"T of x squared is minus 6x minus 15 and it"},{"Start":"04:31.430 ","End":"04:37.220","Text":"also works with this formula because for x squared a is 1,"},{"Start":"04:37.220 ","End":"04:38.810","Text":"b is 0, c is 0,"},{"Start":"04:38.810 ","End":"04:41.120","Text":"if you plug it in here, when a is 1,"},{"Start":"04:41.120 ","End":"04:47.290","Text":"you get minus 6x minus 15, so that\u0027s okay."},{"Start":"04:47.290 ","End":"04:50.280","Text":"That\u0027s what I call the prologue."},{"Start":"04:50.280 ","End":"04:56.430","Text":"Now, let\u0027s get started or rather let\u0027s take a break first and continue after."}],"ID":10208},{"Watched":false,"Name":"Exercise 5 Part b","Duration":"4m 10s","ChapterTopicVideoID":9659,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.310","Text":"Here we are after the break."},{"Start":"00:02.310 ","End":"00:05.955","Text":"This was a transformation."},{"Start":"00:05.955 ","End":"00:11.000","Text":"It was from P_2 to P_2 of"},{"Start":"00:11.000 ","End":"00:17.970","Text":"R. But we preferred to work with regular vectors in R^3."},{"Start":"00:17.970 ","End":"00:25.785","Text":"We showed that T looks like this if we use the standard basis of x squared,"},{"Start":"00:25.785 ","End":"00:28.065","Text":"x, and 1."},{"Start":"00:28.065 ","End":"00:29.880","Text":"Then this is coordinates a, b,"},{"Start":"00:29.880 ","End":"00:33.465","Text":"c. This coordinate is this."},{"Start":"00:33.465 ","End":"00:35.610","Text":"Let\u0027s see what is the kernel?"},{"Start":"00:35.610 ","End":"00:39.300","Text":"The kernel is everything that goes to 0,"},{"Start":"00:39.300 ","End":"00:42.390","Text":"that T sends to 0."},{"Start":"00:42.390 ","End":"00:49.070","Text":"We equate this vector to the 0 vector which is 0, 0, 0 in our"},{"Start":"00:49.070 ","End":"00:54.900","Text":"3 and that gives us 2 equations in 3 unknowns."},{"Start":"00:54.900 ","End":"00:56.000","Text":"Because the first component,"},{"Start":"00:56.000 ","End":"00:58.355","Text":"0 equals 0, it doesn\u0027t give us anything,"},{"Start":"00:58.355 ","End":"01:04.385","Text":"so we have minus 6a is 0 and minus 15a minus 3b is 0."},{"Start":"01:04.385 ","End":"01:08.415","Text":"You can see that c is the free variable,"},{"Start":"01:08.415 ","End":"01:13.240","Text":"to get a basis, we could let say, c equal 1."},{"Start":"01:13.240 ","End":"01:16.115","Text":"Actually, regardless of c,"},{"Start":"01:16.115 ","End":"01:19.490","Text":"we get that from here, a is 0."},{"Start":"01:19.490 ","End":"01:22.560","Text":"If a is 0, then b is 0."},{"Start":"01:22.690 ","End":"01:29.160","Text":"A basis for the solution space if I rearrange"},{"Start":"01:29.160 ","End":"01:36.805","Text":"this would be the vector 0, 0, 1 in R^3."},{"Start":"01:36.805 ","End":"01:40.325","Text":"But now we want to go back to the world of polynomials,"},{"Start":"01:40.325 ","End":"01:47.460","Text":"so it\u0027s 0x squared plus 0x plus 1."},{"Start":"01:47.460 ","End":"01:50.055","Text":"This is the polynomial 1,"},{"Start":"01:50.055 ","End":"01:55.460","Text":"that\u0027s a basis for the kernel of T. Since it only has 1 member,"},{"Start":"01:55.460 ","End":"01:58.960","Text":"the dimension of the kernel is 1."},{"Start":"01:58.960 ","End":"02:03.020","Text":"This is also called the nullity of T, which is 1."},{"Start":"02:03.020 ","End":"02:05.210","Text":"That takes care of the kernel."},{"Start":"02:05.210 ","End":"02:10.850","Text":"Now let\u0027s move on to the image"},{"Start":"02:10.850 ","End":"02:17.450","Text":"of T. Then we also want a basis and the dimension."},{"Start":"02:17.450 ","End":"02:23.740","Text":"Once again, let\u0027s go back to the vector form from R^3 to R^3."},{"Start":"02:23.740 ","End":"02:27.100","Text":"This was the formula we had."},{"Start":"02:27.100 ","End":"02:31.940","Text":"We apply the transformation T to a basis."},{"Start":"02:31.940 ","End":"02:35.005","Text":"We take the standard basis for R^3,"},{"Start":"02:35.005 ","End":"02:36.900","Text":"which are these 3 vectors."},{"Start":"02:36.900 ","End":"02:39.315","Text":"These actually correspond to the x^2,"},{"Start":"02:39.315 ","End":"02:42.515","Text":"x, and 1 in polynomials."},{"Start":"02:42.515 ","End":"02:48.435","Text":"Anyway, what we get are these 3 vectors."},{"Start":"02:48.435 ","End":"02:50.790","Text":"These will span the image."},{"Start":"02:50.790 ","End":"02:55.920","Text":"But we have to make them linearly independent,"},{"Start":"02:56.030 ","End":"03:00.410","Text":"so we take these 3 and put them in a matrix."},{"Start":"03:00.410 ","End":"03:02.150","Text":"We don\u0027t need to take the last 1,"},{"Start":"03:02.150 ","End":"03:03.335","Text":"we throw out 0,"},{"Start":"03:03.335 ","End":"03:07.050","Text":"so we just take these two: 0 minus 6 minus 15,"},{"Start":"03:07.050 ","End":"03:09.015","Text":"and 0, 0 minus 3."},{"Start":"03:09.015 ","End":"03:17.800","Text":"It already is in row echelon form but perhaps we could just divide everything by 3,"},{"Start":"03:17.800 ","End":"03:19.980","Text":"make that minus 3,"},{"Start":"03:19.980 ","End":"03:22.155","Text":"and we\u0027d have this."},{"Start":"03:22.155 ","End":"03:29.520","Text":"These 2 vectors now would be linearly independent and span the image,"},{"Start":"03:29.520 ","End":"03:32.050","Text":"they\u0027re a basis for the image."},{"Start":"03:32.110 ","End":"03:37.060","Text":"As before, we want to go back to polynomials."},{"Start":"03:37.060 ","End":"03:39.420","Text":"The first vector 0, 2,"},{"Start":"03:39.420 ","End":"03:44.970","Text":"5 is 0x squared plus 2x plus 5 is the first 1, 2x plus 5."},{"Start":"03:44.970 ","End":"03:50.680","Text":"The last 1, 0x squared plus 0x plus 1, is just 1."},{"Start":"03:53.330 ","End":"03:55.490","Text":"Did I write the kernel?"},{"Start":"03:55.490 ","End":"03:57.310","Text":"I meant the image."},{"Start":"03:57.310 ","End":"03:59.700","Text":"Yeah, sorry about that typo."},{"Start":"03:59.700 ","End":"04:03.870","Text":"The dimension, count them 1, 2,"},{"Start":"04:03.870 ","End":"04:11.130","Text":"is 2 which is also called the rank of T. So we are done."}],"ID":10209},{"Watched":false,"Name":"Exercise 6 Part a","Duration":"4m 1s","ChapterTopicVideoID":9644,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.270","Text":"In this exercise, we have a linear transformation from the space"},{"Start":"00:06.270 ","End":"00:12.945","Text":"of polynomials of degree 3 or less over the reals and to itself."},{"Start":"00:12.945 ","End":"00:15.825","Text":"But for a change,"},{"Start":"00:15.825 ","End":"00:22.875","Text":"we\u0027re using the letter d rather than t because d,"},{"Start":"00:22.875 ","End":"00:28.755","Text":"what it does to a polynomial is it sends it to its derivative."},{"Start":"00:28.755 ","End":"00:36.320","Text":"D for derivative, you shouldn\u0027t get used to all the time just using letter t. As usual,"},{"Start":"00:36.320 ","End":"00:41.810","Text":"we have to find the basis and the dimension of the kernel of D and likewise for"},{"Start":"00:41.810 ","End":"00:47.750","Text":"the image of D. Let\u0027s start with some general,"},{"Start":"00:47.750 ","End":"00:52.630","Text":"I call the prologue, some preliminary calculations and explanations."},{"Start":"00:52.630 ","End":"00:56.310","Text":"Let\u0027s figure out what t does more specifically."},{"Start":"00:56.310 ","End":"00:58.670","Text":"A typical polynomial of degree 3 or"},{"Start":"00:58.670 ","End":"01:02.240","Text":"less would be of this form x cubed plus bx squared plus"},{"Start":"01:02.240 ","End":"01:08.330","Text":"cx plus d and I\u0027m assuming you know basic derivatives,"},{"Start":"01:08.330 ","End":"01:11.285","Text":"the derivative of this would be as follows;"},{"Start":"01:11.285 ","End":"01:14.400","Text":"3 times a and you lower the degree by 1,"},{"Start":"01:14.400 ","End":"01:19.980","Text":"3ax squared plus 2bx plus c and d gives nothing."},{"Start":"01:20.180 ","End":"01:24.440","Text":"As before, it would be more convenient not to work with polynomials,"},{"Start":"01:24.440 ","End":"01:25.940","Text":"but to work with vectors."},{"Start":"01:25.940 ","End":"01:33.490","Text":"If we use the basis x cubed, x squared, x1."},{"Start":"01:33.490 ","End":"01:35.835","Text":"There are 4 elements in the basis,"},{"Start":"01:35.835 ","End":"01:43.760","Text":"then we can convert this to a formula in R4 and then d would take a, b, c,"},{"Start":"01:43.760 ","End":"01:51.540","Text":"d, add to this and its coordinates are as follows because it\u0027s 0x cubed,"},{"Start":"01:51.540 ","End":"01:53.460","Text":"3ax squared 2bx,"},{"Start":"01:53.460 ","End":"01:57.950","Text":"and c. It\u0027s not really the same d and if I was being pedantic,"},{"Start":"01:57.950 ","End":"02:01.655","Text":"I would call it D prime or D bar or some other letter,"},{"Start":"02:01.655 ","End":"02:07.470","Text":"but we could just call it D and you can tell from the context which D,"},{"Start":"02:07.470 ","End":"02:10.245","Text":"I mean with polynomials or vectors."},{"Start":"02:10.245 ","End":"02:14.760","Text":"Basically, instead of p3 of a,"},{"Start":"02:14.760 ","End":"02:18.125","Text":"we\u0027ve switched to the space R4 which is"},{"Start":"02:18.125 ","End":"02:22.250","Text":"practically equivalent or at least relative to this basis."},{"Start":"02:22.250 ","End":"02:27.990","Text":"Now, let\u0027s start with the kernel and later we\u0027ll do the image."},{"Start":"02:29.450 ","End":"02:32.845","Text":"Forgive the typo, I\u0027m so used to t,"},{"Start":"02:32.845 ","End":"02:34.390","Text":"this is d in our case."},{"Start":"02:34.390 ","End":"02:39.010","Text":"The kernel of d is everything that descends to 0,"},{"Start":"02:39.010 ","End":"02:44.715","Text":"0 the vector is the 000000 components."},{"Start":"02:44.715 ","End":"02:48.820","Text":"We have to just compare these 4 components to these."},{"Start":"02:48.820 ","End":"02:51.360","Text":"Of course the first 1 gives us 0 equals 0,"},{"Start":"02:51.360 ","End":"02:54.300","Text":"doesn\u0027t give anything so we get 3 equations,"},{"Start":"02:54.300 ","End":"02:56.520","Text":"which is that 3a, 2b,"},{"Start":"02:56.520 ","End":"03:02.790","Text":"and c are all 0 and the variable d is free."},{"Start":"03:02.790 ","End":"03:06.280","Text":"We\u0027ve learned the technique if you want a basis for the solution space,"},{"Start":"03:06.280 ","End":"03:09.290","Text":"we can let d equal 1,"},{"Start":"03:09.290 ","End":"03:11.550","Text":"and then a, b,"},{"Start":"03:11.550 ","End":"03:13.110","Text":"and c will be 0."},{"Start":"03:13.110 ","End":"03:14.930","Text":"That\u0027s actually regardless of d. a,"},{"Start":"03:14.930 ","End":"03:16.850","Text":"b and c will always be 0."},{"Start":"03:16.850 ","End":"03:19.970","Text":"If I put them in the right order, a, b, c,"},{"Start":"03:19.970 ","End":"03:23.490","Text":"d I\u0027d get 0, 0, 0,"},{"Start":"03:23.490 ","End":"03:31.350","Text":"1 but it\u0027s time to go back to the world of polynomials and this is the polynomial 1,"},{"Start":"03:31.350 ","End":"03:33.240","Text":"0x cubed, 0x squared,"},{"Start":"03:33.240 ","End":"03:35.085","Text":"0x plus 1 it\u0027s just 1."},{"Start":"03:35.085 ","End":"03:43.790","Text":"That\u0027s the basis for the kernel of d and there\u0027s only 1 member in the basis,"},{"Start":"03:43.790 ","End":"03:46.910","Text":"so it\u0027s dimension is 1,"},{"Start":"03:46.910 ","End":"03:53.270","Text":"and this is also called the nullity of the linear transformation"},{"Start":"03:53.270 ","End":"04:02.350","Text":"d. We\u0027ll do the image after the break. We\u0027ll take a break now."}],"ID":10210},{"Watched":false,"Name":"Exercise 6 Part b","Duration":"2m 51s","ChapterTopicVideoID":9645,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:03.150","Text":"Back from the break, we just did the kernel."},{"Start":"00:03.150 ","End":"00:05.040","Text":"Now you want the image,"},{"Start":"00:05.040 ","End":"00:07.110","Text":"basis, and dimension."},{"Start":"00:07.110 ","End":"00:15.195","Text":"Our linear transformation was called D and in vector form from our 4 to our 4,"},{"Start":"00:15.195 ","End":"00:19.920","Text":"this was the formula for D that we computed."},{"Start":"00:19.920 ","End":"00:25.280","Text":"Now to find the image,"},{"Start":"00:25.280 ","End":"00:32.045","Text":"we want to apply D first of all to a basis of our 4."},{"Start":"00:32.045 ","End":"00:35.380","Text":"Let\u0027s just go ahead with the standard basis."},{"Start":"00:35.380 ","End":"00:38.700","Text":"The standard basis of these 4 vectors,"},{"Start":"00:38.700 ","End":"00:40.755","Text":"1, 0, 0, 0, and so on."},{"Start":"00:40.755 ","End":"00:43.670","Text":"Actually, if you went back to polynomials,"},{"Start":"00:43.670 ","End":"00:45.320","Text":"this would be x cubed,"},{"Start":"00:45.320 ","End":"00:46.760","Text":"this would be x squared,"},{"Start":"00:46.760 ","End":"00:49.580","Text":"this would be x and this would be 1."},{"Start":"00:49.580 ","End":"00:52.625","Text":"Plugging in here to each of these,"},{"Start":"00:52.625 ","End":"00:56.215","Text":"we get these 4 vectors."},{"Start":"00:56.215 ","End":"00:59.780","Text":"These 4 vectors will span the image."},{"Start":"00:59.780 ","End":"01:04.265","Text":"But we want a basis for the image,"},{"Start":"01:04.265 ","End":"01:08.750","Text":"so we want to get rid of 1s that are not linearly independent."},{"Start":"01:08.750 ","End":"01:10.665","Text":"Well, for a start,"},{"Start":"01:10.665 ","End":"01:12.090","Text":"we can get rid of the 0s."},{"Start":"01:12.090 ","End":"01:17.060","Text":"Let\u0027s just take these 3 and bring to row echelon form in the matrix."},{"Start":"01:17.060 ","End":"01:22.760","Text":"This is the matrix, yeah, I should have just deleted the 0 row right away."},{"Start":"01:22.760 ","End":"01:24.545","Text":"I didn\u0027t even need to put it in."},{"Start":"01:24.545 ","End":"01:26.900","Text":"This is the matrix we got it already is in"},{"Start":"01:26.900 ","End":"01:31.265","Text":"row echelon form and there are no extra 0 rows."},{"Start":"01:31.265 ","End":"01:35.835","Text":"We could have actually also divided this row by 3, this by 2,"},{"Start":"01:35.835 ","End":"01:40.340","Text":"so we get 1s here and that makes it simpler so that"},{"Start":"01:40.340 ","End":"01:45.980","Text":"these 3 rows would now span the image and they are linearly independent,"},{"Start":"01:45.980 ","End":"01:49.315","Text":"so these 3 are a basis for the image,"},{"Start":"01:49.315 ","End":"01:53.185","Text":"and if we go back to the world of polynomials,"},{"Start":"01:53.185 ","End":"01:55.540","Text":"this 1 is the polynomial x squared."},{"Start":"01:55.540 ","End":"01:58.120","Text":"It\u0027s like 0 x cubed plus 1 x squared and so on."},{"Start":"01:58.120 ","End":"02:01.150","Text":"This 1 is x and this 1 is 1."},{"Start":"02:01.150 ","End":"02:06.700","Text":"That\u0027s a basis for the image."},{"Start":"02:07.100 ","End":"02:10.180","Text":"We also wanted the dimension of the image,"},{"Start":"02:10.180 ","End":"02:11.580","Text":"also known as the rank,"},{"Start":"02:11.580 ","End":"02:14.595","Text":"and just by counting 1, 2, 3."},{"Start":"02:14.595 ","End":"02:22.045","Text":"Let\u0027s just note that there is that rank nullity theorem and use it as a check."},{"Start":"02:22.045 ","End":"02:27.805","Text":"If I take the dimension of the image of D, which is 3."},{"Start":"02:27.805 ","End":"02:31.690","Text":"The dimension of the kernel of D from the previous part was 1,"},{"Start":"02:31.690 ","End":"02:33.580","Text":"together they give us 4,"},{"Start":"02:33.580 ","End":"02:38.310","Text":"and 4 is the dimension of polynomials of degree 3 or less."},{"Start":"02:38.310 ","End":"02:43.460","Text":"It\u0027s always the 3 plus 1 and we also saw this is equivalent to our 4."},{"Start":"02:43.460 ","End":"02:52.120","Text":"Everything works out fine and the 3 and the 1 plus 4 is extra check and we are done."}],"ID":10211},{"Watched":false,"Name":"Exercise 7","Duration":"4m 20s","ChapterTopicVideoID":9646,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:08.895","Text":"In this exercise, we have to find a linear transformation T that goes from R^3 to R^3,"},{"Start":"00:08.895 ","End":"00:15.210","Text":"such that the image is spanned by these 2 vectors."},{"Start":"00:15.210 ","End":"00:18.675","Text":"Let\u0027s start with a basis for R^3."},{"Start":"00:18.675 ","End":"00:23.010","Text":"I\u0027m talking about this R^3 on the left, the source space."},{"Start":"00:23.010 ","End":"00:25.940","Text":"We\u0027ll take the standard basis,"},{"Start":"00:25.940 ","End":"00:27.665","Text":"which is this,"},{"Start":"00:27.665 ","End":"00:30.920","Text":"and a transformation is uniquely"},{"Start":"00:30.920 ","End":"00:37.265","Text":"determined by specifying what it does to each basis member."},{"Start":"00:37.265 ","End":"00:41.795","Text":"Now if I wanted the image to be spanned by these 2,"},{"Start":"00:41.795 ","End":"00:45.890","Text":"what I could do would be to send, say,"},{"Start":"00:45.890 ","End":"00:49.730","Text":"the first 1 to this vector here,"},{"Start":"00:49.730 ","End":"00:56.055","Text":"and the second basis member to the second vector here."},{"Start":"00:56.055 ","End":"01:01.460","Text":"Any T that satisfies this will already have these 2 in the image,"},{"Start":"01:01.460 ","End":"01:04.550","Text":"and it will contain the span of these 2."},{"Start":"01:04.550 ","End":"01:08.585","Text":"But I don\u0027t wanted to get any larger than that."},{"Start":"01:08.585 ","End":"01:15.380","Text":"So when we apply T to the third basis member,"},{"Start":"01:15.380 ","End":"01:18.950","Text":"it should already be in the span of these 2."},{"Start":"01:18.950 ","End":"01:22.220","Text":"1 way we could do this is by duplication,"},{"Start":"01:22.220 ","End":"01:24.710","Text":"by choosing 1 of these again."},{"Start":"01:24.710 ","End":"01:28.005","Text":"Let\u0027s say the second 1 again."},{"Start":"01:28.005 ","End":"01:31.730","Text":"Then the image is spanned by these 3,"},{"Start":"01:31.730 ","End":"01:33.425","Text":"which is really these 2,"},{"Start":"01:33.425 ","End":"01:37.185","Text":"because just doesn\u0027t add anything."},{"Start":"01:37.185 ","End":"01:39.550","Text":"This is not the only possibility,"},{"Start":"01:39.550 ","End":"01:43.520","Text":"we could choose anything that\u0027s spanned by these 2."},{"Start":"01:43.520 ","End":"01:47.960","Text":"For example, we could have chosen the other 1 instead,"},{"Start":"01:47.960 ","End":"01:51.460","Text":"or we could even choose 0, 0, 0."},{"Start":"01:51.460 ","End":"01:56.190","Text":"The 0 vector if you add it to a set,"},{"Start":"01:56.190 ","End":"01:59.880","Text":"it\u0027s not going to increase the span."},{"Start":"01:59.880 ","End":"02:06.280","Text":"But we\u0027ll stick with 1 of them and we\u0027ll stick with our choice of minus 1, 4, 1."},{"Start":"02:06.280 ","End":"02:12.530","Text":"The question is, how do we find the T, sorry,"},{"Start":"02:12.530 ","End":"02:21.800","Text":"which satisfies these conditions that T sends these 3 basis vectors to these 3 vectors."},{"Start":"02:21.800 ","End":"02:26.210","Text":"If we specify T in matrix form,"},{"Start":"02:26.210 ","End":"02:32.570","Text":"what will do the trick would be this matrix which we get from these vectors"},{"Start":"02:32.570 ","End":"02:41.015","Text":"by taking this 1 and putting it here vertically,"},{"Start":"02:41.015 ","End":"02:43.675","Text":"this 1 here,"},{"Start":"02:43.675 ","End":"02:46.875","Text":"and this 1 over here,"},{"Start":"02:46.875 ","End":"02:48.980","Text":"and defining T of x, y,"},{"Start":"02:48.980 ","End":"02:58.030","Text":"z, which is a general vector in R^3 to be the matrix here times this."},{"Start":"02:58.030 ","End":"03:00.529","Text":"This method is guaranteed,"},{"Start":"03:00.529 ","End":"03:03.920","Text":"but let\u0027s check that it really does the job."},{"Start":"03:03.920 ","End":"03:07.790","Text":"Let\u0027s check that T of 1, 0,"},{"Start":"03:07.790 ","End":"03:10.010","Text":"0, 0, 1, 0, and 0, 0,"},{"Start":"03:10.010 ","End":"03:12.695","Text":"1 are really what we would expect."},{"Start":"03:12.695 ","End":"03:14.940","Text":"For the first 1, 1, 0, 0,"},{"Start":"03:14.940 ","End":"03:17.010","Text":"we write it as a column vector,"},{"Start":"03:17.010 ","End":"03:20.415","Text":"and then in matrix form, it\u0027s this."},{"Start":"03:20.415 ","End":"03:22.940","Text":"If you do this multiplication,"},{"Start":"03:22.940 ","End":"03:27.820","Text":"we get this row with this column just gives us the 4,"},{"Start":"03:27.820 ","End":"03:30.740","Text":"this row with this column gives us the 1."},{"Start":"03:30.740 ","End":"03:35.360","Text":"You\u0027ll notice that we just get the entries in the first column."},{"Start":"03:35.360 ","End":"03:38.755","Text":"The result is 4, 1, 4."},{"Start":"03:38.755 ","End":"03:41.460","Text":"Next, let\u0027s try 0, 1,"},{"Start":"03:41.460 ","End":"03:44.990","Text":"0, so we do the matrix multiplication,"},{"Start":"03:44.990 ","End":"03:51.390","Text":"and you\u0027ll see that what we get each time is just the middle entry in each of the rows,"},{"Start":"03:51.390 ","End":"03:53.070","Text":"and we\u0027ll get minus 1, 4,"},{"Start":"03:53.070 ","End":"03:56.480","Text":"1 as our second,"},{"Start":"03:56.480 ","End":"03:58.600","Text":"which is what we want."},{"Start":"03:58.600 ","End":"04:00.990","Text":"The last 1, 0,"},{"Start":"04:00.990 ","End":"04:03.320","Text":"0, 1, if you do the multiplication,"},{"Start":"04:03.320 ","End":"04:06.680","Text":"this row with this column gives us the minus 1,"},{"Start":"04:06.680 ","End":"04:08.645","Text":"then we get the 4 of them, we get the 1."},{"Start":"04:08.645 ","End":"04:12.175","Text":"Everything is just like we planned it."},{"Start":"04:12.175 ","End":"04:17.700","Text":"This is the answer in matrix form,"},{"Start":"04:17.700 ","End":"04:20.890","Text":"and so we are done."}],"ID":10212},{"Watched":false,"Name":"Exercise 8","Duration":"7m 53s","ChapterTopicVideoID":9647,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:09.295","Text":"In this exercise, we have to find a linear transformation T from R^4 to R^3,"},{"Start":"00:09.295 ","End":"00:13.500","Text":"whose kernel is spanned by these 2 vectors,"},{"Start":"00:13.500 ","End":"00:16.590","Text":"which are vectors in R^4 of course."},{"Start":"00:16.590 ","End":"00:20.550","Text":"I would like to mention that this is"},{"Start":"00:20.550 ","End":"00:25.290","Text":"a slightly more difficult exercise than one would usually get."},{"Start":"00:25.290 ","End":"00:27.555","Text":"I don\u0027t know if this would appear in exam or not,"},{"Start":"00:27.555 ","End":"00:33.275","Text":"but we\u0027ll study this one closely because it\u0027s bit more difficult."},{"Start":"00:33.275 ","End":"00:36.620","Text":"Now the idea is to do it in steps."},{"Start":"00:36.620 ","End":"00:41.730","Text":"The first step is to solve a related problem,"},{"Start":"00:41.730 ","End":"00:45.320","Text":"and that is to find a system of linear equations"},{"Start":"00:45.320 ","End":"00:50.925","Text":"whose solution space is spanned by these two vectors."},{"Start":"00:50.925 ","End":"00:56.524","Text":"When we\u0027ve got that, that will help us with the solution to the original problem."},{"Start":"00:56.524 ","End":"01:02.405","Text":"The way we do that is we put these inside a matrix."},{"Start":"01:02.405 ","End":"01:04.850","Text":"It suits me to switch the order."},{"Start":"01:04.850 ","End":"01:10.745","Text":"I\u0027ll put this one in the second place and this one in the first place."},{"Start":"01:10.745 ","End":"01:12.800","Text":"The reason for that is I want to bring it into"},{"Start":"01:12.800 ","End":"01:15.035","Text":"row echelon form and it will be easier that way,"},{"Start":"01:15.035 ","End":"01:16.685","Text":"you can switch the order."},{"Start":"01:16.685 ","End":"01:22.070","Text":"Then we add a last row which is just x,"},{"Start":"01:22.070 ","End":"01:27.640","Text":"y, z, t, a general vector in R^4."},{"Start":"01:27.640 ","End":"01:30.770","Text":"Like I said, we want to bring this to row echelon form."},{"Start":"01:30.770 ","End":"01:32.030","Text":"There was a first step."},{"Start":"01:32.030 ","End":"01:33.575","Text":"I wanted a 0 here."},{"Start":"01:33.575 ","End":"01:39.725","Text":"I\u0027ll take row 3 and subtract x times row 1,"},{"Start":"01:39.725 ","End":"01:41.870","Text":"but keep that in row 3."},{"Start":"01:41.870 ","End":"01:44.495","Text":"In other words, this minus x times this,"},{"Start":"01:44.495 ","End":"01:48.080","Text":"and that will give us this matrix."},{"Start":"01:48.080 ","End":"01:53.200","Text":"Next step, I want to put a 0 here."},{"Start":"01:53.200 ","End":"02:00.520","Text":"I\u0027ll subtract y minus 2x times the second row from the third row,"},{"Start":"02:00.520 ","End":"02:03.215","Text":"and we\u0027ll get this really should have written it out."},{"Start":"02:03.215 ","End":"02:09.170","Text":"What we did was row 3 minus"},{"Start":"02:09.170 ","End":"02:16.655","Text":"y minus 2x times row 2,"},{"Start":"02:16.655 ","End":"02:19.280","Text":"and put that into row 3."},{"Start":"02:19.280 ","End":"02:23.600","Text":"If we take this minus y minus 2x times this, of course we get 0."},{"Start":"02:23.600 ","End":"02:29.645","Text":"This subtract y minus 2x times 1 gives me the minus y plus 2x."},{"Start":"02:29.645 ","End":"02:37.190","Text":"This similarly, if I subtract y minus 2x times this x minus y plus 2x."},{"Start":"02:37.190 ","End":"02:40.990","Text":"Perhaps I can simplify this a bit."},{"Start":"02:40.990 ","End":"02:51.275","Text":"This entry here, the minus 3x plus 2x gives me minus x and then minus y plus z."},{"Start":"02:51.275 ","End":"02:57.630","Text":"What we do is we put that equal 0 here also we have minus 4x plus"},{"Start":"02:57.630 ","End":"03:04.350","Text":"2x is minus 2x and still minus y plus t equals 0."},{"Start":"03:04.350 ","End":"03:09.020","Text":"If I set these two to be equal to 0,"},{"Start":"03:09.020 ","End":"03:16.150","Text":"then a solution space for this system of equations."},{"Start":"03:16.150 ","End":"03:26.685","Text":"This is the SLA We\u0027re looking for whose solutions are spanned by 0, 1, 1,"},{"Start":"03:26.685 ","End":"03:32.475","Text":"1, and 1, 2, 3, 4,"},{"Start":"03:32.475 ","End":"03:37.160","Text":"I\u0027ve to still show you this that the solution space of"},{"Start":"03:37.160 ","End":"03:42.995","Text":"this is the space spanned by these two which we had in the beginning."},{"Start":"03:42.995 ","End":"03:46.410","Text":"Let me just get some more space here."},{"Start":"03:47.130 ","End":"03:50.800","Text":"We\u0027ll continue to step 2 to build the transformation."},{"Start":"03:50.800 ","End":"03:53.500","Text":"We could have checked already at this stage that"},{"Start":"03:53.500 ","End":"03:57.780","Text":"these two vectors both satisfy these equations."},{"Start":"03:57.780 ","End":"04:00.130","Text":"If they do so does the span."},{"Start":"04:00.130 ","End":"04:04.225","Text":"But let\u0027s just leave that for the next step."},{"Start":"04:04.225 ","End":"04:10.420","Text":"In the next step, we build a transformation from R^4 to"},{"Start":"04:10.420 ","End":"04:17.935","Text":"R^3 by taking each of these and putting it in a 3D vector."},{"Start":"04:17.935 ","End":"04:20.920","Text":"We put this expression on the left,"},{"Start":"04:20.920 ","End":"04:24.340","Text":"minus x minus y plus z first, then minus 2,"},{"Start":"04:24.340 ","End":"04:29.395","Text":"x minus y plus t. Anything missing because we need 3 components,"},{"Start":"04:29.395 ","End":"04:31.355","Text":"we put a 0 in."},{"Start":"04:31.355 ","End":"04:34.550","Text":"This is going to be our transformation."},{"Start":"04:34.550 ","End":"04:37.190","Text":"Now, the explanation is this,"},{"Start":"04:37.190 ","End":"04:39.170","Text":"if x, y, z,"},{"Start":"04:39.170 ","End":"04:43.745","Text":"t are in the kernel of this transformation,"},{"Start":"04:43.745 ","End":"04:45.500","Text":"then all these three are 0."},{"Start":"04:45.500 ","End":"04:47.285","Text":"The last one of course is 0,"},{"Start":"04:47.285 ","End":"04:50.210","Text":"means these two are 0 means that x, y, z,"},{"Start":"04:50.210 ","End":"04:57.660","Text":"t satisfy this system of linear equations there in the solution space."},{"Start":"04:58.640 ","End":"05:05.210","Text":"Conversely, if something isn\u0027t a solution space of this SLA,"},{"Start":"05:05.210 ","End":"05:09.380","Text":"then these two are 0, so it must belong to the kernel."},{"Start":"05:09.380 ","End":"05:18.610","Text":"Now all we have to do is show that the solution space is spanned by these 2 vectors."},{"Start":"05:18.770 ","End":"05:20.810","Text":"Like I said before,"},{"Start":"05:20.810 ","End":"05:23.525","Text":"we can actually substitute these in here,"},{"Start":"05:23.525 ","End":"05:26.270","Text":"each of these and show that these gives 0."},{"Start":"05:26.270 ","End":"05:29.120","Text":"For example, if I put in 0, 1, 1, 1,"},{"Start":"05:29.120 ","End":"05:32.110","Text":"then I\u0027ve got minus 0,"},{"Start":"05:32.110 ","End":"05:35.025","Text":"minus 1 plus 1 is 0."},{"Start":"05:35.025 ","End":"05:43.185","Text":"Here I\u0027ve got minus 2 plus 4 is also 0."},{"Start":"05:43.185 ","End":"05:47.960","Text":"If these two satisfy it and anything in their span satisfies it,"},{"Start":"05:47.960 ","End":"05:51.295","Text":"any linear combination will satisfy."},{"Start":"05:51.295 ","End":"05:56.460","Text":"Conversely, if x,"},{"Start":"05:56.460 ","End":"06:00.395","Text":"y, z, t are spanned by these 2 vectors,"},{"Start":"06:00.395 ","End":"06:02.210","Text":"then when I go back,"},{"Start":"06:02.210 ","End":"06:05.090","Text":"if this is in the span of these two,"},{"Start":"06:05.090 ","End":"06:10.535","Text":"that must mean that these rows are linearly dependent."},{"Start":"06:10.535 ","End":"06:13.940","Text":"That means that when I put into row echelon form,"},{"Start":"06:13.940 ","End":"06:16.645","Text":"I get a row of zeroes."},{"Start":"06:16.645 ","End":"06:19.310","Text":"Which means that since these 2 are non-zero,"},{"Start":"06:19.310 ","End":"06:21.395","Text":"that this is a row of zeroes,"},{"Start":"06:21.395 ","End":"06:23.735","Text":"which means that these two are 0."},{"Start":"06:23.735 ","End":"06:25.580","Text":"Anyway, if you think about it,"},{"Start":"06:25.580 ","End":"06:27.680","Text":"you\u0027ll see that it works out and,"},{"Start":"06:27.680 ","End":"06:33.830","Text":"or you can just accept it as a technique that what we do is the following system."},{"Start":"06:33.830 ","End":"06:37.830","Text":"We start with the 2 vectors and complete with an x, y, z,"},{"Start":"06:37.830 ","End":"06:41.000","Text":"t. Then take into row echelon form,"},{"Start":"06:41.000 ","End":"06:43.820","Text":"anything that\u0027s not 0 here assigned to 0."},{"Start":"06:43.820 ","End":"06:45.755","Text":"This is the SLA,"},{"Start":"06:45.755 ","End":"06:47.390","Text":"and then from the SLA,"},{"Start":"06:47.390 ","End":"06:52.729","Text":"we build the linear transformation as we showed."},{"Start":"06:52.729 ","End":"06:56.620","Text":"We could do one more step of verification."},{"Start":"06:56.620 ","End":"06:59.720","Text":"I did this already, but I didn\u0027t actually write it down."},{"Start":"06:59.720 ","End":"07:00.950","Text":"And that\u0027s what we should do,"},{"Start":"07:00.950 ","End":"07:04.190","Text":"is to put these 2 vectors,"},{"Start":"07:04.190 ","End":"07:05.810","Text":"0, 1, 1, 1, 2, 3,"},{"Start":"07:05.810 ","End":"07:10.835","Text":"4 into the system and show that they\u0027re 0 or rather,"},{"Start":"07:10.835 ","End":"07:14.445","Text":"which is equivalent to put them into"},{"Start":"07:14.445 ","End":"07:21.890","Text":"this formula into the transformation T and show that if I put 0,"},{"Start":"07:21.890 ","End":"07:26.880","Text":"1, 1, 1, I really do get all 0s."},{"Start":"07:26.880 ","End":"07:28.650","Text":"Similarly, if I put in 1, 2, 3,"},{"Start":"07:28.650 ","End":"07:30.560","Text":"4, I get all zeroes."},{"Start":"07:30.560 ","End":"07:33.990","Text":"It\u0027s pretty much the same what we showed already."},{"Start":"07:35.690 ","End":"07:41.240","Text":"We are done and perhaps I should just highlight the answer,"},{"Start":"07:41.240 ","End":"07:45.455","Text":"that the answer transformation is this here."},{"Start":"07:45.455 ","End":"07:53.040","Text":"That\u0027s the transformation of R^4 to or R^3 that we are looking for We\u0027re done."}],"ID":10213},{"Watched":false,"Name":"Exercise 9","Duration":"1m 38s","ChapterTopicVideoID":9648,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.810","Text":"In this exercise, we are given a linear transformation T from V to U."},{"Start":"00:06.810 ","End":"00:08.730","Text":"This is a more theoretical exercise,"},{"Start":"00:08.730 ","End":"00:11.340","Text":"but it\u0027s actually quite easy."},{"Start":"00:11.340 ","End":"00:15.180","Text":"We have to prove that if the dimension of"},{"Start":"00:15.180 ","End":"00:19.860","Text":"the image of T is the same as the dimension of the kernel of T, i.e."},{"Start":"00:19.860 ","End":"00:23.700","Text":"the rank is equal to the nullity of T,"},{"Start":"00:23.700 ","End":"00:29.895","Text":"then the dimension of the space V is an even number."},{"Start":"00:29.895 ","End":"00:34.809","Text":"What we\u0027re going to use here is the rank nullity theorem."},{"Start":"00:34.809 ","End":"00:37.460","Text":"The rank nullity theorem says that,"},{"Start":"00:37.460 ","End":"00:42.640","Text":"whenever we have a linear transformation T from a space V to a space U,"},{"Start":"00:42.640 ","End":"00:44.535","Text":"then the dimension of V,"},{"Start":"00:44.535 ","End":"00:50.740","Text":"the source space is equal to the dimension of the image plus the dimension of the kernel."},{"Start":"00:50.740 ","End":"00:53.925","Text":"You can put brackets if you want or without."},{"Start":"00:53.925 ","End":"00:58.130","Text":"In other words, it\u0027s equal to the rank plus the nullity."},{"Start":"00:58.130 ","End":"01:00.350","Text":"Now, in our case,"},{"Start":"01:00.350 ","End":"01:04.370","Text":"we\u0027re given that this is equal to this as written here,"},{"Start":"01:04.370 ","End":"01:05.615","Text":"so if they\u0027re equal,"},{"Start":"01:05.615 ","End":"01:12.845","Text":"we can give them both the same letter n. I say that this is equal to this,"},{"Start":"01:12.845 ","End":"01:14.270","Text":"or rather from here,"},{"Start":"01:14.270 ","End":"01:19.565","Text":"and we\u0027ll call it n. Now if I plug n here and here,"},{"Start":"01:19.565 ","End":"01:23.870","Text":"then we get the dimension of V is n plus n,"},{"Start":"01:23.870 ","End":"01:25.965","Text":"and n plus n is 2n,"},{"Start":"01:25.965 ","End":"01:27.990","Text":"and I guess I didn\u0027t say it,"},{"Start":"01:27.990 ","End":"01:32.300","Text":"but it\u0027s clear that this means that this is an even number."},{"Start":"01:32.300 ","End":"01:36.380","Text":"The dimension of V is even because 2n is always even."},{"Start":"01:36.380 ","End":"01:39.000","Text":"That\u0027s it, we\u0027re done."}],"ID":10214},{"Watched":false,"Name":"Exercise 10","Duration":"4m 2s","ChapterTopicVideoID":9649,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:10.095","Text":"In this exercise, we have a linear transformation T from R^4 to R^3,"},{"Start":"00:10.095 ","End":"00:12.000","Text":"and the question is,"},{"Start":"00:12.000 ","End":"00:18.330","Text":"is it possible that this transformation is 1-to-1?"},{"Start":"00:18.330 ","End":"00:21.960","Text":"I\u0027ll tell you now already that the answer is no,"},{"Start":"00:21.960 ","End":"00:23.595","Text":"it is not possible,"},{"Start":"00:23.595 ","End":"00:28.365","Text":"and we\u0027re going to use a proposition and a theorem to help us here,"},{"Start":"00:28.365 ","End":"00:30.970","Text":"ones that we\u0027ve seen before."},{"Start":"00:31.130 ","End":"00:37.410","Text":"One proposition that we\u0027re familiar with is that a linear transformation T is"},{"Start":"00:37.410 ","End":"00:43.445","Text":"one-to-1 if and only if its kernel is the trivial subspace,"},{"Start":"00:43.445 ","End":"00:47.285","Text":"in other words, it contains just the 0 vector."},{"Start":"00:47.285 ","End":"00:49.325","Text":"This will help us here."},{"Start":"00:49.325 ","End":"00:53.060","Text":"Also, let\u0027s summarize some other results."},{"Start":"00:53.060 ","End":"00:58.700","Text":"In general, if you have a linear transformation T from V to U,"},{"Start":"00:58.700 ","End":"01:03.700","Text":"we mentioned that the kernel of T is a subspace of V. I"},{"Start":"01:03.700 ","End":"01:05.240","Text":"wrote the mathematical symbol for"},{"Start":"01:05.240 ","End":"01:08.615","Text":"subset because I don\u0027t know that the symbol for subspace,"},{"Start":"01:08.615 ","End":"01:13.620","Text":"and the image of T is a subspace of U,"},{"Start":"01:13.620 ","End":"01:18.380","Text":"and the rank-nullity theorem says that the dimension of V,"},{"Start":"01:18.380 ","End":"01:24.175","Text":"the source space, the domain, if you like,"},{"Start":"01:24.175 ","End":"01:28.670","Text":"the dimension of it is equal to the dimension of the kernel,"},{"Start":"01:28.670 ","End":"01:31.205","Text":"plus the dimension of the image,"},{"Start":"01:31.205 ","End":"01:37.390","Text":"also called the nullity plus the rank."},{"Start":"01:37.390 ","End":"01:39.320","Text":"Armed with all of these,"},{"Start":"01:39.320 ","End":"01:41.815","Text":"we shouldn\u0027t have too much difficulty."},{"Start":"01:41.815 ","End":"01:48.360","Text":"We\u0027ll use the mathematical method of proof by contradiction."},{"Start":"01:48.850 ","End":"01:53.135","Text":"We\u0027ll suppose that T is 1-to-1,"},{"Start":"01:53.135 ","End":"01:56.000","Text":"and if we get into a contradiction,"},{"Start":"01:56.000 ","End":"02:01.700","Text":"that will mean that this supposition is false and therefore T is not 1-to-1."},{"Start":"02:01.700 ","End":"02:06.020","Text":"Well meanwhile, let\u0027s suppose that it is and see how far that gets us."},{"Start":"02:06.020 ","End":"02:07.865","Text":"If it is 1-to-1,"},{"Start":"02:07.865 ","End":"02:10.040","Text":"then by the proposition,"},{"Start":"02:10.040 ","End":"02:17.645","Text":"the kernel is just the trivial subspace,"},{"Start":"02:17.645 ","End":"02:20.600","Text":"which means that its dimension is 0."},{"Start":"02:20.600 ","End":"02:26.580","Text":"The dimension of the space with just 0 in it is 0."},{"Start":"02:26.660 ","End":"02:31.910","Text":"Next, let\u0027s take a transformation from R^4 to R^3 and"},{"Start":"02:31.910 ","End":"02:38.060","Text":"apply the rank-nullity theorem to it,"},{"Start":"02:38.060 ","End":"02:41.790","Text":"which is essentially this line here."},{"Start":"02:41.860 ","End":"02:45.830","Text":"The V here is R^4."},{"Start":"02:45.830 ","End":"02:49.730","Text":"The dimension of R^4 is the dimension of the kernel"},{"Start":"02:49.730 ","End":"02:53.525","Text":"of this T plus the dimension of the image of this T. Now,"},{"Start":"02:53.525 ","End":"02:58.460","Text":"we already know some of the things because we\u0027ve already concluded that the dimension"},{"Start":"02:58.460 ","End":"03:04.135","Text":"of the kernel is 0 and we know the dimension of R^4 is 4."},{"Start":"03:04.135 ","End":"03:08.255","Text":"We have an equation that 4 is equal to"},{"Start":"03:08.255 ","End":"03:14.005","Text":"0 plus the dimension of the image of T. This is of course the rank of T,"},{"Start":"03:14.005 ","End":"03:19.190","Text":"which means that the rank of T or dimension of the image is 4."},{"Start":"03:19.190 ","End":"03:26.950","Text":"Now, this means that we have a subspace of R^3 with dimension 4."},{"Start":"03:26.950 ","End":"03:32.220","Text":"That is not possible because when we take a subspace,"},{"Start":"03:32.220 ","End":"03:34.590","Text":"the dimension can only decrease."},{"Start":"03:34.590 ","End":"03:37.005","Text":"The dimension of R^3 is 3,"},{"Start":"03:37.005 ","End":"03:38.870","Text":"so if we have a subspace,"},{"Start":"03:38.870 ","End":"03:41.735","Text":"it has to be less than or equal to 3."},{"Start":"03:41.735 ","End":"03:45.320","Text":"It\u0027s a contradiction because essentially what we\u0027ve proved,"},{"Start":"03:45.320 ","End":"03:47.750","Text":"is that 4 is less than or equal to 3,"},{"Start":"03:47.750 ","End":"03:50.015","Text":"which is certainly not true,"},{"Start":"03:50.015 ","End":"03:55.445","Text":"and this contradiction stemmed from the supposition that T is 1-to-1,"},{"Start":"03:55.445 ","End":"03:58.115","Text":"and so the answer has to be no,"},{"Start":"03:58.115 ","End":"04:00.470","Text":"T is not 1-to-1,"},{"Start":"04:00.470 ","End":"04:03.300","Text":"and we are done."}],"ID":10215},{"Watched":false,"Name":"Exercise 11","Duration":"12m 37s","ChapterTopicVideoID":27885,"CourseChapterTopicPlaylistID":7315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.220","Text":"Hi."},{"Start":"00:02.220 ","End":"00:06.134","Text":"We\u0027re returning back to our linear algebra module."},{"Start":"00:06.134 ","End":"00:11.550","Text":"Today, we\u0027re going to be looking at the following linear transformation T,"},{"Start":"00:11.550 ","End":"00:16.890","Text":"which as you can see maps a 2 by 3 matrix with"},{"Start":"00:16.890 ","End":"00:23.160","Text":"integer inputs to a 3 by 2 matrix with integer inputs."},{"Start":"00:23.160 ","End":"00:27.180","Text":"Now we need to find the kernel of T and the image of"},{"Start":"00:27.180 ","End":"00:32.350","Text":"T and then determine whether T is injective or surjective."},{"Start":"00:32.350 ","End":"00:40.705","Text":"It might be useful to quickly write down the definitions of the kernel and the image."},{"Start":"00:40.705 ","End":"00:47.390","Text":"We have our definition so let\u0027s just translate this mathematics here."},{"Start":"00:47.390 ","End":"00:55.080","Text":"We have the kernel of the transformation T is the set of matrices in the 2 by"},{"Start":"00:55.080 ","End":"00:58.880","Text":"3 matrix dimension such that when you apply"},{"Start":"00:58.880 ","End":"01:04.470","Text":"the transformation T to this matrix you get the matrix 0."},{"Start":"01:04.470 ","End":"01:07.890","Text":"For clarification, what this means is,"},{"Start":"01:07.890 ","End":"01:11.620","Text":"once we apply the transformation then we should be"},{"Start":"01:11.620 ","End":"01:17.810","Text":"expecting the matrix where all entries are 0."},{"Start":"01:18.860 ","End":"01:25.150","Text":"Now, the image of T are all the matrices you get when"},{"Start":"01:25.150 ","End":"01:30.909","Text":"you apply the transformation to the matrix that\u0027s in the original sets,"},{"Start":"01:30.909 ","End":"01:33.660","Text":"the 2 by 3 matrices."},{"Start":"01:33.660 ","End":"01:41.185","Text":"One way to imagine the image is to look at it in a more simple example."},{"Start":"01:41.185 ","End":"01:47.825","Text":"Say we had f(x) is equal to x^2,"},{"Start":"01:47.825 ","End":"01:52.655","Text":"then the image would be basically what the plot looks like."},{"Start":"01:52.655 ","End":"01:56.810","Text":"What we get in our function after we put in all values of x,"},{"Start":"01:56.810 ","End":"02:00.305","Text":"so everything that we can expect to get out."},{"Start":"02:00.305 ","End":"02:05.725","Text":"Let\u0027s go about finding the kernel and the image of this transformation."},{"Start":"02:05.725 ","End":"02:07.915","Text":"To find the kernel,"},{"Start":"02:07.915 ","End":"02:12.500","Text":"we require a matrix such that when we"},{"Start":"02:12.500 ","End":"02:18.425","Text":"apply the transformation to it then we get 0 as our output."},{"Start":"02:18.425 ","End":"02:22.775","Text":"If we recall what our transformation T was,"},{"Start":"02:22.775 ","End":"02:25.550","Text":"which we\u0027ll just quickly write down now,"},{"Start":"02:25.550 ","End":"02:32.550","Text":"then we need to find a matrix such that we get all 0 entries after applying our"},{"Start":"02:32.550 ","End":"02:40.805","Text":"T. Now it\u0027s quite easy to see that 1 matrix that will work is if we just have the matrix."},{"Start":"02:40.805 ","End":"02:47.112","Text":"Let\u0027s say take M is equal to 0,"},{"Start":"02:47.112 ","End":"02:48.765","Text":"0, 0,"},{"Start":"02:48.765 ","End":"02:51.435","Text":"0, 0, 0,"},{"Start":"02:51.435 ","End":"02:53.445","Text":"or in other words,"},{"Start":"02:53.445 ","End":"02:57.810","Text":"A_i is equal to 0,"},{"Start":"02:57.810 ","End":"03:02.625","Text":"for i is equal from 1,"},{"Start":"03:02.625 ","End":"03:07.285","Text":"2, and 3 so that\u0027s just the more simple way of expressing it."},{"Start":"03:07.285 ","End":"03:10.195","Text":"Then what happens if we apply T to this?"},{"Start":"03:10.195 ","End":"03:14.862","Text":"Then we\u0027re going to get T applied to 0,"},{"Start":"03:14.862 ","End":"03:17.298","Text":"0, 0, 0,"},{"Start":"03:17.298 ","End":"03:18.720","Text":"0, 0,"},{"Start":"03:18.720 ","End":"03:22.314","Text":"which is equal to just 0,"},{"Start":"03:22.314 ","End":"03:23.340","Text":"0, 0,"},{"Start":"03:23.340 ","End":"03:25.305","Text":"0, 0, 0."},{"Start":"03:25.305 ","End":"03:29.020","Text":"I.e we have found a matrix that belongs"},{"Start":"03:29.020 ","End":"03:32.920","Text":"to the kernel of T. Now we have to ask the question,"},{"Start":"03:32.920 ","End":"03:37.720","Text":"are there any other matrices that when we apply this transformation"},{"Start":"03:37.720 ","End":"03:43.315","Text":"to it do we get just the matrix with zeros in every entry?"},{"Start":"03:43.315 ","End":"03:50.075","Text":"Now, you can convince yourself that no there aren\u0027t because let\u0027s say we had"},{"Start":"03:50.075 ","End":"03:57.685","Text":"another matrix where maybe for example we had the entry was maybe 2,"},{"Start":"03:57.685 ","End":"04:00.155","Text":"then what we\u0027re going to get is,"},{"Start":"04:00.155 ","End":"04:05.240","Text":"well that 2 here which was in A_12 is going to map over"},{"Start":"04:05.240 ","End":"04:10.940","Text":"to here and then you can see we no longer have the zero matrix as we required."},{"Start":"04:10.940 ","End":"04:18.480","Text":"The only matrix that belongs to the kernel of T is the matrix that contains all zeros."},{"Start":"04:18.480 ","End":"04:22.020","Text":"Let\u0027s write that up a bit more neatly."},{"Start":"04:22.020 ","End":"04:26.090","Text":"We have our kernel and that is basically"},{"Start":"04:26.090 ","End":"04:31.835","Text":"just the matrix which contains all 0 entries and remember,"},{"Start":"04:31.835 ","End":"04:35.800","Text":"we were looking at matrices with only integer inputs."},{"Start":"04:35.800 ","End":"04:40.520","Text":"That\u0027s the only matrix that when we apply the transformation T will we"},{"Start":"04:40.520 ","End":"04:45.735","Text":"get back the 0 matrix in this required form."},{"Start":"04:45.735 ","End":"04:49.730","Text":"Now, let\u0027s look at the image of T and remember,"},{"Start":"04:49.730 ","End":"04:55.030","Text":"the image of T as we defined it which is everything that can"},{"Start":"04:55.030 ","End":"05:02.440","Text":"occur once we map any matrix following the required linear transformation."},{"Start":"05:02.440 ","End":"05:05.230","Text":"Let\u0027s start. Well, we already know that"},{"Start":"05:05.230 ","End":"05:09.355","Text":"the matrix containing all zeros belongs in the image."},{"Start":"05:09.355 ","End":"05:11.035","Text":"But what else do we get?"},{"Start":"05:11.035 ","End":"05:17.245","Text":"Well, if you think about it a little bit then it\u0027s clear to see that if we just change"},{"Start":"05:17.245 ","End":"05:20.830","Text":"our entries in the original matrix then that will"},{"Start":"05:20.830 ","End":"05:24.910","Text":"follow after a mapping into the mapped matrix."},{"Start":"05:24.910 ","End":"05:29.020","Text":"What that means is from observation we can already"},{"Start":"05:29.020 ","End":"05:34.115","Text":"see that all matrices that are in this form,"},{"Start":"05:34.115 ","End":"05:38.070","Text":"3, 2, so 3 rows,"},{"Start":"05:38.070 ","End":"05:47.130","Text":"2 columns can be created following a sufficient mapping from a matrix in the 2 by 3 form."},{"Start":"05:47.130 ","End":"05:49.100","Text":"Let\u0027s just prove this."},{"Start":"05:49.100 ","End":"05:55.524","Text":"Let\u0027s just say we have a matrix M and we\u0027ll call this a,"},{"Start":"05:55.524 ","End":"05:57.315","Text":"b, c,"},{"Start":"05:57.315 ","End":"05:58.956","Text":"d, e,"},{"Start":"05:58.956 ","End":"06:04.640","Text":"f, and then if we apply our transformation to M,"},{"Start":"06:04.640 ","End":"06:09.260","Text":"so T(M) then what are we going to get?"},{"Start":"06:09.260 ","End":"06:12.965","Text":"Well, we just follow the form that we have here."},{"Start":"06:12.965 ","End":"06:19.980","Text":"The things in the top row will go into the first column so then we\u0027ll just have a,"},{"Start":"06:19.980 ","End":"06:22.770","Text":"b, c and then we\u0027ll have d,"},{"Start":"06:22.770 ","End":"06:24.750","Text":"e, f here."},{"Start":"06:24.750 ","End":"06:27.015","Text":"Remember a, b,"},{"Start":"06:27.015 ","End":"06:28.800","Text":"c, d, e,"},{"Start":"06:28.800 ","End":"06:33.435","Text":"and f these are just real numbers."},{"Start":"06:33.435 ","End":"06:38.960","Text":"We can basically make this matrix whatever we want depending on our choice of a,"},{"Start":"06:38.960 ","End":"06:40.640","Text":"b, c, d, e,"},{"Start":"06:40.640 ","End":"06:43.665","Text":"and f. What does that mean?"},{"Start":"06:43.665 ","End":"06:51.175","Text":"Well, that means that we can define the kernel of the image now so let\u0027s write this up."},{"Start":"06:51.175 ","End":"06:55.860","Text":"We have defined our image of T now as"},{"Start":"06:55.860 ","End":"07:00.100","Text":"all matrices that belong to the set of matrices with"},{"Start":"07:00.100 ","End":"07:08.560","Text":"3 rows and 2 columns and here we\u0027ve said that the entries can be any integer values."},{"Start":"07:08.560 ","End":"07:15.120","Text":"Because remember, we were mapping to the matrices with integer inputs."},{"Start":"07:15.120 ","End":"07:21.280","Text":"We\u0027ve now defined our image and our kernel so let\u0027s go on to determining if"},{"Start":"07:21.280 ","End":"07:27.600","Text":"this transformation is injective and surjective or both."},{"Start":"07:27.600 ","End":"07:33.235","Text":"It will be useful again to write down our definitions"},{"Start":"07:33.235 ","End":"07:39.410","Text":"of injectivity and surjectivity so let\u0027s do that quickly."},{"Start":"07:39.410 ","End":"07:46.130","Text":"Here, we have our definitions of injectivity and surjectivity."},{"Start":"07:46.130 ","End":"07:50.030","Text":"It might look a bit daunting all this mathematical notation"},{"Start":"07:50.030 ","End":"07:54.080","Text":"so let\u0027s just go through it a little bit slowly."},{"Start":"07:54.080 ","End":"08:01.460","Text":"The condition for injectivity says if we apply our transformation to"},{"Start":"08:01.460 ","End":"08:08.715","Text":"a matrix say M_1 and we get M star which is a matrix in the image,"},{"Start":"08:08.715 ","End":"08:14.580","Text":"and then if we apply the same transformation to another matrix M_2 which gives us"},{"Start":"08:14.580 ","End":"08:20.685","Text":"M star then that would imply that M_1 is equal to M_2."},{"Start":"08:20.685 ","End":"08:28.415","Text":"Another way of saying this is there exists only 1 matrix that will give us M star."},{"Start":"08:28.415 ","End":"08:32.615","Text":"There are not 2 distinct matrices that after applying"},{"Start":"08:32.615 ","End":"08:37.745","Text":"our transformation will give us M star as the outputs."},{"Start":"08:37.745 ","End":"08:40.700","Text":"One way you might like to think of this and this is"},{"Start":"08:40.700 ","End":"08:44.077","Text":"another way people call injective things,"},{"Start":"08:44.077 ","End":"08:50.530","Text":"you say there is a 1 to 1 correspondence."},{"Start":"08:50.530 ","End":"08:54.620","Text":"Because basically what we mean here is that there\u0027s"},{"Start":"08:54.620 ","End":"08:56.990","Text":"only 1 matrix that after we apply"},{"Start":"08:56.990 ","End":"09:00.905","Text":"the transformation that will give us this other matrix M star,"},{"Start":"09:00.905 ","End":"09:04.085","Text":"there\u0027s not another matrix that will allow us to do this."},{"Start":"09:04.085 ","End":"09:07.935","Text":"Now let\u0027s consider the surjective case."},{"Start":"09:07.935 ","End":"09:16.080","Text":"This says for all matrices M star in the 3 by 2 matrix sets which"},{"Start":"09:16.080 ","End":"09:19.565","Text":"remember is our image then there exists"},{"Start":"09:19.565 ","End":"09:24.900","Text":"a matrix M in our domain so the set of matrices 2,"},{"Start":"09:24.900 ","End":"09:30.985","Text":"3 such that after applying the transformation M we get M star."},{"Start":"09:30.985 ","End":"09:34.820","Text":"An easier way to think of this is that if you consider"},{"Start":"09:34.820 ","End":"09:39.060","Text":"all the possible matrices that exist in the codomain so where we\u0027re"},{"Start":"09:39.060 ","End":"09:43.130","Text":"mapping to then there\u0027s a matrix in the domain that when we"},{"Start":"09:43.130 ","End":"09:47.990","Text":"apply the transformation to will give us that matrix in the image."},{"Start":"09:47.990 ","End":"09:51.820","Text":"Let\u0027s see if this is injective."},{"Start":"09:51.820 ","End":"09:55.520","Text":"We can tell already just by considering"},{"Start":"09:55.520 ","End":"09:59.810","Text":"the transformation that this is an injective case."},{"Start":"09:59.810 ","End":"10:07.220","Text":"Because if we change any of these entries in the transformation then that\u0027s going to"},{"Start":"10:07.220 ","End":"10:11.735","Text":"upset the column and row entries in our outputs"},{"Start":"10:11.735 ","End":"10:16.675","Text":"and therefore it won\u0027t map to the same thing that we have before."},{"Start":"10:16.675 ","End":"10:20.795","Text":"Let\u0027s just demonstrate that with a concrete example."},{"Start":"10:20.795 ","End":"10:24.215","Text":"Now, you can see here that if we change"},{"Start":"10:24.215 ","End":"10:28.940","Text":"only 2 entries in the original matrix then the output that we"},{"Start":"10:28.940 ","End":"10:32.600","Text":"get is different to what we would"},{"Start":"10:32.600 ","End":"10:38.030","Text":"have if we took the original matrix with the unaltered entries,"},{"Start":"10:38.030 ","End":"10:41.035","Text":"which remember was equal to,"},{"Start":"10:41.035 ","End":"10:45.395","Text":"it was just A_11, A_12,"},{"Start":"10:45.395 ","End":"10:49.325","Text":"A_13 and then A_21,"},{"Start":"10:49.325 ","End":"10:53.940","Text":"A_22, and A_23."},{"Start":"10:53.940 ","End":"10:57.770","Text":"You can see that there is in fact only 1 matrix that"},{"Start":"10:57.770 ","End":"11:01.380","Text":"when we apply our transformation T_2 we will get"},{"Start":"11:01.380 ","End":"11:10.800","Text":"this distinct output and therefore we can conclude that the matrix is in fact injective."},{"Start":"11:10.800 ","End":"11:16.935","Text":"We\u0027ve established injectivity so we are happy with that now."},{"Start":"11:16.935 ","End":"11:20.840","Text":"Now, we just need to consider whether this matrix or"},{"Start":"11:20.840 ","End":"11:26.045","Text":"this transformation is surjective and then we have completed the question."},{"Start":"11:26.045 ","End":"11:29.045","Text":"Remember what we said surjectivity meant."},{"Start":"11:29.045 ","End":"11:34.365","Text":"It means that everything in the co-domain gets mapped to."},{"Start":"11:34.365 ","End":"11:43.700","Text":"Again, this is quite easy to see that if we had a particular matrix say M star equals a,"},{"Start":"11:43.700 ","End":"11:44.865","Text":"b, d,"},{"Start":"11:44.865 ","End":"11:46.290","Text":"e, c, f,"},{"Start":"11:46.290 ","End":"11:52.585","Text":"in our set of matrices with 3 rows and 2 columns then"},{"Start":"11:52.585 ","End":"11:59.450","Text":"there is a matrix in our original domain that we can apply T to,"},{"Start":"11:59.450 ","End":"12:00.905","Text":"to give us this."},{"Start":"12:00.905 ","End":"12:02.300","Text":"What would that be? Well,"},{"Start":"12:02.300 ","End":"12:03.785","Text":"let\u0027s think about it."},{"Start":"12:03.785 ","End":"12:08.883","Text":"If we do the transformation T applied to a,"},{"Start":"12:08.883 ","End":"12:11.334","Text":"b, c, d, e,"},{"Start":"12:11.334 ","End":"12:19.815","Text":"f, then following are defined transformation then this gives us exactly what we want."},{"Start":"12:19.815 ","End":"12:21.825","Text":"So a, b, c,"},{"Start":"12:21.825 ","End":"12:23.565","Text":"d, e,"},{"Start":"12:23.565 ","End":"12:28.280","Text":"f. We have proven that everything in the co-domain gets mapped"},{"Start":"12:28.280 ","End":"12:33.350","Text":"to and we have established surjectivity as well."},{"Start":"12:33.350 ","End":"12:35.360","Text":"I hope that video was helpful,"},{"Start":"12:35.360 ","End":"12:37.650","Text":"thank you very much."}],"ID":29068}],"Thumbnail":null,"ID":7315},{"Name":"Isomorphism and Inverse","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1 Part a","Duration":"5m 21s","ChapterTopicVideoID":9621,"CourseChapterTopicPlaylistID":7316,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9621.jpeg","UploadDate":"2017-07-26T08:44:02.1970000","DurationForVideoObject":"PT5M21S","Description":null,"MetaTitle":"Exercise 1 Part a - Isomorphism and Inverse: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Isomorphism and Inverse practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/linear-algebra/linear-transformations/isomorphism-and-inverse/vid10216","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.050","Text":"In this exercise, we have a linear transformation T from R^3-R^3,"},{"Start":"00:07.050 ","End":"00:10.500","Text":"and it\u0027s defined with the following formula,"},{"Start":"00:10.500 ","End":"00:13.665","Text":"and we have a few true or false,"},{"Start":"00:13.665 ","End":"00:19.410","Text":"yes or no questions about T, is it 1-1?"},{"Start":"00:19.410 ","End":"00:21.225","Text":"Is it onto?"},{"Start":"00:21.225 ","End":"00:24.150","Text":"Is it an isomorphism?"},{"Start":"00:24.150 ","End":"00:26.760","Text":"Does it have an inverse?"},{"Start":"00:26.760 ","End":"00:28.680","Text":"In the last case, if it has an inverse,"},{"Start":"00:28.680 ","End":"00:31.785","Text":"we have to actually find it computed."},{"Start":"00:31.785 ","End":"00:36.550","Text":"Let\u0027s start with the first 1, is T 1-1?"},{"Start":"00:36.770 ","End":"00:42.500","Text":"For the 1-1, we\u0027ll use the proposition that T is 1-1"},{"Start":"00:42.500 ","End":"00:51.030","Text":"if and only if the kernel of T contains just the 0 of R^3."},{"Start":"00:51.110 ","End":"00:53.599","Text":"Let\u0027s find the kernel."},{"Start":"00:53.599 ","End":"00:56.120","Text":"The kernel is everything that T sends to 0,"},{"Start":"00:56.120 ","End":"00:58.370","Text":"so T of x, y, z is this,"},{"Start":"00:58.370 ","End":"01:04.320","Text":"this has to be 0, but 0 of R^3 is 0, 0, 0."},{"Start":"01:04.320 ","End":"01:09.600","Text":"That gives us 3 equations and 3 unknowns, x, y, z"},{"Start":"01:09.600 ","End":"01:13.680","Text":"and that is this system of linear equations."},{"Start":"01:13.680 ","End":"01:15.030","Text":"Well, it\u0027s homogeneous,"},{"Start":"01:15.030 ","End":"01:17.385","Text":"3 equations and 3 unknowns."},{"Start":"01:17.385 ","End":"01:20.920","Text":"We\u0027ll solve it with matrices."},{"Start":"01:20.920 ","End":"01:23.195","Text":"We take the coefficients,"},{"Start":"01:23.195 ","End":"01:26.570","Text":"we just need the left-hand side because it\u0027s homogeneous,"},{"Start":"01:26.570 ","End":"01:28.895","Text":"we don\u0027t need the augmented matrix,"},{"Start":"01:28.895 ","End":"01:33.350","Text":"1 minus 1, 1, 0, 1, 1 and so on."},{"Start":"01:33.350 ","End":"01:36.935","Text":"We want to bring this to row echelon form."},{"Start":"01:36.935 ","End":"01:44.390","Text":"I guess first thing to do is to add the first row to the third row"},{"Start":"01:44.390 ","End":"01:47.550","Text":"and that will bring us to this."},{"Start":"01:47.550 ","End":"01:49.370","Text":"Now, we want 0 here,"},{"Start":"01:49.370 ","End":"01:53.710","Text":"so we\u0027ll add the second row to the third row."},{"Start":"01:53.710 ","End":"01:56.030","Text":"That will give us this,"},{"Start":"01:56.030 ","End":"01:58.670","Text":"which is already echelon form,"},{"Start":"01:58.670 ","End":"02:01.310","Text":"but we can divide the last row by 3,"},{"Start":"02:01.310 ","End":"02:04.535","Text":"and make the numbers nicer."},{"Start":"02:04.535 ","End":"02:13.000","Text":"Continuing now, from here back to system of linear equations."},{"Start":"02:13.000 ","End":"02:18.355","Text":"Actually, because this has no 0 rows,"},{"Start":"02:18.355 ","End":"02:23.490","Text":"the only solution is going to be everything is 0, x, y, and z are 0"},{"Start":"02:23.490 ","End":"02:26.005","Text":"but you can also see this by back substitution."},{"Start":"02:26.005 ","End":"02:27.850","Text":"Z is 0, plug it in here,"},{"Start":"02:27.850 ","End":"02:29.470","Text":"it gives y equals 0."},{"Start":"02:29.470 ","End":"02:33.045","Text":"Z and y are 0, gives x is 0."},{"Start":"02:33.045 ","End":"02:36.755","Text":"As I said, x, y, and z are all 0,"},{"Start":"02:36.755 ","End":"02:46.610","Text":"so the kernel just contains the vector 0, 0, 0 and this is the 0 element of R^3."},{"Start":"02:47.150 ","End":"02:51.220","Text":"Thus we said if the kernel just contains the 0,"},{"Start":"02:51.220 ","End":"02:54.974","Text":"then the transformation T is 1-1."},{"Start":"02:54.974 ","End":"02:56.845","Text":"That\u0027s the first part."},{"Start":"02:56.845 ","End":"03:02.630","Text":"The next part is to decide whether or not T is onto."},{"Start":"03:02.630 ","End":"03:06.100","Text":"For this, we\u0027ll use the proposition that T is onto"},{"Start":"03:06.100 ","End":"03:11.410","Text":"if and only if the dimension of the image of T is the dimension of U,"},{"Start":"03:11.410 ","End":"03:13.255","Text":"I guess it didn\u0027t say what U is,"},{"Start":"03:13.255 ","End":"03:18.180","Text":"but we always have T going from V to U."},{"Start":"03:18.180 ","End":"03:21.165","Text":"U, in this case, is R^3."},{"Start":"03:21.165 ","End":"03:25.980","Text":"Well, V and U are both R^3."},{"Start":"03:25.980 ","End":"03:29.610","Text":"Dimension of R^3 is 3."},{"Start":"03:29.610 ","End":"03:34.070","Text":"This proposition in our case says that T is onto"},{"Start":"03:34.070 ","End":"03:38.935","Text":"if and only if the dimension of its image is 3."},{"Start":"03:38.935 ","End":"03:41.550","Text":"Now, we\u0027ll use the rank-nullity theorem."},{"Start":"03:41.550 ","End":"03:45.185","Text":"Remember that just moments ago we showed"},{"Start":"03:45.185 ","End":"03:51.965","Text":"that T is 1-1 by showing that the kernel of T is the 0 subspace,"},{"Start":"03:51.965 ","End":"03:54.155","Text":"and so has dimension 0,"},{"Start":"03:54.155 ","End":"04:02.710","Text":"so the rank-nullity theorem becomes at the dimension of the image of T plus 0 equals 3."},{"Start":"04:02.780 ","End":"04:06.935","Text":"That means that the dimension of the image is 3 minus 0."},{"Start":"04:06.935 ","End":"04:10.340","Text":"This is 3, and that\u0027s what we wanted here."},{"Start":"04:10.340 ","End":"04:12.560","Text":"That was the condition for T to be onto,"},{"Start":"04:12.560 ","End":"04:13.790","Text":"so the answer is yes,"},{"Start":"04:13.790 ","End":"04:16.820","Text":"T is onto also."},{"Start":"04:16.820 ","End":"04:22.865","Text":"The next question, true or false T is an isomorphism."},{"Start":"04:22.865 ","End":"04:32.240","Text":"The answer to that is true because isomorphism is by definition, 1-1 and onto."},{"Start":"04:32.240 ","End":"04:37.550","Text":"Now, we\u0027ve already shown in the first 2 sections that 1-1"},{"Start":"04:37.550 ","End":"04:41.890","Text":"and onto so T is an isomorphism."},{"Start":"04:41.890 ","End":"04:45.690","Text":"The last question is T invertible?"},{"Start":"04:45.690 ","End":"04:48.080","Text":"If it is, we have to find the inverse."},{"Start":"04:48.080 ","End":"04:53.480","Text":"The answer is yes, T is invertible because if it\u0027s an isomorphism,"},{"Start":"04:53.480 ","End":"04:57.285","Text":"then it\u0027s always invertible and vice versa."},{"Start":"04:57.285 ","End":"04:59.475","Text":"It has an inverse."},{"Start":"04:59.475 ","End":"05:01.250","Text":"Invertible means has an inverse,"},{"Start":"05:01.250 ","End":"05:05.970","Text":"and we\u0027ll call the inverse T to the power of minus 1,"},{"Start":"05:05.970 ","End":"05:07.800","Text":"that\u0027s the symbol for inverse."},{"Start":"05:07.800 ","End":"05:11.510","Text":"We have to not just say that it exists,"},{"Start":"05:11.510 ","End":"05:15.600","Text":"we have to actually compute it or find it."},{"Start":"05:15.730 ","End":"05:17.870","Text":"Let\u0027s take a break now,"},{"Start":"05:17.870 ","End":"05:21.900","Text":"and we\u0027ll compute T to the minus 1 after the break."}],"ID":10216},{"Watched":false,"Name":"Exercise 1 Part b","Duration":"5m 52s","ChapterTopicVideoID":9622,"CourseChapterTopicPlaylistID":7316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.535","Text":"Back from the break."},{"Start":"00:02.535 ","End":"00:08.340","Text":"We were just about to find the inverse of the transformation T."},{"Start":"00:08.340 ","End":"00:11.520","Text":"What does it mean to be an inverse?"},{"Start":"00:11.520 ","End":"00:15.270","Text":"It means that if T takes x, y, z,"},{"Start":"00:15.270 ","End":"00:18.900","Text":"the vector, to the vector r, s, t,"},{"Start":"00:18.900 ","End":"00:21.360","Text":"remember we are from R^3 to R^3,"},{"Start":"00:21.360 ","End":"00:24.510","Text":"then the inverse will take us back."},{"Start":"00:24.510 ","End":"00:31.830","Text":"It will take in r, s, t and give us back our x, y, z."},{"Start":"00:31.830 ","End":"00:36.900","Text":"The formula for T was given as this."},{"Start":"00:36.900 ","End":"00:39.720","Text":"We assign this now to r, s, t."},{"Start":"00:39.720 ","End":"00:42.800","Text":"Our task is to compute the inverse."},{"Start":"00:42.800 ","End":"00:48.110","Text":"In other words, we want to compute x, y, z in terms of r, s, t."},{"Start":"00:48.110 ","End":"00:52.620","Text":"We have 3 equations and 3 unknowns,"},{"Start":"00:52.620 ","End":"00:55.125","Text":"x plus y plus z is r, and so on."},{"Start":"00:55.125 ","End":"00:57.810","Text":"The r, s, t are given,"},{"Start":"00:57.810 ","End":"00:59.235","Text":"they are like constants."},{"Start":"00:59.235 ","End":"01:03.810","Text":"We have to find x, y, and z in terms of them."},{"Start":"01:03.950 ","End":"01:06.465","Text":"We\u0027ll do it with matrices."},{"Start":"01:06.465 ","End":"01:10.505","Text":"We need an augmented matrix because it\u0027s not homogeneous."},{"Start":"01:10.505 ","End":"01:13.700","Text":"Here we put the coefficients of x, y, z"},{"Start":"01:13.700 ","End":"01:16.760","Text":"and on the right column for r, s, t."},{"Start":"01:16.760 ","End":"01:21.785","Text":"We need to do row operations to bring it to echelon form."},{"Start":"01:21.785 ","End":"01:26.690","Text":"First, let\u0027s add the first row to the third row."},{"Start":"01:26.690 ","End":"01:29.315","Text":"That will give us a 0 here."},{"Start":"01:29.315 ","End":"01:36.090","Text":"Don\u0027t forget to do it also on the right-hand column, add the r to the t."},{"Start":"01:36.090 ","End":"01:37.800","Text":"We got here t plus r."},{"Start":"01:37.800 ","End":"01:47.375","Text":"Then I got to here by adding the second row to the third row to get a 0 here,"},{"Start":"01:47.375 ","End":"01:50.720","Text":"and then we get 1 plus 2 is 3, s plus t plus r."},{"Start":"01:50.720 ","End":"01:52.730","Text":"This is what we have now."},{"Start":"01:52.730 ","End":"01:55.330","Text":"This is an echelon form."},{"Start":"01:55.330 ","End":"01:59.720","Text":"Back from here to a system of linear equations,"},{"Start":"01:59.720 ","End":"02:02.750","Text":"it\u0027s going to have a unique solution because"},{"Start":"02:02.750 ","End":"02:09.180","Text":"the matrix here has no 0s in it."},{"Start":"02:09.180 ","End":"02:13.100","Text":"We just have to do what we call backward substitution."},{"Start":"02:13.100 ","End":"02:15.320","Text":"From the last equation,"},{"Start":"02:15.320 ","End":"02:21.750","Text":"we can get z which will be 1/3 of t plus r plus s."},{"Start":"02:21.750 ","End":"02:27.760","Text":"Now we want to plug z into the second equation."},{"Start":"02:27.760 ","End":"02:31.260","Text":"Then we plug in z to the second equation here."},{"Start":"02:31.260 ","End":"02:39.000","Text":"We get y is s minus z which is s minus and this is z."},{"Start":"02:39.000 ","End":"02:43.935","Text":"After simplification, we get this."},{"Start":"02:43.935 ","End":"02:51.505","Text":"Then the first equation which is x equals,"},{"Start":"02:51.505 ","End":"02:53.570","Text":"bring the y and the z to the other side,"},{"Start":"02:53.570 ","End":"02:56.555","Text":"so it\u0027s r plus y minus z."},{"Start":"02:56.555 ","End":"02:59.900","Text":"But we have already y and z here,"},{"Start":"02:59.900 ","End":"03:02.930","Text":"we plug them in and simplify."},{"Start":"03:02.930 ","End":"03:05.360","Text":"This is what we get for x."},{"Start":"03:05.360 ","End":"03:07.880","Text":"We now have x, y, and z."},{"Start":"03:07.880 ","End":"03:12.880","Text":"I just have to tidy this up a bit."},{"Start":"03:13.280 ","End":"03:21.045","Text":"We saw earlier that T inverse of r, s, t is the x, y, and z."},{"Start":"03:21.045 ","End":"03:24.565","Text":"We just copy the x is this,"},{"Start":"03:24.565 ","End":"03:26.245","Text":"that goes here,"},{"Start":"03:26.245 ","End":"03:29.630","Text":"the y from here is now here,"},{"Start":"03:29.630 ","End":"03:33.790","Text":"and the z from here is here."},{"Start":"03:33.790 ","End":"03:36.420","Text":"The variables r, s, and t"},{"Start":"03:36.420 ","End":"03:40.490","Text":"were dummy intermediate variables because"},{"Start":"03:40.490 ","End":"03:44.060","Text":"we didn\u0027t want to use x, y, and z for the computations."},{"Start":"03:44.060 ","End":"03:47.620","Text":"But now that we have a closed formula for T inverse,"},{"Start":"03:47.620 ","End":"03:51.130","Text":"we go back from r, s, t to x, y, z."},{"Start":"03:51.130 ","End":"03:52.970","Text":"They are just dummy variables,"},{"Start":"03:52.970 ","End":"03:56.390","Text":"could be any 3 but we prefer x, y, and z."},{"Start":"03:56.390 ","End":"04:01.250","Text":"Just replace everywhere with x, s with y, and t with z."},{"Start":"04:01.250 ","End":"04:06.400","Text":"This is now the formula for T inverse."},{"Start":"04:06.590 ","End":"04:11.240","Text":"Here was our original T of x, y, z."},{"Start":"04:11.240 ","End":"04:16.810","Text":"The reason I brought it is that we can do some verification."},{"Start":"04:16.810 ","End":"04:23.865","Text":"Let\u0027s test our result on 1, 2, 3 instead of x, y, z."},{"Start":"04:23.865 ","End":"04:28.620","Text":"If we apply T to 1, 2, 3 from this formula,"},{"Start":"04:28.620 ","End":"04:30.930","Text":"we get 2, 5, 2."},{"Start":"04:30.930 ","End":"04:37.680","Text":"What I expect is if I apply T inverse to 2, 5, 2,"},{"Start":"04:37.680 ","End":"04:39.390","Text":"if we didn\u0027t make a mistake,"},{"Start":"04:39.390 ","End":"04:42.760","Text":"we should get back to 1, 2, 3."},{"Start":"04:42.880 ","End":"04:46.415","Text":"T inverse of 2, 5, 2,"},{"Start":"04:46.415 ","End":"04:49.915","Text":"this time using this formula,"},{"Start":"04:49.915 ","End":"04:52.140","Text":"let\u0027s just check 1 of them,"},{"Start":"04:52.140 ","End":"04:53.910","Text":"say the first 1."},{"Start":"04:53.910 ","End":"04:57.615","Text":"We need 1/3 of x plus y minus 2z."},{"Start":"04:57.615 ","End":"05:07.520","Text":"X plus y is 7, 7 minus 2z is 7 minus 4 is 3, 3 over 3 is 1."},{"Start":"05:07.520 ","End":"05:08.510","Text":"Yes."},{"Start":"05:08.510 ","End":"05:16.760","Text":"I\u0027ll leave you to check the other 2 that we get also 2 and 3 here using this."},{"Start":"05:16.760 ","End":"05:18.380","Text":"This formula gives us 2,"},{"Start":"05:18.380 ","End":"05:20.900","Text":"this formula will give us 3."},{"Start":"05:20.900 ","End":"05:25.325","Text":"For this particular sample 1, 2, 3,"},{"Start":"05:25.325 ","End":"05:29.750","Text":"we do get that T inverse of T of 1, 2, 3 is 1, 2, 3"},{"Start":"05:29.750 ","End":"05:33.140","Text":"and it should work for every x, y, z"},{"Start":"05:33.140 ","End":"05:39.950","Text":"but we just took an example just to see that we didn\u0027t go wrong basically."},{"Start":"05:39.950 ","End":"05:47.520","Text":"It\u0027s worthwhile if you have the time to do at least 1 value too, numerically check it."},{"Start":"05:47.590 ","End":"05:49.820","Text":"That was the last section."},{"Start":"05:49.820 ","End":"05:52.200","Text":"We are done."}],"ID":10217},{"Watched":false,"Name":"Exercise 2","Duration":"5m 3s","ChapterTopicVideoID":9623,"CourseChapterTopicPlaylistID":7316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.155","Text":"In this exercise, we have a linear transformation T from R^3 to R^3."},{"Start":"00:07.155 ","End":"00:09.435","Text":"This is its definition."},{"Start":"00:09.435 ","End":"00:15.270","Text":"We have 4, true or false yes or no questions."},{"Start":"00:15.270 ","End":"00:16.860","Text":"Is T is 1 to 1,"},{"Start":"00:16.860 ","End":"00:19.380","Text":"is it onto, is it an isomorphism?"},{"Start":"00:19.380 ","End":"00:21.195","Text":"Does it have an inverse?"},{"Start":"00:21.195 ","End":"00:23.310","Text":"The last case, if it does have an inverse,"},{"Start":"00:23.310 ","End":"00:25.275","Text":"we actually we have to find it."},{"Start":"00:25.275 ","End":"00:31.320","Text":"Let\u0027s start with the first 1 is T 1 to 1."},{"Start":"00:31.320 ","End":"00:36.560","Text":"For this, we\u0027ll remember there is a proposition that in general"},{"Start":"00:36.560 ","End":"00:43.310","Text":"a linear transformation is 1 to 1 if and only if its kernel contains just the 0 vector."},{"Start":"00:43.310 ","End":"00:47.135","Text":"In our case, let\u0027s check what the kernel is."},{"Start":"00:47.135 ","End":"00:50.540","Text":"The kernel would be those vectors x, y, z,"},{"Start":"00:50.540 ","End":"00:55.190","Text":"such that this expression is equal to the 0 vector."},{"Start":"00:55.190 ","End":"00:57.455","Text":"In other words, each component is 0."},{"Start":"00:57.455 ","End":"01:02.000","Text":"That would give us a system of 3 equations and 3 unknowns x,"},{"Start":"01:02.000 ","End":"01:04.850","Text":"y, z. It\u0027s homogeneous."},{"Start":"01:04.850 ","End":"01:09.800","Text":"Let\u0027s try and solve it using matrices."},{"Start":"01:09.800 ","End":"01:11.540","Text":"Because it\u0027s homogeneous."},{"Start":"01:11.540 ","End":"01:13.280","Text":"We don\u0027t need the right-hand side."},{"Start":"01:13.280 ","End":"01:14.750","Text":"We don\u0027t need an augmented matrix,"},{"Start":"01:14.750 ","End":"01:17.630","Text":"just copy the coefficients into a matrix."},{"Start":"01:17.630 ","End":"01:20.150","Text":"Now we want to bring it to row echelon form."},{"Start":"01:20.150 ","End":"01:23.245","Text":"Let\u0027s see, at the first row,"},{"Start":"01:23.245 ","End":"01:27.015","Text":"now subtract the first row from the last row."},{"Start":"01:27.015 ","End":"01:29.475","Text":"That will give us this."},{"Start":"01:29.475 ","End":"01:34.810","Text":"Now we want to subtract the second row from the third row."},{"Start":"01:34.810 ","End":"01:37.005","Text":"That will give us this."},{"Start":"01:37.005 ","End":"01:41.260","Text":"Notice that we have a row of all 0s."},{"Start":"01:41.260 ","End":"01:45.080","Text":"From here let\u0027s go back to system of linear equations."},{"Start":"01:45.080 ","End":"01:49.805","Text":"We get just 2 equations from these 2 rows. We get these equations."},{"Start":"01:49.805 ","End":"01:56.750","Text":"Notice that z is the free variable because the pivot terms in each row,"},{"Start":"01:56.750 ","End":"01:58.835","Text":"this and this and z is free."},{"Start":"01:58.835 ","End":"02:00.680","Text":"Which means that to get a bases,"},{"Start":"02:00.680 ","End":"02:04.015","Text":"we could let say z equal 1."},{"Start":"02:04.015 ","End":"02:09.770","Text":"Then if we plug in z equals 1 here we get that y is minus 1."},{"Start":"02:09.770 ","End":"02:13.270","Text":"Then from here we get x is minus 2."},{"Start":"02:13.270 ","End":"02:16.610","Text":"The kernel is spanned by the vectors."},{"Start":"02:16.610 ","End":"02:17.690","Text":"Get it in the right order,"},{"Start":"02:17.690 ","End":"02:21.985","Text":"minus 2 minus 1,1, which is this."},{"Start":"02:21.985 ","End":"02:28.350","Text":"The dimension of the kernel therefore is 1 because it\u0027s just 1 member in the bases,"},{"Start":"02:28.350 ","End":"02:35.075","Text":"in any event, the kernel has dimension bigger than 0."},{"Start":"02:35.075 ","End":"02:38.330","Text":"T is not 1 to 1."},{"Start":"02:38.330 ","End":"02:43.220","Text":"As a matter of fact, as soon as we got to this point where we got a row of zeros,"},{"Start":"02:43.220 ","End":"02:47.990","Text":"we could actually stop and conclude that T is not 1 to 1,"},{"Start":"02:47.990 ","End":"02:50.170","Text":"which is just the details."},{"Start":"02:50.170 ","End":"02:54.060","Text":"Let\u0027s go on to the next question."},{"Start":"02:54.060 ","End":"02:59.265","Text":"Here we want to know if T is onto."},{"Start":"02:59.265 ","End":"03:05.630","Text":"We\u0027ll use a proposition which says that T is onto if and"},{"Start":"03:05.630 ","End":"03:08.690","Text":"only if the dimension of the image of T is the"},{"Start":"03:08.690 ","End":"03:11.810","Text":"same as the dimension of U. U is the target space."},{"Start":"03:11.810 ","End":"03:14.765","Text":"We have T from V to U."},{"Start":"03:14.765 ","End":"03:17.275","Text":"In this case it\u0027s our R^3."},{"Start":"03:17.275 ","End":"03:20.590","Text":"Because U is R^3 whose dimension is 3."},{"Start":"03:20.590 ","End":"03:27.120","Text":"Basically we\u0027re asking if the dimension of the image of T is 3."},{"Start":"03:27.120 ","End":"03:30.590","Text":"If so then we\u0027re onto and otherwise we\u0027re not."},{"Start":"03:30.590 ","End":"03:32.660","Text":"At this point, we would use"},{"Start":"03:32.660 ","End":"03:35.780","Text":"the rank nullity theorem to say that"},{"Start":"03:35.780 ","End":"03:39.320","Text":"the dimension of image of T plus dimension of kernel of T,"},{"Start":"03:39.320 ","End":"03:42.110","Text":"this is the rank and this is the nullity,"},{"Start":"03:42.110 ","End":"03:47.325","Text":"is the dimension of V actually."},{"Start":"03:47.325 ","End":"03:48.960","Text":"But that\u0027s also our R^3."},{"Start":"03:48.960 ","End":"03:50.310","Text":"This is 3."},{"Start":"03:50.310 ","End":"03:54.480","Text":"This is 1 from the previous question."},{"Start":"03:54.480 ","End":"04:02.200","Text":"This would have to equal 2 and 2 is not equal to 3."},{"Start":"04:02.630 ","End":"04:06.009","Text":"T is not onto."},{"Start":"04:06.140 ","End":"04:09.435","Text":"The next question or true or false,"},{"Start":"04:09.435 ","End":"04:12.400","Text":"T is an isomorphism."},{"Start":"04:12.400 ","End":"04:18.300","Text":"Now an isomorphism has to be both 1 to 1 and onto."},{"Start":"04:19.820 ","End":"04:24.485","Text":"In fact, we are neither."},{"Start":"04:24.485 ","End":"04:30.830","Text":"But failing either 1 of them would be good enough to say that T is not an isomorphism."},{"Start":"04:30.830 ","End":"04:38.155","Text":"Just for example, the not being onto already means is not an isomorphism."},{"Start":"04:38.155 ","End":"04:42.365","Text":"The last question is T invertible?"},{"Start":"04:42.365 ","End":"04:44.465","Text":"The answer is also no,"},{"Start":"04:44.465 ","End":"04:49.970","Text":"because invertible means it has an inverse and that\u0027s if and only if it\u0027s an isomorphism."},{"Start":"04:49.970 ","End":"04:52.160","Text":"Since it is not an isomorphism,"},{"Start":"04:52.160 ","End":"04:54.995","Text":"it\u0027s not going to be invertible."},{"Start":"04:54.995 ","End":"04:57.200","Text":"Here\u0027s that in writing."},{"Start":"04:57.200 ","End":"05:03.090","Text":"We\u0027ve answered all the questions and we are done."}],"ID":10218},{"Watched":false,"Name":"Exercise 3 Part a","Duration":"5m 35s","ChapterTopicVideoID":9620,"CourseChapterTopicPlaylistID":7316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.310","Text":"In this exercise, we have a linear transformation from this space,"},{"Start":"00:05.310 ","End":"00:09.420","Text":"which is the polynomials of degree 2 or less,"},{"Start":"00:09.420 ","End":"00:18.030","Text":"to this space which is R^3 and just regular 3D space."},{"Start":"00:18.030 ","End":"00:20.370","Text":"It\u0027s defined as follows,"},{"Start":"00:20.370 ","End":"00:25.425","Text":"T of general polynomial here is a plus bx plus cx squared."},{"Start":"00:25.425 ","End":"00:29.700","Text":"It goes to the vector in R^3 as follows;"},{"Start":"00:29.700 ","End":"00:34.170","Text":"a plus b plus c, a minus b, b minus 2c."},{"Start":"00:34.170 ","End":"00:41.590","Text":"We have to check which of these 4 is true or false, yes or no."},{"Start":"00:41.630 ","End":"00:43.890","Text":"See if T is one-to-one,"},{"Start":"00:43.890 ","End":"00:46.135","Text":"if it\u0027s onto, if it\u0027s an isomorphism,"},{"Start":"00:46.135 ","End":"00:47.690","Text":"and if it has an inverse,"},{"Start":"00:47.690 ","End":"00:48.940","Text":"and the case it has an inverse,"},{"Start":"00:48.940 ","End":"00:50.885","Text":"and we have to actually find it."},{"Start":"00:50.885 ","End":"00:54.220","Text":"First start with the one to one."},{"Start":"00:54.220 ","End":"00:58.805","Text":"For this we use the proposition that something"},{"Start":"00:58.805 ","End":"01:05.060","Text":"is T is one-to-one if and only if the kernel is just 0."},{"Start":"01:05.060 ","End":"01:09.020","Text":"That\u0027s in general, let\u0027s see what it means in our case."},{"Start":"01:09.020 ","End":"01:10.880","Text":"Let\u0027s see what\u0027s in the kernel,"},{"Start":"01:10.880 ","End":"01:12.050","Text":"something\u0027s in the kernel,"},{"Start":"01:12.050 ","End":"01:13.850","Text":"then T of it is 0."},{"Start":"01:13.850 ","End":"01:16.250","Text":"T of this polynomial is this,"},{"Start":"01:16.250 ","End":"01:24.845","Text":"and this is 0 so we just equate components and we get 3 equations in 3 unknowns,"},{"Start":"01:24.845 ","End":"01:31.025","Text":"a, b, c. Let\u0027s solve it with matrices."},{"Start":"01:31.025 ","End":"01:34.730","Text":"Here is the matrix corresponding to this system."},{"Start":"01:34.730 ","End":"01:39.290","Text":"We don\u0027t need an augmented matrix because we\u0027re dealing in the homogeneous system."},{"Start":"01:39.290 ","End":"01:43.950","Text":"Let\u0027s see, you want to bring this to row echelon form."},{"Start":"01:44.140 ","End":"01:50.180","Text":"Subtract the first row from the second row,"},{"Start":"01:50.180 ","End":"01:52.650","Text":"and then we get this."},{"Start":"01:53.900 ","End":"02:00.885","Text":"Next we take twice the last row plus the second row."},{"Start":"02:00.885 ","End":"02:03.500","Text":"That will give us 0 here,"},{"Start":"02:03.500 ","End":"02:06.560","Text":"0 here, and minus 5 here."},{"Start":"02:06.560 ","End":"02:12.500","Text":"Anyway, this is already echelon form and there are no 0 rows."},{"Start":"02:12.500 ","End":"02:17.335","Text":"Back to the system of linear equations."},{"Start":"02:17.335 ","End":"02:22.180","Text":"What we have is this."},{"Start":"02:22.340 ","End":"02:26.180","Text":"Actually already here we can see that because there are"},{"Start":"02:26.180 ","End":"02:30.870","Text":"no 0 rows we\u0027ll only get the 0 solution."},{"Start":"02:30.870 ","End":"02:32.495","Text":"Then lets continue anyway,"},{"Start":"02:32.495 ","End":"02:37.070","Text":"we have this system and if we do back substitution from the last row,"},{"Start":"02:37.070 ","End":"02:39.785","Text":"we got c is 0."},{"Start":"02:39.785 ","End":"02:43.160","Text":"Plug-in here, c is 0, you get b is 0,"},{"Start":"02:43.160 ","End":"02:44.690","Text":"plug in b and c,"},{"Start":"02:44.690 ","End":"02:47.390","Text":"0, and you\u0027ll get that a is 0,"},{"Start":"02:47.390 ","End":"02:50.980","Text":"so a, b, and c are all 0,"},{"Start":"02:50.980 ","End":"02:53.330","Text":"which means that our polynomial,"},{"Start":"02:53.330 ","End":"02:59.315","Text":"which is a plus bx plus cx squared is the 0 polynomial."},{"Start":"02:59.315 ","End":"03:06.690","Text":"The kernel just contains the 0 vector."},{"Start":"03:06.690 ","End":"03:09.345","Text":"In this case it\u0027s a polynomial."},{"Start":"03:09.345 ","End":"03:11.520","Text":"If that\u0027s the case,"},{"Start":"03:11.520 ","End":"03:18.275","Text":"then T is certainly one to one because of that proposition we mentioned above."},{"Start":"03:18.275 ","End":"03:20.345","Text":"That\u0027s the first part."},{"Start":"03:20.345 ","End":"03:22.805","Text":"Let\u0027s move on to the next."},{"Start":"03:22.805 ","End":"03:26.170","Text":"Is T onto?"},{"Start":"03:26.170 ","End":"03:29.980","Text":"There is that proposition that T is onto if and only"},{"Start":"03:29.980 ","End":"03:34.915","Text":"if the dimension of the image of T is the dimension of U."},{"Start":"03:34.915 ","End":"03:38.740","Text":"Perhaps I didn\u0027t say U is the target space."},{"Start":"03:38.740 ","End":"03:41.405","Text":"We have T from V-U."},{"Start":"03:41.405 ","End":"03:47.290","Text":"In this case U is 3 and its dimension is 3."},{"Start":"03:47.420 ","End":"03:53.140","Text":"What we have to check is if the dimension of the image of T is 3,"},{"Start":"03:53.140 ","End":"03:56.960","Text":"then we\u0027re okay, we\u0027re onto otherwise not."},{"Start":"03:57.020 ","End":"04:03.920","Text":"We\u0027ll use the rank nullity theorem to show this."},{"Start":"04:04.000 ","End":"04:09.680","Text":"We have that the dimension of the image of T plus the dimension of"},{"Start":"04:09.680 ","End":"04:14.720","Text":"kernel of T is the dimension of the space V,"},{"Start":"04:14.720 ","End":"04:24.200","Text":"which in our case was P_2 over R. The dimension of this is 2 plus 1 is 3."},{"Start":"04:24.200 ","End":"04:28.875","Text":"The dimension of this is 0 from the previous part."},{"Start":"04:28.875 ","End":"04:34.510","Text":"That leaves this as being 3 minus 0 is 3."},{"Start":"04:35.630 ","End":"04:39.750","Text":"T is indeed onto."},{"Start":"04:39.750 ","End":"04:44.520","Text":"In the previous part we showed it\u0027s one-to-one."},{"Start":"04:44.520 ","End":"04:46.770","Text":"The next question,"},{"Start":"04:46.770 ","End":"04:49.245","Text":"is T an isomorphism?"},{"Start":"04:49.245 ","End":"04:51.195","Text":"Well, definitely yes."},{"Start":"04:51.195 ","End":"04:55.535","Text":"We\u0027ve already showed that it\u0027s one-to-one and onto,"},{"Start":"04:55.535 ","End":"05:00.410","Text":"which is actually the definition of isomorphism, one-to-one and onto."},{"Start":"05:00.410 ","End":"05:02.930","Text":"The answer is yes."},{"Start":"05:02.930 ","End":"05:05.585","Text":"Now is T invertible?"},{"Start":"05:05.585 ","End":"05:09.485","Text":"Yes, because it\u0027s an isomorphism it\u0027s invertible."},{"Start":"05:09.485 ","End":"05:11.850","Text":"That\u0027s always true."},{"Start":"05:11.980 ","End":"05:15.635","Text":"An invertible means it has an inverse."},{"Start":"05:15.635 ","End":"05:18.490","Text":"The inverse we write like this T^minus 1,"},{"Start":"05:18.490 ","End":"05:24.820","Text":"T inverse and we have to do now is to actually compute it."},{"Start":"05:24.820 ","End":"05:29.765","Text":"Because the question said if it is invertible, find its inverse."},{"Start":"05:29.765 ","End":"05:35.220","Text":"Let\u0027s take a break first and then we\u0027ll compute the inverse of T."}],"ID":10219},{"Watched":false,"Name":"Exercise 3 Part b","Duration":"6m 21s","ChapterTopicVideoID":9624,"CourseChapterTopicPlaylistID":7316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:02.670","Text":"We\u0027re back from the break,"},{"Start":"00:02.670 ","End":"00:06.360","Text":"and we\u0027re about to find the inverse of T,"},{"Start":"00:06.360 ","End":"00:09.089","Text":"where T is given by this formula."},{"Start":"00:09.089 ","End":"00:12.900","Text":"It goes from the space of polynomials of degree 2 or"},{"Start":"00:12.900 ","End":"00:20.235","Text":"less to 3 by this formula."},{"Start":"00:20.235 ","End":"00:24.690","Text":"We want T inverse or T to the minus 1,"},{"Start":"00:24.690 ","End":"00:32.760","Text":"which means that if T takes polynomial a plus bx plus cx squared to the vector r, s, t,"},{"Start":"00:32.760 ","End":"00:36.060","Text":"then T inverse goes in the opposite direction,"},{"Start":"00:36.060 ","End":"00:42.370","Text":"takes r, s, t back to a plus bx plus cx squared."},{"Start":"00:42.520 ","End":"00:46.655","Text":"Now, T, using this definition, is this."},{"Start":"00:46.655 ","End":"00:51.485","Text":"On the other hand, it\u0027s going to be equal r, s, t, so we\u0027ll have now 3 equations."},{"Start":"00:51.485 ","End":"00:56.270","Text":"Each component, we compare a plus b plus c will equal r,"},{"Start":"00:56.270 ","End":"00:59.130","Text":"and there will be 2 other equalities."},{"Start":"00:59.180 ","End":"01:01.350","Text":"This is what we get,"},{"Start":"01:01.350 ","End":"01:06.720","Text":"this system of 3 equations and 3 unknowns, a, b, and c."},{"Start":"01:06.720 ","End":"01:13.090","Text":"R, s, and t are given, and then a, b, and c are the variables."},{"Start":"01:13.390 ","End":"01:20.260","Text":"Here\u0027s the augmented matrix corresponding to this system."},{"Start":"01:20.260 ","End":"01:24.020","Text":"We want to bring it to row echelon form. Let see."},{"Start":"01:24.020 ","End":"01:28.925","Text":"Subtract the first row from the second row."},{"Start":"01:28.925 ","End":"01:31.220","Text":"That will give us this."},{"Start":"01:31.220 ","End":"01:36.035","Text":"Now let\u0027s see how do we get a 0 here?"},{"Start":"01:36.035 ","End":"01:43.490","Text":"How about twice the last row plus the second row?"},{"Start":"01:43.490 ","End":"01:49.740","Text":"That will give us this, and this part is in row echelon form."},{"Start":"01:50.450 ","End":"01:54.785","Text":"Back to system of linear equations."},{"Start":"01:54.785 ","End":"01:57.330","Text":"This is the system."},{"Start":"01:57.560 ","End":"02:00.225","Text":"Let\u0027s work in decimals."},{"Start":"02:00.225 ","End":"02:01.500","Text":"From the last equation,"},{"Start":"02:01.500 ","End":"02:06.135","Text":"if we divide by minus 5,"},{"Start":"02:06.135 ","End":"02:22.710","Text":"then we will get that c is equal to 2t divided by minus 5 is minus 0.4t."},{"Start":"02:22.780 ","End":"02:27.980","Text":"I put it in the order r, s, t. This is what we get for c."},{"Start":"02:27.980 ","End":"02:38.470","Text":"Now plug in here, b is minus a 1/2."},{"Start":"02:38.470 ","End":"02:40.980","Text":"We bring the c over to the other side,"},{"Start":"02:40.980 ","End":"02:42.900","Text":"so we go s minus r plus c here."},{"Start":"02:42.900 ","End":"02:45.690","Text":"Then we divide by the minus 2,"},{"Start":"02:45.690 ","End":"02:48.190","Text":"which gives us minus a 1/2."},{"Start":"02:52.460 ","End":"02:58.540","Text":"Let\u0027s see. We need to replace c by this expression here."},{"Start":"02:58.540 ","End":"03:05.540","Text":"Then after we simplify it and multiply it by minus 0.5, we get this for b."},{"Start":"03:05.630 ","End":"03:10.750","Text":"Then finally we get to a which is r minus b minus c,"},{"Start":"03:10.750 ","End":"03:13.610","Text":"but we have b and c from here,"},{"Start":"03:13.610 ","End":"03:15.750","Text":"and it simplifies to this."},{"Start":"03:15.750 ","End":"03:20.890","Text":"Now we have a, b, and c in terms of r, s, and t."},{"Start":"03:20.890 ","End":"03:27.865","Text":"The inverse of r, s, t is the a, b, c that we have."},{"Start":"03:27.865 ","End":"03:32.060","Text":"This is our a which goes here,"},{"Start":"03:33.170 ","End":"03:38.550","Text":"and this is b which goes here."},{"Start":"03:38.550 ","End":"03:44.620","Text":"Don\u0027t forget that what we\u0027re building is a plus bx plus cx squared."},{"Start":"03:44.620 ","End":"03:49.250","Text":"I just put an extra 1 here because a is a time 1."},{"Start":"03:49.250 ","End":"03:56.710","Text":"Then c, from here, goes with the x squared."},{"Start":"03:56.780 ","End":"04:03.500","Text":"So the inverse of vector r, s, t is the polynomial that\u0027s written here."},{"Start":"04:03.500 ","End":"04:09.380","Text":"Now we used r, s, and t because we were also using a, b, c,"},{"Start":"04:09.380 ","End":"04:13.070","Text":"but now there\u0027s no conflict anymore."},{"Start":"04:13.070 ","End":"04:16.950","Text":"We just have everything in terms of 3 dummy variables r, s, t,"},{"Start":"04:16.950 ","End":"04:19.980","Text":"so let\u0027s switch back to a, b, c."},{"Start":"04:19.980 ","End":"04:23.795","Text":"Just switch that, if r, s, t, put a, b, c everywhere."},{"Start":"04:23.795 ","End":"04:30.240","Text":"This is the formula for the inverse of T, in general, in a, b, c."},{"Start":"04:30.240 ","End":"04:35.195","Text":"We\u0027re done, but it\u0027s a good idea to do some checking."},{"Start":"04:35.195 ","End":"04:39.560","Text":"Let\u0027s see that T and T to the minus 1 really are inverses,"},{"Start":"04:39.560 ","End":"04:43.301","Text":"so let\u0027s test T out on the polynomial 1,"},{"Start":"04:43.301 ","End":"04:47.780","Text":"which is 1 plus 0x plus 0x squared."},{"Start":"04:47.780 ","End":"04:56.400","Text":"Using the formula above for T, this goes to 1, 1, 0."},{"Start":"04:56.400 ","End":"05:00.170","Text":"The formula is scrolled off, but this is it."},{"Start":"05:00.170 ","End":"05:06.605","Text":"If we apply T-inverse, we still have that here, 2, 1, 1, 0."},{"Start":"05:06.605 ","End":"05:14.835","Text":"Plug it in. If you plug in, a is 1, and b is 1, and c is 0 here, we get 1."},{"Start":"05:14.835 ","End":"05:19.740","Text":"Then here if a is 1, and b is 1, and c is 0, this is 0."},{"Start":"05:19.740 ","End":"05:24.160","Text":"This comes out 0. In short, we get back to the 1."},{"Start":"05:25.790 ","End":"05:29.120","Text":"We go from the polynomial to the vector,"},{"Start":"05:29.120 ","End":"05:31.510","Text":"then back to the polynomial."},{"Start":"05:31.510 ","End":"05:37.530","Text":"Similarly, with the 2 other bases members with x,"},{"Start":"05:37.530 ","End":"05:40.935","Text":"we\u0027ve applied T to x, which is this,"},{"Start":"05:40.935 ","End":"05:45.035","Text":"and used the formula for T, which is just off-screen at the moment,"},{"Start":"05:45.035 ","End":"05:46.910","Text":"we get this vector."},{"Start":"05:46.910 ","End":"05:51.220","Text":"If we apply T minus 1 to this vector here,"},{"Start":"05:51.220 ","End":"05:54.540","Text":"this one, then we get back to x."},{"Start":"05:54.540 ","End":"05:56.475","Text":"Similarly, with x squared,"},{"Start":"05:56.475 ","End":"06:01.820","Text":"x squared goes to this vector and T inverse of that vector is back to x squared."},{"Start":"06:01.820 ","End":"06:04.345","Text":"These are just a few checks."},{"Start":"06:04.345 ","End":"06:08.745","Text":"You can do some of these or many as you want."},{"Start":"06:08.745 ","End":"06:11.780","Text":"In any event, we didn\u0027t make any mistakes."},{"Start":"06:11.780 ","End":"06:17.840","Text":"Then this is the answer for the formula for the inverse of T."},{"Start":"06:17.840 ","End":"06:20.730","Text":"We are done."}],"ID":10220},{"Watched":false,"Name":"Exercise 4 Part a","Duration":"6m 15s","ChapterTopicVideoID":9618,"CourseChapterTopicPlaylistID":7316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.880","Text":"In this exercise, we have a linear transformation from this space to this space."},{"Start":"00:05.880 ","End":"00:11.190","Text":"This is the 2 by 2 matrices over the reals."},{"Start":"00:11.190 ","End":"00:18.375","Text":"This is the polynomials of degree 3 or less also over the real numbers."},{"Start":"00:18.375 ","End":"00:23.580","Text":"It\u0027s given by the formula that T of the 2 by 2 matrix a, b, c,"},{"Start":"00:23.580 ","End":"00:27.015","Text":"d gives us this polynomial,"},{"Start":"00:27.015 ","End":"00:30.390","Text":"which is of degree 3 or less."},{"Start":"00:30.390 ","End":"00:33.630","Text":"We have to check if T is 1 to 1,"},{"Start":"00:33.630 ","End":"00:38.105","Text":"if it\u0027s onto, if it\u0027s an isomorphism."},{"Start":"00:38.105 ","End":"00:42.830","Text":"If it has an inverse and if it does have an inverse,"},{"Start":"00:42.830 ","End":"00:46.510","Text":"we have to actually find it to compute it."},{"Start":"00:46.510 ","End":"00:49.005","Text":"Let\u0027s start to pull it up."},{"Start":"00:49.005 ","End":"00:54.615","Text":"Prologue, the preliminary computations."},{"Start":"00:54.615 ","End":"01:02.280","Text":"What we\u0027ll do is instead of looking at the space of 2 by 2 matrices in polynomials,"},{"Start":"01:02.280 ","End":"01:05.555","Text":"we\u0027ll just flatten everything out into vectors."},{"Start":"01:05.555 ","End":"01:09.410","Text":"Like this matrix will be the vector a, b, c,"},{"Start":"01:09.410 ","End":"01:17.490","Text":"d. This polynomial also we\u0027ll just look at it as a vector."},{"Start":"01:17.530 ","End":"01:20.090","Text":"Like taking the basis 1,"},{"Start":"01:20.090 ","End":"01:22.625","Text":"x, x squared and x cubed."},{"Start":"01:22.625 ","End":"01:25.475","Text":"It\u0027s not really the same."},{"Start":"01:25.475 ","End":"01:28.100","Text":"T should call it T prime,"},{"Start":"01:28.100 ","End":"01:29.330","Text":"T bar or something,"},{"Start":"01:29.330 ","End":"01:33.685","Text":"but there\u0027s no confusion,"},{"Start":"01:33.685 ","End":"01:36.780","Text":"we\u0027ll just keep calling it T. T of a b,"},{"Start":"01:36.780 ","End":"01:39.570","Text":"c, d is a minus b,"},{"Start":"01:39.570 ","End":"01:41.505","Text":"then c plus d,"},{"Start":"01:41.505 ","End":"01:49.140","Text":"a minus c and then d. Now we come to the question is T 1 to 1?"},{"Start":"01:49.460 ","End":"02:00.500","Text":"The usual way of showing this is to show that the kernel is just the vector 0 or not."},{"Start":"02:00.500 ","End":"02:04.735","Text":"Depending if we were looking for 1 to 1 or not."},{"Start":"02:04.735 ","End":"02:13.195","Text":"I mean, there\u0027s the preposition that T is 1 to 1 if and only if the kernel is just the 0."},{"Start":"02:13.195 ","End":"02:16.295","Text":"Let\u0027s see what is the kernel."},{"Start":"02:16.295 ","End":"02:20.160","Text":"Like I said, we\u0027ll use the vector form."},{"Start":"02:20.160 ","End":"02:22.470","Text":"If T of a, b, c,"},{"Start":"02:22.470 ","End":"02:26.510","Text":"d is 0 vector being this,"},{"Start":"02:26.510 ","End":"02:30.820","Text":"that means that by the formula that this is 0."},{"Start":"02:30.820 ","End":"02:34.220","Text":"There are 4 components just equate each of"},{"Start":"02:34.220 ","End":"02:38.840","Text":"them and then we get this system of linear equations,"},{"Start":"02:38.840 ","End":"02:42.455","Text":"a minus b is 0 and so on."},{"Start":"02:42.455 ","End":"02:44.930","Text":"We\u0027ll do it with matrices."},{"Start":"02:44.930 ","End":"02:50.480","Text":"The matrix for this homogeneous system is this 1 minus 1,"},{"Start":"02:50.480 ","End":"02:52.875","Text":"0, 0 and so on."},{"Start":"02:52.875 ","End":"02:55.660","Text":"I want to bring it to row echelon form."},{"Start":"02:55.660 ","End":"03:00.835","Text":"Subtract the first row from the second row."},{"Start":"03:00.835 ","End":"03:07.150","Text":"That gives us this matrix which is in echelon form and there are no rows of zeros."},{"Start":"03:07.150 ","End":"03:11.740","Text":"Actually, we could already say that it is 1 to 1."},{"Start":"03:11.740 ","End":"03:18.550","Text":"But let\u0027s continue from the matrix back to the SLA."},{"Start":"03:18.550 ","End":"03:20.125","Text":"This is what we get."},{"Start":"03:20.125 ","End":"03:21.850","Text":"If we use back substitution,"},{"Start":"03:21.850 ","End":"03:23.200","Text":"you can see that a, b,"},{"Start":"03:23.200 ","End":"03:26.240","Text":"c and d are all 0."},{"Start":"03:26.570 ","End":"03:28.980","Text":"The 0 plug it in here."},{"Start":"03:28.980 ","End":"03:30.120","Text":"I get c is 0."},{"Start":"03:30.120 ","End":"03:31.230","Text":"Plug c is 0,"},{"Start":"03:31.230 ","End":"03:35.325","Text":"you got b is 0 and a is 0."},{"Start":"03:35.325 ","End":"03:40.125","Text":"At the end let\u0027s go back into the 2 by 2 matrix form a,"},{"Start":"03:40.125 ","End":"03:43.290","Text":"b, c, d, they\u0027re all 0."},{"Start":"03:43.290 ","End":"03:46.545","Text":"The kernel is just the 0 matrix."},{"Start":"03:46.545 ","End":"03:49.965","Text":"In short I can just call it 0."},{"Start":"03:49.965 ","End":"03:53.960","Text":"When the kernel contains just the 0 element,"},{"Start":"03:53.960 ","End":"03:56.990","Text":"the 0 vector, in this case 0 matrix,"},{"Start":"03:56.990 ","End":"03:59.600","Text":"then we are 1 to 1."},{"Start":"03:59.600 ","End":"04:02.975","Text":"I mean, the transformation is 1 to 1."},{"Start":"04:02.975 ","End":"04:09.940","Text":"Now onto the next is T onto."},{"Start":"04:09.940 ","End":"04:15.470","Text":"As usually we\u0027re using the same proposition that T is"},{"Start":"04:15.470 ","End":"04:21.680","Text":"onto if and only if the dimension of the image of T is the dimension of U,"},{"Start":"04:21.680 ","End":"04:25.460","Text":"U is the target."},{"Start":"04:25.460 ","End":"04:31.345","Text":"We have T from V to U, this is the U."},{"Start":"04:31.345 ","End":"04:37.740","Text":"In our case, it\u0027s the space of polynomials of degree 3 or less."},{"Start":"04:38.240 ","End":"04:44.610","Text":"The dimension of this space is 3 plus 1 is 4."},{"Start":"04:44.610 ","End":"04:52.320","Text":"All we have to check to see if T is onto or not is if the dimension of the image is 4."},{"Start":"04:52.320 ","End":"04:55.595","Text":"Now we have the rank nullity theorem,"},{"Start":"04:55.595 ","End":"04:58.880","Text":"which says that the dimension of the image plus the dimension of"},{"Start":"04:58.880 ","End":"05:02.540","Text":"the kernel is the dimension of this V,"},{"Start":"05:02.540 ","End":"05:06.955","Text":"which is 2 by 2 matrices and this has dimension 4."},{"Start":"05:06.955 ","End":"05:10.805","Text":"We already found out that the kernel has dimension 0."},{"Start":"05:10.805 ","End":"05:18.415","Text":"The dimension of the image is 4 minus 0 is 4 and that\u0027s what we wanted."},{"Start":"05:18.415 ","End":"05:22.560","Text":"That means that T really is onto."},{"Start":"05:22.560 ","End":"05:25.145","Text":"Yeah, just wrote it down."},{"Start":"05:25.145 ","End":"05:30.350","Text":"Now that we know that T is 1 to 1 and onto, the next question,"},{"Start":"05:30.350 ","End":"05:38.860","Text":"is T an isomorphism has to be yes because isomorphism means 1 to 1 and onto."},{"Start":"05:38.860 ","End":"05:42.995","Text":"That 1 is also true or yes."},{"Start":"05:42.995 ","End":"05:49.500","Text":"The next question is T invertible or does T have an inverse?"},{"Start":"05:49.500 ","End":"05:52.725","Text":"The answer also is yes so this is true."},{"Start":"05:52.725 ","End":"05:58.445","Text":"Because it\u0027s an isomorphism, it\u0027s automatically invertible."},{"Start":"05:58.445 ","End":"06:01.010","Text":"The inverse is written like this,"},{"Start":"06:01.010 ","End":"06:03.050","Text":"T to the minus 1."},{"Start":"06:03.050 ","End":"06:05.630","Text":"The question said that if it is invertible,"},{"Start":"06:05.630 ","End":"06:07.955","Text":"we have to actually find the inverse."},{"Start":"06:07.955 ","End":"06:12.335","Text":"We\u0027ll do that after the break."},{"Start":"06:12.335 ","End":"06:14.880","Text":"Let\u0027s just take a break here."}],"ID":10221},{"Watched":false,"Name":"Exercise 4 Part b","Duration":"5m 10s","ChapterTopicVideoID":9619,"CourseChapterTopicPlaylistID":7316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:02.970","Text":"Here we are after the break,"},{"Start":"00:02.970 ","End":"00:06.000","Text":"we were about to compute the inverse of T."},{"Start":"00:06.000 ","End":"00:11.170","Text":"Let\u0027s go back to vector form instead of the 2 by 2 matrices and polynomials,"},{"Start":"00:11.170 ","End":"00:17.070","Text":"we\u0027ll use the vectors as if they\u0027re both 4."},{"Start":"00:17.070 ","End":"00:20.655","Text":"Now, by definition of inverse,"},{"Start":"00:20.655 ","End":"00:27.019","Text":"if a, b, c, d goes to r, s, t, u, and if T takes this to this,"},{"Start":"00:27.019 ","End":"00:30.360","Text":"and T inverse takes this 1 back to this,"},{"Start":"00:30.360 ","End":"00:32.530","Text":"so we have this."},{"Start":"00:32.530 ","End":"00:37.860","Text":"Now, we already have the formula for T of a, b, c, d is this,"},{"Start":"00:37.860 ","End":"00:39.930","Text":"and this is going to be r, s, t, u."},{"Start":"00:39.930 ","End":"00:44.610","Text":"Then we get 4 equations and 4 unknowns."},{"Start":"00:44.610 ","End":"00:46.650","Text":"This is the system we get."},{"Start":"00:46.650 ","End":"00:51.915","Text":"R, s, t, u are given and a, b, c, d are the unknowns."},{"Start":"00:51.915 ","End":"00:56.560","Text":"From this system, we get this augmented matrix,"},{"Start":"00:56.560 ","End":"01:00.260","Text":"and we want to bring this to row echelon form."},{"Start":"01:00.260 ","End":"01:02.480","Text":"Let\u0027s see."},{"Start":"01:02.480 ","End":"01:08.660","Text":"We just subtract the first row from the second row."},{"Start":"01:08.660 ","End":"01:10.565","Text":"Then we get this,"},{"Start":"01:10.565 ","End":"01:13.760","Text":"which is in row echelon form,"},{"Start":"01:13.760 ","End":"01:22.265","Text":"and now go back again from matrix to an SLA system."},{"Start":"01:22.265 ","End":"01:28.010","Text":"The solution of this is just that c."},{"Start":"01:28.010 ","End":"01:37.170","Text":"D is u, so that\u0027s here, and from here we get that c is s minus d,"},{"Start":"01:37.170 ","End":"01:39.000","Text":"but we plug in that there\u0027s u,"},{"Start":"01:39.000 ","End":"01:41.490","Text":"so we\u0027ve got c is minus u."},{"Start":"01:41.490 ","End":"01:45.185","Text":"Then b is c plus t minus r,"},{"Start":"01:45.185 ","End":"01:49.400","Text":"which gives us in terms of r, s, t, u, this expression,"},{"Start":"01:49.400 ","End":"01:53.030","Text":"and a which is b plus r, gives us this."},{"Start":"01:53.030 ","End":"01:54.740","Text":"Now, we have a, b, c,"},{"Start":"01:54.740 ","End":"01:59.030","Text":"and d in terms of r, s, t, u,"},{"Start":"01:59.030 ","End":"02:05.320","Text":"and that will give us the inverse."},{"Start":"02:05.870 ","End":"02:17.280","Text":"T inverse of r, s, t, u is the a, b, c,"},{"Start":"02:17.280 ","End":"02:20.250","Text":"and d which are written here."},{"Start":"02:20.250 ","End":"02:24.780","Text":"So this is it. Now, r, s, t,"},{"Start":"02:24.780 ","End":"02:29.550","Text":"and u are just temporary dummy variables."},{"Start":"02:29.550 ","End":"02:32.340","Text":"We couldn\u0027t use a, b, c, d."},{"Start":"02:32.340 ","End":"02:35.450","Text":"But now that we\u0027ve got our result,"},{"Start":"02:35.450 ","End":"02:39.080","Text":"we can rewrite it in terms of a, b, c, d,"},{"Start":"02:39.080 ","End":"02:45.000","Text":"which is any 4 variables that we want, a, b, c, d,"},{"Start":"02:45.000 ","End":"02:47.115","Text":"which is pretty standard."},{"Start":"02:47.115 ","End":"02:52.405","Text":"Now, we have this is T inverse."},{"Start":"02:52.405 ","End":"02:59.000","Text":"As a reminder, the original T was given by this formula."},{"Start":"02:59.000 ","End":"03:05.950","Text":"Let\u0027s do some checks just to see that we\u0027re on the right track with this inverse."},{"Start":"03:05.950 ","End":"03:07.790","Text":"We\u0027ll take as an example,"},{"Start":"03:07.790 ","End":"03:10.565","Text":"a b, c, d is 1, 2, 3, 4."},{"Start":"03:10.565 ","End":"03:15.475","Text":"Then T of 1, 2, 3, 4 using this formula gives us this."},{"Start":"03:15.475 ","End":"03:22.010","Text":"If we use this formula for T inverse and apply it to this vector,"},{"Start":"03:22.010 ","End":"03:23.270","Text":"if you do the computations,"},{"Start":"03:23.270 ","End":"03:27.605","Text":"you\u0027ll see that we really do get back to 1, 2, 3, 4."},{"Start":"03:27.605 ","End":"03:30.175","Text":"So that\u0027s this example."},{"Start":"03:30.175 ","End":"03:34.580","Text":"Now, it\u0027s 1 thing we have to do that if we just left it like this,"},{"Start":"03:34.580 ","End":"03:35.660","Text":"it wouldn\u0027t be right."},{"Start":"03:35.660 ","End":"03:39.920","Text":"We originally had matrices and polynomials."},{"Start":"03:39.920 ","End":"03:42.320","Text":"We didn\u0027t just have vectors in our 4,"},{"Start":"03:42.320 ","End":"03:49.790","Text":"so we have to go back to the original language so that T is not this."},{"Start":"03:49.790 ","End":"03:51.845","Text":"T is given by this,"},{"Start":"03:51.845 ","End":"03:54.170","Text":"the a, b, c, d is this way,"},{"Start":"03:54.170 ","End":"04:00.670","Text":"and these are just the coefficients of the polynomial,"},{"Start":"04:00.670 ","End":"04:03.615","Text":"1x, x squared, x cubed."},{"Start":"04:03.615 ","End":"04:09.435","Text":"That\u0027s T, which goes from matrices to polynomials."},{"Start":"04:09.435 ","End":"04:13.220","Text":"T inverse, we\u0027ll go the other way around."},{"Start":"04:13.220 ","End":"04:14.810","Text":"Well, the formulas here,"},{"Start":"04:14.810 ","End":"04:21.690","Text":"but now we interpret as the a, b, c, d here is the polynomial,"},{"Start":"04:21.690 ","End":"04:25.094","Text":"a plus bx plus cx plus dx cubed."},{"Start":"04:25.094 ","End":"04:28.670","Text":"Sorry, I said it wrong,"},{"Start":"04:28.670 ","End":"04:31.470","Text":"a plus bx plus cx squared plus dx cubed."},{"Start":"04:31.470 ","End":"04:36.430","Text":"The vector written here,"},{"Start":"04:36.430 ","End":"04:42.350","Text":"just put it in matrix form this is top left,"},{"Start":"04:42.350 ","End":"04:44.060","Text":"top right, bottom left,"},{"Start":"04:44.060 ","End":"04:45.935","Text":"bottom right is this,"},{"Start":"04:45.935 ","End":"04:52.630","Text":"and it\u0027s the other way around from polynomials to matrices."},{"Start":"04:54.050 ","End":"04:57.004","Text":"This would be the formula,"},{"Start":"04:57.004 ","End":"05:04.260","Text":"the correct 1 for T inverse, and that\u0027s it."},{"Start":"05:04.260 ","End":"05:09.990","Text":"We\u0027ve answered all the parts of the question and we are done."}],"ID":10222}],"Thumbnail":null,"ID":7316},{"Name":"Composition of Linear Transformations","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"3m 42s","ChapterTopicVideoID":9630,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.575","Text":"In this exercise, we have 2 linear transformations,"},{"Start":"00:04.575 ","End":"00:09.825","Text":"T and S. T goes from R^3 to R^3,"},{"Start":"00:09.825 ","End":"00:13.890","Text":"but S goes from R^3 to R^2."},{"Start":"00:13.890 ","End":"00:16.350","Text":"Now, each of them has its formula."},{"Start":"00:16.350 ","End":"00:19.425","Text":"T is defined as follows."},{"Start":"00:19.425 ","End":"00:25.200","Text":"Notice that it takes a 3D vector and gives out a 3D vector,"},{"Start":"00:25.200 ","End":"00:30.250","Text":"whereas S takes a 3D vector and gives us a 2D vector."},{"Start":"00:30.250 ","End":"00:36.110","Text":"Our task is to find a formula, if possible,"},{"Start":"00:36.110 ","End":"00:39.005","Text":"for S plus T,"},{"Start":"00:39.005 ","End":"00:42.920","Text":"the addition of 2 functions of transformations."},{"Start":"00:42.920 ","End":"00:46.490","Text":"Now, if possible, because it might not be possible,"},{"Start":"00:46.490 ","End":"00:49.020","Text":"it might not make sense."},{"Start":"00:49.960 ","End":"00:53.510","Text":"In fact, this will be a case where it will not make"},{"Start":"00:53.510 ","End":"00:57.750","Text":"sense to add these 2 transformations."},{"Start":"00:57.820 ","End":"01:00.230","Text":"Let\u0027s start out naively,"},{"Start":"01:00.230 ","End":"01:02.825","Text":"just like with the addition of functions,"},{"Start":"01:02.825 ","End":"01:06.320","Text":"the sum of 2 functions applied to, in this case,"},{"Start":"01:06.320 ","End":"01:10.900","Text":"a vector would be to apply each 1 separately and then to add,"},{"Start":"01:10.900 ","End":"01:12.850","Text":"so we\u0027d have S of x,"},{"Start":"01:12.850 ","End":"01:16.485","Text":"y, z plus T of x, y, z."},{"Start":"01:16.485 ","End":"01:23.330","Text":"This part is not a problem because both S and T are defined on R^3."},{"Start":"01:23.330 ","End":"01:26.970","Text":"The source or domain is R^3,"},{"Start":"01:26.970 ","End":"01:30.540","Text":"so we can apply S and T to XYZ,"},{"Start":"01:30.540 ","End":"01:34.160","Text":"and using these 2 formulas for S,"},{"Start":"01:34.160 ","End":"01:35.770","Text":"we get this expression,"},{"Start":"01:35.770 ","End":"01:37.290","Text":"and it\u0027s a 2D vector,"},{"Start":"01:37.290 ","End":"01:39.500","Text":"notice, and for T we get this,"},{"Start":"01:39.500 ","End":"01:41.300","Text":"it\u0027s a 3D vector."},{"Start":"01:41.300 ","End":"01:47.215","Text":"Now, how are we supposed to add a 2D vector and the 3D vector?"},{"Start":"01:47.215 ","End":"01:49.740","Text":"Well, we can\u0027t, it\u0027s not defined,"},{"Start":"01:49.740 ","End":"01:53.770","Text":"and this expression is nonsense."},{"Start":"01:53.890 ","End":"01:58.840","Text":"To make sense of what\u0027s going on and to find the general rule,"},{"Start":"01:58.840 ","End":"02:04.565","Text":"what we can say is that if you want to add or subtract 2 transformations,"},{"Start":"02:04.565 ","End":"02:08.825","Text":"and let\u0027s say we have spaces U and V,"},{"Start":"02:08.825 ","End":"02:10.220","Text":"let\u0027s make it the other way around,"},{"Start":"02:10.220 ","End":"02:12.340","Text":"from V to U."},{"Start":"02:12.340 ","End":"02:20.670","Text":"If we have S from V to U and T from the same V to the same U,"},{"Start":"02:20.670 ","End":"02:27.050","Text":"then we can define S plus T or S minus T and otherwise not."},{"Start":"02:27.050 ","End":"02:28.775","Text":"That is the rule."},{"Start":"02:28.775 ","End":"02:30.350","Text":"In other words, a defined if and only if"},{"Start":"02:30.350 ","End":"02:34.400","Text":"both transformations are from the same V to the same U."},{"Start":"02:34.400 ","End":"02:36.499","Text":"Why does this make sense?"},{"Start":"02:36.499 ","End":"02:39.740","Text":"We saw here, if we\u0027re applying S plus T,"},{"Start":"02:39.740 ","End":"02:43.880","Text":"Each of them has to be applied to the same vector so they both have"},{"Start":"02:43.880 ","End":"02:48.619","Text":"to start off on the same domain or source."},{"Start":"02:48.619 ","End":"02:50.900","Text":"Now, when we get to the addition part,"},{"Start":"02:50.900 ","End":"02:54.605","Text":"we have to be able to add them and we can\u0027t always"},{"Start":"02:54.605 ","End":"03:00.620","Text":"add 2 vectors if they\u0027re in different vector spaces."},{"Start":"03:00.620 ","End":"03:04.265","Text":"If 1 went to U and the other went to W,"},{"Start":"03:04.265 ","End":"03:06.020","Text":"you wouldn\u0027t be able to add them"},{"Start":"03:06.020 ","End":"03:10.055","Text":"because you can\u0027t generally add things in different spaces,"},{"Start":"03:10.055 ","End":"03:12.800","Text":"that\u0027s why this makes sense."},{"Start":"03:12.800 ","End":"03:16.490","Text":"In our case, we did have the same V,"},{"Start":"03:16.490 ","End":"03:18.860","Text":"they both started from R^3,"},{"Start":"03:18.860 ","End":"03:28.040","Text":"which is why we\u0027re even able to get the process started to apply S and T to the same XYZ."},{"Start":"03:28.040 ","End":"03:31.260","Text":"But when we got to U, it was a different U."},{"Start":"03:31.260 ","End":"03:35.055","Text":"In 1 case it was R^2 and in 1 case it was R^3,"},{"Start":"03:35.055 ","End":"03:38.410","Text":"so we weren\u0027t able to add them."},{"Start":"03:38.840 ","End":"03:42.820","Text":"We are done with this ."}],"ID":10188},{"Watched":false,"Name":"Exercise 2","Duration":"2m 31s","ChapterTopicVideoID":9631,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.770","Text":"In this exercise, we have a linear transformation, not t,"},{"Start":"00:07.770 ","End":"00:11.670","Text":"it\u0027s called S from R3 to R2,"},{"Start":"00:11.670 ","End":"00:15.900","Text":"and it\u0027s defined that S takes a 3D vector x, y,"},{"Start":"00:15.900 ","End":"00:20.655","Text":"z, sends it to the 2D vector x minus z,y."},{"Start":"00:20.655 ","End":"00:28.510","Text":"That\u0027s S. Our task is to find a formula that defines the transformation 4S."},{"Start":"00:29.450 ","End":"00:32.275","Text":"Now, when we say 4S,"},{"Start":"00:32.275 ","End":"00:33.800","Text":"S is a function,"},{"Start":"00:33.800 ","End":"00:40.250","Text":"a transformation is a function and a scalar times a function."},{"Start":"00:40.250 ","End":"00:45.845","Text":"We define that to be pointwise or for each vector,"},{"Start":"00:45.845 ","End":"00:50.220","Text":"the function 4S or the transformation 4S,"},{"Start":"00:50.220 ","End":"00:51.675","Text":"when you apply to it,"},{"Start":"00:51.675 ","End":"01:00.080","Text":"is 4 times what you get if you apply just S. This is the way we define 4S."},{"Start":"01:00.080 ","End":"01:03.440","Text":"Now we have the formula for S up here."},{"Start":"01:03.440 ","End":"01:05.210","Text":"S of this is this."},{"Start":"01:05.210 ","End":"01:08.960","Text":"Now we just have to multiply it by 4,"},{"Start":"01:08.960 ","End":"01:11.535","Text":"and we do that component-wise,"},{"Start":"01:11.535 ","End":"01:14.960","Text":"and this would be our answer."},{"Start":"01:14.960 ","End":"01:20.105","Text":"Notice that there were no problems in the previous exercise,"},{"Start":"01:20.105 ","End":"01:23.990","Text":"we saw that with addition and subtraction of transformations,"},{"Start":"01:23.990 ","End":"01:25.655","Text":"there could be problems."},{"Start":"01:25.655 ","End":"01:29.510","Text":"But whenever we have a transformation,"},{"Start":"01:29.510 ","End":"01:30.830","Text":"let\u0027s not use S again,"},{"Start":"01:30.830 ","End":"01:37.205","Text":"lets say T from a vector space V to a vector space U,"},{"Start":"01:37.205 ","End":"01:41.475","Text":"there\u0027s no problem defining K times T,"},{"Start":"01:41.475 ","End":"01:45.465","Text":"and it will also be from v to u,"},{"Start":"01:45.465 ","End":"01:53.270","Text":"and we just take T of whatever vector it is and multiply by the scalar K. I"},{"Start":"01:53.270 ","End":"01:55.495","Text":"guess the only thing that"},{"Start":"01:55.495 ","End":"02:01.040","Text":"the only restriction is that both vector spaces have to be over the same field."},{"Start":"02:01.040 ","End":"02:03.425","Text":"I mean the constant,"},{"Start":"02:03.425 ","End":"02:06.080","Text":"the scalar k has to come from some field."},{"Start":"02:06.080 ","End":"02:11.915","Text":"We\u0027re assuming that v and u are over the same field,"},{"Start":"02:11.915 ","End":"02:13.280","Text":"F, which would be say,"},{"Start":"02:13.280 ","End":"02:15.955","Text":"reals or complex or whatever."},{"Start":"02:15.955 ","End":"02:21.030","Text":"Anyway, this is the answer for this problem."},{"Start":"02:21.030 ","End":"02:25.640","Text":"In general, you can always take a linear transformation and multiply it"},{"Start":"02:25.640 ","End":"02:31.350","Text":"by a scalar. We\u0027re done."}],"ID":10189},{"Watched":false,"Name":"Exercise 3","Duration":"3m 26s","ChapterTopicVideoID":9632,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.060","Text":"In this exercise, we have 2 linear transformations, T and S."},{"Start":"00:06.060 ","End":"00:12.690","Text":"T is from R^3-R^3 and defined by this formula."},{"Start":"00:12.690 ","End":"00:20.145","Text":"S is from R^3-R^2 and defined by this formula."},{"Start":"00:20.145 ","End":"00:24.075","Text":"The question is, can we find a formula,"},{"Start":"00:24.075 ","End":"00:26.910","Text":"because it may not be possible,"},{"Start":"00:26.910 ","End":"00:36.460","Text":"a formula which defines the expression 4S minus 10T."},{"Start":"00:36.610 ","End":"00:41.840","Text":"I\u0027ll just say that if you were paying attention in the previous exercises,"},{"Start":"00:41.840 ","End":"00:47.975","Text":"the problem in general is not the part where you multiply by a scalar,"},{"Start":"00:47.975 ","End":"00:52.070","Text":"but the addition and subtraction are a problem,"},{"Start":"00:52.070 ","End":"00:58.190","Text":"especially when the transformations don\u0027t apply to the same spaces."},{"Start":"00:58.190 ","End":"01:02.060","Text":"For example, here they both start from R^3,"},{"Start":"01:02.060 ","End":"01:05.760","Text":"sorry but 1 of them ends up in R^3"},{"Start":"01:05.760 ","End":"01:09.045","Text":"and the other ends up in R^2 so we can expect problems."},{"Start":"01:09.045 ","End":"01:14.975","Text":"Let\u0027s start, 4S minus 10T."},{"Start":"01:14.975 ","End":"01:19.750","Text":"The first step would be to break it into a subtraction."},{"Start":"01:19.750 ","End":"01:23.320","Text":"We apply 4S to our vector x, y,"},{"Start":"01:23.320 ","End":"01:27.995","Text":"z, and then we apply 10T to it and subtract."},{"Start":"01:27.995 ","End":"01:31.290","Text":"How do we do 4S?"},{"Start":"01:31.290 ","End":"01:33.359","Text":"We\u0027ve seen this before,"},{"Start":"01:33.359 ","End":"01:38.840","Text":"4S times something just means apply S to it and then multiply by 4."},{"Start":"01:38.840 ","End":"01:42.470","Text":"This is what we get, 4 times S of x, y, z,"},{"Start":"01:42.470 ","End":"01:45.380","Text":"and here minus 10,"},{"Start":"01:45.380 ","End":"01:47.960","Text":"and then T of x, y, z,"},{"Start":"01:47.960 ","End":"01:49.505","Text":"which is this."},{"Start":"01:49.505 ","End":"01:51.960","Text":"Yeah, here\u0027s the formula."},{"Start":"01:53.840 ","End":"02:01.039","Text":"Now, here we get stuck because this is a 2D vector and this is a 3D vector."},{"Start":"02:01.039 ","End":"02:04.320","Text":"You know we could have gone 1 more step."},{"Start":"02:04.320 ","End":"02:09.510","Text":"We could have said that this is 4x minus 4z, 4y."},{"Start":"02:09.510 ","End":"02:15.470","Text":"The second part we could have said"},{"Start":"02:15.470 ","End":"02:23.000","Text":"is 10x comma 40x minus 10y from here,"},{"Start":"02:23.000 ","End":"02:31.580","Text":"and then 10 times this is 10x plus 40y minus 10z."},{"Start":"02:31.580 ","End":"02:34.460","Text":"The main thing to notice is not the actual expressions,"},{"Start":"02:34.460 ","End":"02:38.015","Text":"but that here it\u0027s a 2D vector,"},{"Start":"02:38.015 ","End":"02:39.890","Text":"and here it\u0027s a 3D vector."},{"Start":"02:39.890 ","End":"02:41.610","Text":"There\u0027s 2 commas in it,"},{"Start":"02:41.610 ","End":"02:43.215","Text":"there\u0027s 3 components,"},{"Start":"02:43.215 ","End":"02:46.999","Text":"and this is what is nonsense."},{"Start":"02:46.999 ","End":"02:52.170","Text":"You can\u0027t take a 2D vector and subtract a 3D vector."},{"Start":"02:52.760 ","End":"02:59.520","Text":"We saw this impossibility earlier when we tried to add S plus T."},{"Start":"02:59.520 ","End":"03:02.780","Text":"The rule there applies here also"},{"Start":"03:02.780 ","End":"03:06.950","Text":"is that we can do such additions and combinations"},{"Start":"03:06.950 ","End":"03:11.030","Text":"if we\u0027re going from the same space to the same space,"},{"Start":"03:11.030 ","End":"03:12.530","Text":"but S and T differ."},{"Start":"03:12.530 ","End":"03:15.590","Text":"They start off the same, the same R^3,"},{"Start":"03:15.590 ","End":"03:19.370","Text":"but they go to different places, 1, 2 R^3 and 1, 2 R^2,"},{"Start":"03:19.370 ","End":"03:23.240","Text":"so there\u0027s no reason that it should be possible."},{"Start":"03:23.240 ","End":"03:26.220","Text":"Done with this 1."}],"ID":10190},{"Watched":false,"Name":"Exercise 4","Duration":"4m ","ChapterTopicVideoID":9633,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:04.995","Text":"In this exercise, we have 2 linear transformations."},{"Start":"00:04.995 ","End":"00:13.560","Text":"We have T from R^3 to R^3 and is given by this formula."},{"Start":"00:13.560 ","End":"00:18.945","Text":"We also have S from R^3 to R^2,"},{"Start":"00:18.945 ","End":"00:21.420","Text":"and it\u0027s given by this formula."},{"Start":"00:21.420 ","End":"00:24.075","Text":"Notice 3 components, x, y, z,"},{"Start":"00:24.075 ","End":"00:28.305","Text":"and here 2 components, 2 by 2."},{"Start":"00:28.305 ","End":"00:32.180","Text":"Our task is to find the formula if possible,"},{"Start":"00:32.180 ","End":"00:36.845","Text":"because it might not be, that defines TS."},{"Start":"00:36.845 ","End":"00:42.210","Text":"When we put 2 functions or transformations next to each other,"},{"Start":"00:42.210 ","End":"00:44.390","Text":"we mean function composition."},{"Start":"00:44.390 ","End":"00:51.050","Text":"You can write it also with a little circle T composed S. Actually it means first apply"},{"Start":"00:51.050 ","End":"00:59.230","Text":"S and then apply T. Actually goes from right to left."},{"Start":"00:59.360 ","End":"01:07.330","Text":"For convenience, I just copy the T and S will have them handy."},{"Start":"01:07.330 ","End":"01:13.275","Text":"Now let\u0027s see if we can compute TS of a vector x, y, z,"},{"Start":"01:13.275 ","End":"01:20.945","Text":"or T composed with S. This just means T of S,"},{"Start":"01:20.945 ","End":"01:26.840","Text":"of the innermost 1 is S. S is applied first and then T,"},{"Start":"01:26.840 ","End":"01:28.400","Text":"that\u0027s what I meant over here,"},{"Start":"01:28.400 ","End":"01:32.805","Text":"that S is first and T is second,"},{"Start":"01:32.805 ","End":"01:36.180","Text":"though T and S keep the order TS,"},{"Start":"01:36.180 ","End":"01:38.505","Text":"TS but S is first."},{"Start":"01:38.505 ","End":"01:41.550","Text":"So let\u0027s see, S of x, y, z."},{"Start":"01:41.550 ","End":"01:45.900","Text":"We can refer here and replace S of x,"},{"Start":"01:45.900 ","End":"01:50.550","Text":"y, z by x minus z, y."},{"Start":"01:50.550 ","End":"01:55.425","Text":"Now we have to try and apply T to this."},{"Start":"01:55.425 ","End":"01:59.680","Text":"But this doesn\u0027t make sense because T, yeah,"},{"Start":"01:59.680 ","End":"02:05.680","Text":"you can see it here starts from R^3 and also goes to R^3."},{"Start":"02:05.680 ","End":"02:07.360","Text":"But the point is that T,"},{"Start":"02:07.360 ","End":"02:10.900","Text":"its source or its domain is R^3,"},{"Start":"02:10.900 ","End":"02:14.620","Text":"I don\u0027t know what to do with T of a 2D vector."},{"Start":"02:14.620 ","End":"02:18.205","Text":"It\u0027s not in the formula, it\u0027s meaningless."},{"Start":"02:18.205 ","End":"02:21.670","Text":"Or if you like, call it nonsense."},{"Start":"02:21.690 ","End":"02:26.900","Text":"Now we cannot find a formula for T composed with"},{"Start":"02:26.900 ","End":"02:33.785","Text":"S. Notice though I\u0027m just giving you a bit of extra perspective,"},{"Start":"02:33.785 ","End":"02:38.070","Text":"and even outside of linear algebra in just functions."},{"Start":"02:38.070 ","End":"02:40.760","Text":"If I have 2 functions, f and g,"},{"Start":"02:40.760 ","End":"02:47.690","Text":"and I want to know what is g composed f of something."},{"Start":"02:47.690 ","End":"02:50.015","Text":"For this to make sense,"},{"Start":"02:50.015 ","End":"02:58.345","Text":"the target or range of f has to be the domain or source of g. In other words,"},{"Start":"02:58.345 ","End":"03:05.015","Text":"if f goes from some space x to some space y,"},{"Start":"03:05.015 ","End":"03:09.670","Text":"then g has to go from y to,"},{"Start":"03:09.670 ","End":"03:12.500","Text":"doesn\u0027t matter what are some other space z in"},{"Start":"03:12.500 ","End":"03:15.910","Text":"order to be able to say g of f of something."},{"Start":"03:15.910 ","End":"03:17.925","Text":"To apply, to say,"},{"Start":"03:17.925 ","End":"03:21.840","Text":"g of f of x,"},{"Start":"03:21.840 ","End":"03:24.420","Text":"f of x would be in y,"},{"Start":"03:24.420 ","End":"03:28.490","Text":"and g would have to apply to the space y."},{"Start":"03:28.490 ","End":"03:33.500","Text":"You can only take the composition if they match,"},{"Start":"03:33.500 ","End":"03:40.864","Text":"if this middle space or set is the same,"},{"Start":"03:40.864 ","End":"03:47.440","Text":"that f takes stuff to the place where g takes them from."},{"Start":"03:47.440 ","End":"03:51.470","Text":"I hope that makes some sense."},{"Start":"03:51.470 ","End":"03:54.740","Text":"If not, then just leave it like this."},{"Start":"03:54.740 ","End":"03:56.090","Text":"If you understand this,"},{"Start":"03:56.090 ","End":"04:00.270","Text":"that\u0027s fine. We\u0027re done."}],"ID":10191},{"Watched":false,"Name":"Exercise 5","Duration":"8m 7s","ChapterTopicVideoID":9629,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.865","Text":"In this exercise, we have 2 linear transformations, T and S,"},{"Start":"00:05.865 ","End":"00:12.060","Text":"but T goes from R^3 to R^3 and is given by this formula,"},{"Start":"00:12.060 ","End":"00:18.705","Text":"whereas S goes from R^3 to R^2 and is given by this formula."},{"Start":"00:18.705 ","End":"00:21.585","Text":"Our task is to find a formula,"},{"Start":"00:21.585 ","End":"00:27.960","Text":"if possible because it may not be, that defines ST."},{"Start":"00:27.960 ","End":"00:30.960","Text":"When I just write 2 functions next to each other,"},{"Start":"00:30.960 ","End":"00:35.110","Text":"it means the function composition, S composed with T,"},{"Start":"00:35.110 ","End":"00:39.570","Text":"which means apply T first and then apply S."},{"Start":"00:39.570 ","End":"00:42.090","Text":"Now, in the previous exercise,"},{"Start":"00:42.090 ","End":"00:44.250","Text":"we had to compute TS,"},{"Start":"00:44.250 ","End":"00:47.775","Text":"and we failed, it wasn\u0027t possible."},{"Start":"00:47.775 ","End":"00:50.580","Text":"What we said there,"},{"Start":"00:50.580 ","End":"00:54.965","Text":"basically, is that if we\u0027re going to apply first T then S,"},{"Start":"00:54.965 ","End":"00:59.585","Text":"the destination of T,"},{"Start":"00:59.585 ","End":"01:05.245","Text":"T goes from something to R^3,"},{"Start":"01:05.245 ","End":"01:07.815","Text":"doesn\u0027t matter where it comes from,"},{"Start":"01:07.815 ","End":"01:14.460","Text":"and S starts from R^3 and goes somewhere else."},{"Start":"01:14.460 ","End":"01:15.480","Text":"That was to be R^2,"},{"Start":"01:15.480 ","End":"01:19.160","Text":"but that\u0027s not what I want to emphasize."},{"Start":"01:19.160 ","End":"01:21.920","Text":"As long as the destination of T or"},{"Start":"01:21.920 ","End":"01:26.060","Text":"the target of T is the same as the source or domain of S,"},{"Start":"01:26.060 ","End":"01:30.475","Text":"then we\u0027ll be okay because T will take us somewhere,"},{"Start":"01:30.475 ","End":"01:34.655","Text":"and then we\u0027ll be able to apply S because it\u0027s compatible."},{"Start":"01:34.655 ","End":"01:42.870","Text":"In this case, it\u0027s okay because T goes to R^3 and S goes from R^3."},{"Start":"01:42.870 ","End":"01:49.430","Text":"Let\u0027s actually do it and compute ST. For convenience,"},{"Start":"01:49.430 ","End":"01:53.510","Text":"I copied T and S again with a slight difference."},{"Start":"01:53.510 ","End":"01:58.190","Text":"T, I kept as is with x, y, z,"},{"Start":"01:58.190 ","End":"02:02.070","Text":"but I wanted to use different letters for S,"},{"Start":"02:02.070 ","End":"02:05.750","Text":"so we replace x, y, z with A, B, C."},{"Start":"02:05.750 ","End":"02:10.710","Text":"This will make it easier to understand when we take the composition."},{"Start":"02:10.840 ","End":"02:15.680","Text":"We want to compute ST of x, y, z,"},{"Start":"02:15.680 ","End":"02:18.380","Text":"ST meaning function composition,"},{"Start":"02:18.380 ","End":"02:22.910","Text":"means S of T of x, y, z."},{"Start":"02:22.910 ","End":"02:27.560","Text":"We first apply T. Now, from here to here,"},{"Start":"02:27.560 ","End":"02:32.465","Text":"we just replace T of x, y, z from its formula."},{"Start":"02:32.465 ","End":"02:34.490","Text":"The bracket is optional."},{"Start":"02:34.490 ","End":"02:39.100","Text":"I mean, we could keep the parenthesis in, not necessary."},{"Start":"02:39.100 ","End":"02:41.910","Text":"Now, we have the formula for S."},{"Start":"02:41.910 ","End":"02:48.420","Text":"This is A, this is B, and this is C."},{"Start":"02:48.420 ","End":"02:53.175","Text":"What we want is A minus C, B."},{"Start":"02:53.175 ","End":"02:57.975","Text":"We need x minus x plus 4y minus z,"},{"Start":"02:57.975 ","End":"03:02.325","Text":"comes out to z minus 4y,"},{"Start":"03:02.325 ","End":"03:08.830","Text":"and the second component B is just the 4x minus y."},{"Start":"03:09.260 ","End":"03:12.030","Text":"We have the formula for ST,"},{"Start":"03:12.030 ","End":"03:14.960","Text":"but I feel the answer is not quite complete."},{"Start":"03:14.960 ","End":"03:22.115","Text":"Just like we said that T goes from R^3 to R^3 and S goes from R^3 to R^2,"},{"Start":"03:22.115 ","End":"03:28.865","Text":"I\u0027d like to say what ST goes from and to."},{"Start":"03:28.865 ","End":"03:33.335","Text":"If we look at this diagram here,"},{"Start":"03:33.335 ","End":"03:44.415","Text":"T goes from R^3 to R^3 and S goes from R^3 to R^2."},{"Start":"03:44.415 ","End":"03:52.200","Text":"If we compose them as composed with T or just write it as ST,"},{"Start":"03:52.200 ","End":"03:56.080","Text":"it goes from R^3 to R^2."},{"Start":"03:57.890 ","End":"04:01.545","Text":"That\u0027s just for completeness."},{"Start":"04:01.545 ","End":"04:04.210","Text":"There\u0027s another thing I wanted to say,"},{"Start":"04:04.210 ","End":"04:10.290","Text":"is that there is a technique for doing the computation."},{"Start":"04:10.690 ","End":"04:17.015","Text":"Just like here, we replace the x, y, z with A, B, C,"},{"Start":"04:17.015 ","End":"04:20.375","Text":"and we did some calculations here."},{"Start":"04:20.375 ","End":"04:23.450","Text":"There is a more standard way of doing the computation of"},{"Start":"04:23.450 ","End":"04:27.990","Text":"a composition and that\u0027s what I\u0027d like to show you now."},{"Start":"04:28.180 ","End":"04:33.245","Text":"Here, again, T and S started a new page."},{"Start":"04:33.245 ","End":"04:40.133","Text":"We\u0027re going to use matrices to compute ST."},{"Start":"04:43.133 ","End":"04:45.800","Text":"We\u0027ll start with S."},{"Start":"04:45.800 ","End":"04:48.350","Text":"Notice that S is given as this formula."},{"Start":"04:48.350 ","End":"04:49.490","Text":"Now, the first component"},{"Start":"04:49.490 ","End":"04:58.240","Text":"x minus z could be written as 1x plus 0y minus 1z"},{"Start":"04:58.240 ","End":"05:04.005","Text":"and the second component y could be written as"},{"Start":"05:04.005 ","End":"05:09.930","Text":"0x plus 1y plus 0z."},{"Start":"05:09.930 ","End":"05:11.540","Text":"Now, if I take these numbers,"},{"Start":"05:11.540 ","End":"05:13.100","Text":"1, 0, minus 1,"},{"Start":"05:13.100 ","End":"05:17.350","Text":"0, 1, 0 and put them in a matrix,"},{"Start":"05:17.350 ","End":"05:21.620","Text":"turns out that S of x, y, z"},{"Start":"05:21.620 ","End":"05:28.310","Text":"will exactly be the matrix multiplication of this matrix"},{"Start":"05:28.310 ","End":"05:32.270","Text":"1, 0, minus 1, 0, 1, 0 with x, y, z."},{"Start":"05:32.270 ","End":"05:34.820","Text":"Look, if I take this row with this column,"},{"Start":"05:34.820 ","End":"05:38.985","Text":"I get x minus z like here."},{"Start":"05:38.985 ","End":"05:42.450","Text":"This row with this column gives us y."},{"Start":"05:42.450 ","End":"05:45.550","Text":"It gives us this as a column vector."},{"Start":"05:45.550 ","End":"05:47.985","Text":"Similarly for T,"},{"Start":"05:47.985 ","End":"05:49.590","Text":"I can also get a matrix."},{"Start":"05:49.590 ","End":"05:51.765","Text":"This time it\u0027ll be a 3 by 3."},{"Start":"05:51.765 ","End":"05:55.800","Text":"x is just 1x plus 0y plus 0z,"},{"Start":"05:55.800 ","End":"06:02.895","Text":"the second component is 4x minus 1y plus 0z,"},{"Start":"06:02.895 ","End":"06:09.240","Text":"and the last one is 1x plus 4y minus 1z."},{"Start":"06:09.240 ","End":"06:13.770","Text":"I have these 2 matrices that describe S and T."},{"Start":"06:13.770 ","End":"06:20.030","Text":"Now, here\u0027s how we compute ST of x, y, z."},{"Start":"06:20.030 ","End":"06:26.160","Text":"We simply take the matrix for S,"},{"Start":"06:26.160 ","End":"06:28.410","Text":"the matrix for T,"},{"Start":"06:28.410 ","End":"06:30.620","Text":"and multiply them together,"},{"Start":"06:30.620 ","End":"06:33.380","Text":"and apply that to x, y, z."},{"Start":"06:33.380 ","End":"06:39.370","Text":"What we need to do is a matrix multiplication of this times this,"},{"Start":"06:39.370 ","End":"06:42.335","Text":"and this times this gives the result of this."},{"Start":"06:42.335 ","End":"06:47.480","Text":"For example, 1, 0, minus 1 with 1, 4, 1"},{"Start":"06:47.480 ","End":"06:50.750","Text":"gives, let\u0027s see,"},{"Start":"06:50.750 ","End":"06:54.000","Text":"1 times 1, 0 times 4,"},{"Start":"06:54.000 ","End":"06:56.115","Text":"minus 1 times 1 gives 0."},{"Start":"06:56.115 ","End":"07:01.310","Text":"This with this gives minus 4 and so on."},{"Start":"07:01.310 ","End":"07:06.810","Text":"You can check if the product of this matrix with this matrix is this matrix."},{"Start":"07:08.210 ","End":"07:11.465","Text":"If we multiply it out,"},{"Start":"07:11.465 ","End":"07:16.180","Text":"we\u0027ll see that this with this gives minus 4y plus z,"},{"Start":"07:16.180 ","End":"07:19.035","Text":"this with this gives 4x minus y,"},{"Start":"07:19.035 ","End":"07:20.810","Text":"and this is exactly what we got"},{"Start":"07:20.810 ","End":"07:25.920","Text":"except as a row matrix earlier for ST of x, y, z."},{"Start":"07:25.920 ","End":"07:33.735","Text":"I recall we had minus 4y plus z, 4x minus y."},{"Start":"07:33.735 ","End":"07:37.360","Text":"Here, same thing, just a column vector."},{"Start":"07:39.560 ","End":"07:42.060","Text":"I\u0027ll just copy the result,"},{"Start":"07:42.060 ","End":"07:44.775","Text":"ST of x, y, z is this,"},{"Start":"07:44.775 ","End":"07:46.605","Text":"put a box around it."},{"Start":"07:46.605 ","End":"07:49.070","Text":"Like I said, it\u0027s the same as what we had before"},{"Start":"07:49.070 ","End":"07:52.220","Text":"except that here we\u0027re working on column vectors"},{"Start":"07:52.220 ","End":"07:55.070","Text":"and previously we used row vectors,"},{"Start":"07:55.070 ","End":"08:04.240","Text":"and the way we got this was by doing a matrix multiplication."},{"Start":"08:04.880 ","End":"08:08.470","Text":"That\u0027s it for this question."}],"ID":10192},{"Watched":false,"Name":"Exercise 6","Duration":"3m 51s","ChapterTopicVideoID":9625,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"In this exercise,"},{"Start":"00:01.620 ","End":"00:05.745","Text":"we have a linear transformation T from R^3 to R^3,"},{"Start":"00:05.745 ","End":"00:10.290","Text":"and it\u0027s given by this formula, T of x, y, z"},{"Start":"00:10.290 ","End":"00:17.445","Text":"and got 3 components here, x, 4x minus y, x plus 4y minus z."},{"Start":"00:17.445 ","End":"00:21.540","Text":"Our task is to find a formula if possible,"},{"Start":"00:21.540 ","End":"00:24.720","Text":"for the transformation T squared,"},{"Start":"00:24.720 ","End":"00:29.475","Text":"which means T times T or T composed with T rather."},{"Start":"00:29.475 ","End":"00:32.670","Text":"For convenience I\u0027d also like to write T"},{"Start":"00:32.670 ","End":"00:36.300","Text":"in terms of 3 other dummy variables, x, y, and z,"},{"Start":"00:36.300 ","End":"00:38.430","Text":"just any 3 letters really."},{"Start":"00:38.430 ","End":"00:42.480","Text":"We could write T of A, B, C and replace x, y, z by A, B, C."},{"Start":"00:42.480 ","End":"00:46.180","Text":"You\u0027ll see why I did this in a moment."},{"Start":"00:46.340 ","End":"00:51.870","Text":"Let\u0027s see what does T squared due to x, y, z the vector."},{"Start":"00:51.870 ","End":"00:55.155","Text":"T squared means T of T of."},{"Start":"00:55.155 ","End":"00:58.290","Text":"Now we already have T of x, y, z."},{"Start":"00:58.290 ","End":"01:03.180","Text":"Just have to replace this by this,"},{"Start":"01:03.180 ","End":"01:09.135","Text":"and that gives us T of this vector."},{"Start":"01:09.135 ","End":"01:13.890","Text":"At this point, I\u0027m going to use this formula with A, B, C."},{"Start":"01:13.890 ","End":"01:20.610","Text":"The x will be my A, 4x minus y will B, x plus 4y minus z will be C."},{"Start":"01:20.610 ","End":"01:25.260","Text":"Then to substitute it in here, sorry in here."},{"Start":"01:25.260 ","End":"01:28.200","Text":"This is what we get, A is just x,"},{"Start":"01:28.200 ","End":"01:32.250","Text":"4A minus B is 4 times this minus this, and so on."},{"Start":"01:32.250 ","End":"01:36.285","Text":"A plus 4B minus C is this,"},{"Start":"01:36.285 ","End":"01:38.925","Text":"just need to simplify a bit,"},{"Start":"01:38.925 ","End":"01:41.400","Text":"and this is the answer."},{"Start":"01:41.400 ","End":"01:46.620","Text":"This is the linear transformation T squared."},{"Start":"01:46.620 ","End":"01:50.445","Text":"What it does to x, y, z is it sends it to this."},{"Start":"01:50.445 ","End":"01:53.910","Text":"I\u0027d like to show you another way of doing this."},{"Start":"01:53.910 ","End":"01:59.460","Text":"That\u0027s very useful then you can choose which way you like."},{"Start":"01:59.460 ","End":"02:03.795","Text":"Here again is T of x, y, z,"},{"Start":"02:03.795 ","End":"02:06.050","Text":"and with this method,"},{"Start":"02:06.050 ","End":"02:09.025","Text":"I write it in matrix form."},{"Start":"02:09.025 ","End":"02:11.145","Text":"It comes out to be this."},{"Start":"02:11.145 ","End":"02:16.340","Text":"Although, this is a column vector will just be a bit loose with row vectors,"},{"Start":"02:16.340 ","End":"02:19.294","Text":"column vectors not be too fussy."},{"Start":"02:19.294 ","End":"02:23.435","Text":"I mean, this row with this column gives us x."},{"Start":"02:23.435 ","End":"02:27.335","Text":"This with this gives 4x minus y, which is this."},{"Start":"02:27.335 ","End":"02:31.290","Text":"Strictly speaking, we get this as a column vector."},{"Start":"02:31.790 ","End":"02:37.015","Text":"The transformation T is represented by this matrix."},{"Start":"02:37.015 ","End":"02:43.400","Text":"Applying T to a vector is like multiplying a column vector by this matrix."},{"Start":"02:43.400 ","End":"02:47.000","Text":"Now if I want to apply T twice, T of T,"},{"Start":"02:47.000 ","End":"02:52.160","Text":"it just means that I have to multiply by this matrix once, twice."},{"Start":"02:52.160 ","End":"02:55.580","Text":"In other words, by this matrix squared."},{"Start":"02:55.580 ","End":"02:59.915","Text":"That will be the matrix that will correspond to the transformation T squared."},{"Start":"02:59.915 ","End":"03:03.455","Text":"Now it\u0027s not difficult to square this matrix."},{"Start":"03:03.455 ","End":"03:04.970","Text":"I\u0027ll spare you the details,"},{"Start":"03:04.970 ","End":"03:07.100","Text":"just give you the answer."},{"Start":"03:07.100 ","End":"03:09.785","Text":"Comes out to be this."},{"Start":"03:09.785 ","End":"03:15.470","Text":"I\u0027m sure you could try it and check that this times itself is this."},{"Start":"03:15.470 ","End":"03:21.020","Text":"Which means, if you do the product and go back to row matrix"},{"Start":"03:21.020 ","End":"03:27.170","Text":"1, 0, 0 with x, y, z is just x. 0, 1, 0 with x, y, z is just y,"},{"Start":"03:27.170 ","End":"03:31.830","Text":"and this with this gives me 16x minus 8y plus z."},{"Start":"03:32.660 ","End":"03:36.590","Text":"This gives us the same answer as we got before."},{"Start":"03:36.590 ","End":"03:38.285","Text":"If you scroll back, what do we get?"},{"Start":"03:38.285 ","End":"03:43.475","Text":"X, y, 16x minus 8y plus z, go back there we are."},{"Start":"03:43.475 ","End":"03:47.040","Text":"Same thing, same answer, different method."},{"Start":"03:47.040 ","End":"03:48.020","Text":"There you are."},{"Start":"03:48.020 ","End":"03:52.050","Text":"We have 2 methods of doing this and we\u0027re done."}],"ID":10193},{"Watched":false,"Name":"Exercise 7","Duration":"7m 7s","ChapterTopicVideoID":9626,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.980","Text":"In this exercise, we have a transformation T from R^3 to R^3,"},{"Start":"00:06.980 ","End":"00:11.745","Text":"a linear transformation, and it\u0027s given with this formula."},{"Start":"00:11.745 ","End":"00:16.365","Text":"Our task is to find the formula, if possible,"},{"Start":"00:16.365 ","End":"00:23.930","Text":"to define the inverse of T. We\u0027re assuming that T does have an inverse and if it doesn\u0027t,"},{"Start":"00:23.930 ","End":"00:26.710","Text":"we\u0027ll soon find out that it doesn\u0027t."},{"Start":"00:26.710 ","End":"00:32.730","Text":"Now, T inverse or we write it T to the minus 1,"},{"Start":"00:32.730 ","End":"00:38.180","Text":"what this means for it to be an inverse is that if T takes a vector x,"},{"Start":"00:38.180 ","End":"00:40.660","Text":"y, z to r, s, t,"},{"Start":"00:40.660 ","End":"00:44.625","Text":"the source and the destination are both R^3,"},{"Start":"00:44.625 ","End":"00:47.790","Text":"then T^1 will do the reverse job,"},{"Start":"00:47.790 ","End":"00:49.580","Text":"it will take r, s,"},{"Start":"00:49.580 ","End":"00:53.090","Text":"t and give us back x, y, z."},{"Start":"00:53.090 ","End":"00:55.835","Text":"That\u0027s what T inverse will do."},{"Start":"00:55.835 ","End":"00:59.630","Text":"Now we have the formula for T of x, y, z,"},{"Start":"00:59.630 ","End":"01:01.610","Text":"which is this,"},{"Start":"01:01.610 ","End":"01:05.990","Text":"so what we do is equate this to r, s,"},{"Start":"01:05.990 ","End":"01:08.720","Text":"t. When we equate vectors,"},{"Start":"01:08.720 ","End":"01:10.700","Text":"we equate them component-wise,"},{"Start":"01:10.700 ","End":"01:13.670","Text":"each of them is a vector in R^3,"},{"Start":"01:13.670 ","End":"01:18.979","Text":"so we get 3 equations or if you like a system of linear equations,"},{"Start":"01:18.979 ","End":"01:20.060","Text":"x is equal to r,"},{"Start":"01:20.060 ","End":"01:23.570","Text":"4x minus y is s, and so on."},{"Start":"01:23.570 ","End":"01:27.170","Text":"The r, s, t,"},{"Start":"01:27.170 ","End":"01:33.635","Text":"here are like given and we\u0027ve got to find x, y,"},{"Start":"01:33.635 ","End":"01:35.360","Text":"and z in terms of r,"},{"Start":"01:35.360 ","End":"01:39.065","Text":"s, and t. Now,"},{"Start":"01:39.065 ","End":"01:43.070","Text":"this system is not exactly in row echelon form,"},{"Start":"01:43.070 ","End":"01:46.939","Text":"it kind of just from top-down instead of bottom-up."},{"Start":"01:46.939 ","End":"01:50.290","Text":"We can still use method of,"},{"Start":"01:50.290 ","End":"01:55.080","Text":"and it\u0027s not going to be a reverse substitution it\u0027s going to be a forward substitution."},{"Start":"01:55.080 ","End":"01:57.360","Text":"Look, x is already equal to r,"},{"Start":"01:57.360 ","End":"01:58.870","Text":"we have that already,"},{"Start":"01:58.870 ","End":"02:01.745","Text":"and then we can get y from here."},{"Start":"02:01.745 ","End":"02:05.115","Text":"Y is, what would that be?"},{"Start":"02:05.115 ","End":"02:08.900","Text":"4x minus s, but we have x from here,"},{"Start":"02:08.900 ","End":"02:12.370","Text":"so it\u0027s 4r minus s. Then in the next 1,"},{"Start":"02:12.370 ","End":"02:18.240","Text":"we\u0027ve got z as equaling x plus 4y minus t,"},{"Start":"02:18.240 ","End":"02:26.240","Text":"but we already have x and y so we plug those in so that we get z in terms of r, s,"},{"Start":"02:26.240 ","End":"02:29.190","Text":"t. We\u0027ve got x, y,"},{"Start":"02:29.190 ","End":"02:33.280","Text":"and z in terms of r s, t,"},{"Start":"02:33.280 ","End":"02:35.375","Text":"which means that we can write this,"},{"Start":"02:35.375 ","End":"02:37.100","Text":"that T inverse of r,"},{"Start":"02:37.100 ","End":"02:38.675","Text":"s, t is,"},{"Start":"02:38.675 ","End":"02:40.910","Text":"the x is here, the y,"},{"Start":"02:40.910 ","End":"02:42.350","Text":"4r minus s is here,"},{"Start":"02:42.350 ","End":"02:45.365","Text":"and the z from here is here."},{"Start":"02:45.365 ","End":"02:49.820","Text":"But now we can change the letters that we use."},{"Start":"02:49.820 ","End":"02:51.890","Text":"We don\u0027t need r, s, and t anymore,"},{"Start":"02:51.890 ","End":"02:55.760","Text":"we can go back to x, y, z."},{"Start":"02:55.760 ","End":"02:58.350","Text":"They\u0027re just dummy variables there,"},{"Start":"02:58.350 ","End":"03:00.710","Text":"but it\u0027s customary to use x,"},{"Start":"03:00.710 ","End":"03:02.675","Text":"y, z where possible,"},{"Start":"03:02.675 ","End":"03:05.570","Text":"and r, s, t was an intermediate phase."},{"Start":"03:05.570 ","End":"03:08.455","Text":"Now we have T inverse,"},{"Start":"03:08.455 ","End":"03:11.765","Text":"and it might be a good idea to do a check."},{"Start":"03:11.765 ","End":"03:14.840","Text":"Let\u0027s just take, for example,"},{"Start":"03:14.840 ","End":"03:17.540","Text":"the vector 1, 2, 3,"},{"Start":"03:17.540 ","End":"03:21.945","Text":"and apply T to it so the formula for T,"},{"Start":"03:21.945 ","End":"03:25.670","Text":"well, it\u0027s just scrolled off somewhere."},{"Start":"03:25.670 ","End":"03:30.680","Text":"Let\u0027s see now, where is it? Here it is."},{"Start":"03:30.680 ","End":"03:33.320","Text":"If we apply T to 1, 2, 3,"},{"Start":"03:33.320 ","End":"03:36.865","Text":"then we will get 1, 2, 6."},{"Start":"03:36.865 ","End":"03:41.900","Text":"If that\u0027s correct, if we apply T inverse to 1,"},{"Start":"03:41.900 ","End":"03:45.430","Text":"2, 6, we should get back the 1, 2, 3."},{"Start":"03:45.430 ","End":"03:47.550","Text":"We take this 1, 2, 6,"},{"Start":"03:47.550 ","End":"03:49.850","Text":"and plug it in here and let\u0027s see."},{"Start":"03:49.850 ","End":"03:52.445","Text":"The first component is x,"},{"Start":"03:52.445 ","End":"03:54.340","Text":"which is the 1."},{"Start":"03:54.340 ","End":"03:57.540","Text":"Next, we have 4x minus y,"},{"Start":"03:57.540 ","End":"04:01.755","Text":"4 times 1 minus 2 is 2,"},{"Start":"04:01.755 ","End":"04:04.240","Text":"and the last 1,"},{"Start":"04:04.550 ","End":"04:14.600","Text":"17x minus 4y minus z is 17 minus 8 minus 6 is 3 so all is well,"},{"Start":"04:14.600 ","End":"04:16.250","Text":"this really is the inverse,"},{"Start":"04:16.250 ","End":"04:21.820","Text":"or at least we\u0027ve done 1 check and it\u0027s probably right."},{"Start":"04:24.220 ","End":"04:26.990","Text":"We have solved the question,"},{"Start":"04:26.990 ","End":"04:31.855","Text":"but I\u0027d like to show you another way with matrices."},{"Start":"04:31.855 ","End":"04:34.890","Text":"Let\u0027s start again with T of x,"},{"Start":"04:34.890 ","End":"04:37.850","Text":"y, z, this was the formula that we had."},{"Start":"04:37.850 ","End":"04:41.135","Text":"Now let\u0027s write this in matrix form."},{"Start":"04:41.135 ","End":"04:47.770","Text":"T of x, y, z would be the product of this matrix with this vector."},{"Start":"04:47.770 ","End":"04:50.750","Text":"Now where did I get these numbers from?"},{"Start":"04:50.750 ","End":"04:55.830","Text":"X is like 1x plus 0 y plus 0 z,"},{"Start":"04:55.830 ","End":"04:57.480","Text":"it\u0027s like 1, 0, 0."},{"Start":"04:57.480 ","End":"04:59.850","Text":"Here I have 4 minus 1,"},{"Start":"04:59.850 ","End":"05:01.895","Text":"0, that\u0027s the second row."},{"Start":"05:01.895 ","End":"05:04.820","Text":"Here I have 1, 4 minus 1,"},{"Start":"05:04.820 ","End":"05:06.610","Text":"that\u0027s the third row."},{"Start":"05:06.610 ","End":"05:11.540","Text":"Now, in these computations we\u0027ll lose"},{"Start":"05:11.540 ","End":"05:16.009","Text":"the tolerant with the vertical or horizontal vectors."},{"Start":"05:16.009 ","End":"05:23.645","Text":"Strictly speaking, a vertical vector would give us this in column vector form."},{"Start":"05:23.645 ","End":"05:27.470","Text":"But we\u0027re not going to be too fussy or too pedantic."},{"Start":"05:27.470 ","End":"05:29.810","Text":"I could have written this also as x, y,"},{"Start":"05:29.810 ","End":"05:34.145","Text":"z as a column vector also anyway."},{"Start":"05:34.145 ","End":"05:38.720","Text":"This is T, and the way we compute"},{"Start":"05:38.720 ","End":"05:43.970","Text":"the inverse transformation is simply to take the inverse matrix,"},{"Start":"05:43.970 ","End":"05:48.184","Text":"so I need to take this matrix and compute its inverse."},{"Start":"05:48.184 ","End":"05:51.800","Text":"Now it\u0027s quite tedious to"},{"Start":"05:51.800 ","End":"05:55.685","Text":"compute an inverse and this is not what we\u0027re studying at the moment,"},{"Start":"05:55.685 ","End":"06:01.160","Text":"so I\u0027m just going to give you the inverse."},{"Start":"06:01.160 ","End":"06:05.510","Text":"This is not the inverse, it\u0027s from here to here,"},{"Start":"06:05.510 ","End":"06:06.680","Text":"this is the inverse,"},{"Start":"06:06.680 ","End":"06:10.790","Text":"the minus 1 of this is this."},{"Start":"06:10.790 ","End":"06:13.835","Text":"Like I said, it requires some computation"},{"Start":"06:13.835 ","End":"06:17.285","Text":"and I\u0027m just going to leave that and leave you to do it."},{"Start":"06:17.285 ","End":"06:23.060","Text":"You could of course check that this matrix times this matrix is the identity matrix,"},{"Start":"06:23.060 ","End":"06:24.800","Text":"whichever way you multiply it,"},{"Start":"06:24.800 ","End":"06:27.235","Text":"and that would be at least a check."},{"Start":"06:27.235 ","End":"06:30.440","Text":"Now that we have the inverse matrix,"},{"Start":"06:30.440 ","End":"06:33.230","Text":"we can now just multiply out 1,"},{"Start":"06:33.230 ","End":"06:34.760","Text":"0, 0 times x, y,"},{"Start":"06:34.760 ","End":"06:37.610","Text":"z is just x and this with this is"},{"Start":"06:37.610 ","End":"06:44.070","Text":"4x minus y and this with this gives 17x minus 4y minus z."},{"Start":"06:44.070 ","End":"06:46.370","Text":"If you go back to the previous method,"},{"Start":"06:46.370 ","End":"06:48.020","Text":"this is exactly what we got,"},{"Start":"06:48.020 ","End":"06:51.350","Text":"only we wrote it as a row vector and not a column vector,"},{"Start":"06:51.350 ","End":"06:53.720","Text":"and like I said, we\u0027re not being too fussy about that."},{"Start":"06:53.720 ","End":"06:57.375","Text":"This is another way of doing the problem,"},{"Start":"06:57.375 ","End":"07:01.265","Text":"figuring out the matrix and taking the inverse of"},{"Start":"07:01.265 ","End":"07:07.320","Text":"the matrix to represent the inverse transformation. We\u0027re done."}],"ID":10194},{"Watched":false,"Name":"Exercise 8","Duration":"5m 25s","ChapterTopicVideoID":9627,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.430","Text":"In this exercise, we\u0027re given the linear transformation T from R^3 to R^3,"},{"Start":"00:08.430 ","End":"00:11.925","Text":"given by this formula."},{"Start":"00:11.925 ","End":"00:20.715","Text":"Our task is to find a formula for the transformation T to the minus 2."},{"Start":"00:20.715 ","End":"00:22.200","Text":"Now what does this mean,"},{"Start":"00:22.200 ","End":"00:24.345","Text":"T to the minus 2?"},{"Start":"00:24.345 ","End":"00:26.340","Text":"There\u0027s a couple of ways of defining it."},{"Start":"00:26.340 ","End":"00:34.690","Text":"I can either say the inverse of T squared or I can say it\u0027s T inverse times T inverse."},{"Start":"00:35.570 ","End":"00:38.175","Text":"We\u0027ll stick with 1 of them,"},{"Start":"00:38.175 ","End":"00:41.325","Text":"say it\u0027s T inverse times T inverse."},{"Start":"00:41.325 ","End":"00:46.490","Text":"Now we already computed T inverse in a previous exercise,"},{"Start":"00:46.490 ","End":"00:49.740","Text":"so no need to do that again."},{"Start":"00:49.990 ","End":"00:55.670","Text":"It actually looks pretty similar to T at least in the first 2 components,"},{"Start":"00:55.670 ","End":"00:58.010","Text":"but the last 1 is quite different,"},{"Start":"00:58.010 ","End":"00:59.495","Text":"there\u0027s a 17 here."},{"Start":"00:59.495 ","End":"01:02.310","Text":"Anyway, we did this before."},{"Start":"01:02.320 ","End":"01:04.865","Text":"Now I\u0027d like to write this again,"},{"Start":"01:04.865 ","End":"01:07.010","Text":"but with A, B, C instead of x, y,"},{"Start":"01:07.010 ","End":"01:11.630","Text":"z. I wrote the wrong kind of brackets."},{"Start":"01:11.630 ","End":"01:15.290","Text":"I wrote round brackets instead of square brackets."},{"Start":"01:15.290 ","End":"01:18.210","Text":"That\u0027s easily fixed, sorry."},{"Start":"01:18.350 ","End":"01:20.940","Text":"Instead of x, y, z, I\u0027m putting A,"},{"Start":"01:20.940 ","End":"01:22.610","Text":"B, C everywhere."},{"Start":"01:22.610 ","End":"01:28.500","Text":"You\u0027ll see why I do this in a moment, it\u0027s just convenient."},{"Start":"01:29.900 ","End":"01:32.130","Text":"Now I\u0027m going to do it."},{"Start":"01:32.130 ","End":"01:38.280","Text":"I said before is that T to the minus 2 means T inverse of T inverse."},{"Start":"01:38.280 ","End":"01:44.270","Text":"It\u0027s pretty much like the rules of exponents in regular algebra."},{"Start":"01:44.270 ","End":"01:46.430","Text":"Transformations pretty much follow"},{"Start":"01:46.430 ","End":"01:49.640","Text":"the rules that T to the negative 2 is T to the negative 1,"},{"Start":"01:49.640 ","End":"01:51.080","Text":"T to the negative 1."},{"Start":"01:51.080 ","End":"01:54.610","Text":"Product really means composition of functions."},{"Start":"01:54.610 ","End":"01:58.590","Text":"Now replace T inverse of x, y,"},{"Start":"01:58.590 ","End":"02:06.555","Text":"z, i.e this by what it\u0027s equal to, which is this."},{"Start":"02:06.555 ","End":"02:12.410","Text":"Now I need to apply T inverse to this and that\u0027s what I did this with the ABC."},{"Start":"02:12.410 ","End":"02:14.330","Text":"This I\u0027ll call A,"},{"Start":"02:14.330 ","End":"02:15.860","Text":"this I\u0027ll call B,"},{"Start":"02:15.860 ","End":"02:18.305","Text":"and the third component called C,"},{"Start":"02:18.305 ","End":"02:25.020","Text":"and then we can substitute like instead of A,"},{"Start":"02:26.290 ","End":"02:28.370","Text":"here it is written out,"},{"Start":"02:28.370 ","End":"02:30.860","Text":"and you\u0027ll see that A is just A."},{"Start":"02:30.860 ","End":"02:33.050","Text":"Then what did we have?"},{"Start":"02:33.050 ","End":"02:36.740","Text":"We have 4A minus B,"},{"Start":"02:36.740 ","End":"02:40.370","Text":"so it\u0027s 4x minus y."},{"Start":"02:40.370 ","End":"02:47.720","Text":"The last 1, 17A minus 4B minus C, this."},{"Start":"02:47.720 ","End":"02:50.155","Text":"Now we just need to simplify,"},{"Start":"02:50.155 ","End":"02:53.920","Text":"and it boils down to this."},{"Start":"02:53.920 ","End":"02:59.135","Text":"Just write this equals this and put it in a nice box and that\u0027s the answer,"},{"Start":"02:59.135 ","End":"03:03.080","Text":"but don\u0027t go because I want to show you how we do"},{"Start":"03:03.080 ","End":"03:09.960","Text":"it another way with multiplication or inverse of matrices."},{"Start":"03:10.580 ","End":"03:15.845","Text":"Here we are again with the formula for T inverse,"},{"Start":"03:15.845 ","End":"03:18.990","Text":"which we had already."},{"Start":"03:20.180 ","End":"03:23.670","Text":"Here it is in matrix form."},{"Start":"03:23.670 ","End":"03:26.220","Text":"We\u0027ll use column vectors x,"},{"Start":"03:26.220 ","End":"03:28.865","Text":"y, z, the column vector is here."},{"Start":"03:28.865 ","End":"03:33.810","Text":"The way we get the entries in this matrix is x would be like 1,"},{"Start":"03:33.810 ","End":"03:36.570","Text":"0, 0, as far as x, y, z goes."},{"Start":"03:36.570 ","End":"03:38.280","Text":"Here we have 4x minus y,"},{"Start":"03:38.280 ","End":"03:41.205","Text":"so the coefficients are 4 minus 1, 0."},{"Start":"03:41.205 ","End":"03:45.850","Text":"For this 1, 17 minus 4 minus 1."},{"Start":"03:46.520 ","End":"03:49.050","Text":"If you multiply this out,"},{"Start":"03:49.050 ","End":"03:50.860","Text":"it will give you the same as this,"},{"Start":"03:50.860 ","End":"03:52.835","Text":"but as a column vector."},{"Start":"03:52.835 ","End":"04:00.500","Text":"Now, T to the minus 2 is the square of T inverse."},{"Start":"04:00.500 ","End":"04:03.010","Text":"Like I said, mostly these transformations,"},{"Start":"04:03.010 ","End":"04:07.599","Text":"the exponents and negative exponents work pretty much like an algebra."},{"Start":"04:07.599 ","End":"04:12.100","Text":"Just like we have A to the minus 2 is A to the minus 1"},{"Start":"04:12.100 ","End":"04:17.720","Text":"squared and works pretty much the same with transformations."},{"Start":"04:18.110 ","End":"04:24.510","Text":"T to the power of minus 2 would be this thing squared times x,"},{"Start":"04:24.510 ","End":"04:26.935","Text":"y, z, this matrix."},{"Start":"04:26.935 ","End":"04:31.310","Text":"I\u0027ll just give you the answer for this matrix squared,"},{"Start":"04:31.310 ","End":"04:35.035","Text":"won\u0027t waste time on the actual computations."},{"Start":"04:35.035 ","End":"04:39.910","Text":"This thing squared is this."},{"Start":"04:41.720 ","End":"04:44.539","Text":"This is an answer."},{"Start":"04:44.539 ","End":"04:45.950","Text":"If I wanted component-wise,"},{"Start":"04:45.950 ","End":"04:51.300","Text":"just multiply out and what we get is this,"},{"Start":"04:51.300 ","End":"04:53.190","Text":"this row 1, 0,"},{"Start":"04:53.190 ","End":"04:55.170","Text":"0 with x, y, z is just x."},{"Start":"04:55.170 ","End":"04:59.250","Text":"Similarly, this with this is just y and this with this is"},{"Start":"04:59.250 ","End":"05:05.505","Text":"minus 16x plus 8y plus 1z is this."},{"Start":"05:05.505 ","End":"05:13.145","Text":"This is the answer for T to the minus 2 and I believe it\u0027s the same as we got earlier,"},{"Start":"05:13.145 ","End":"05:15.220","Text":"go and check that."},{"Start":"05:15.220 ","End":"05:18.585","Text":"Yeah, that\u0027s what it was."},{"Start":"05:18.585 ","End":"05:23.280","Text":"We\u0027ve got the same answer with 2 methods,"},{"Start":"05:23.280 ","End":"05:26.020","Text":"and we are done."}],"ID":10195},{"Watched":false,"Name":"Exercise 9","Duration":"2m 35s","ChapterTopicVideoID":9628,"CourseChapterTopicPlaylistID":7317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.010","Text":"In this exercise, we have a linear transformation S. But note that it\u0027s from R^3 to R^2."},{"Start":"00:08.010 ","End":"00:10.845","Text":"The source and destination are not the same."},{"Start":"00:10.845 ","End":"00:13.440","Text":"Here\u0027s the formula for rate S at x,y,z"},{"Start":"00:13.440 ","End":"00:15.675","Text":"is x minus z,y."},{"Start":"00:15.675 ","End":"00:24.060","Text":"We have to see if we can find a formula for S squared, which is SS,"},{"Start":"00:24.060 ","End":"00:29.220","Text":"which really means S composition with S."},{"Start":"00:29.220 ","End":"00:34.760","Text":"I can tell you right away that it\u0027s"},{"Start":"00:34.760 ","End":"00:40.910","Text":"not going to be possible because if S takes us from R^3 to R^2,"},{"Start":"00:40.910 ","End":"00:46.190","Text":"applying S again will not work because S doesn\u0027t apply to R^2."},{"Start":"00:46.190 ","End":"00:48.050","Text":"But let\u0027s say we didn\u0027t notice that."},{"Start":"00:48.050 ","End":"00:50.680","Text":"Let\u0027s just proceed naively."},{"Start":"00:50.680 ","End":"00:52.680","Text":"So we say, okay,"},{"Start":"00:52.680 ","End":"00:54.690","Text":"S squared of x,y,z,"},{"Start":"00:54.690 ","End":"00:57.844","Text":"S squared is S of S,"},{"Start":"00:57.844 ","End":"01:01.795","Text":"its composition of x,y,z."},{"Start":"01:01.795 ","End":"01:06.210","Text":"Now we say, we have a formula for S of x,y,z,"},{"Start":"01:06.210 ","End":"01:09.124","Text":"which is just this."},{"Start":"01:09.124 ","End":"01:12.410","Text":"We take this x minus z,y and put it here."},{"Start":"01:12.410 ","End":"01:15.200","Text":"Now we want to apply S to it."},{"Start":"01:15.200 ","End":"01:17.840","Text":"Here\u0027s where we get stuck."},{"Start":"01:17.840 ","End":"01:21.895","Text":"Because S is defined on R^3,"},{"Start":"01:21.895 ","End":"01:24.615","Text":"it needs 3 components to work on."},{"Start":"01:24.615 ","End":"01:27.360","Text":"From here, we have a vector in R^2,"},{"Start":"01:27.360 ","End":"01:30.530","Text":"there\u0027s only 2 components so there\u0027s no way I can apply this"},{"Start":"01:30.530 ","End":"01:36.335","Text":"S, could also just say that this is nonsense."},{"Start":"01:36.335 ","End":"01:39.140","Text":"Let\u0027s just examine again what went wrong,"},{"Start":"01:39.140 ","End":"01:40.805","Text":"why it was nonsense."},{"Start":"01:40.805 ","End":"01:47.010","Text":"Because look, S is from R^3 to R^2."},{"Start":"01:47.010 ","End":"01:51.420","Text":"I\u0027m applying it twice and S again is from R^3 to R^2."},{"Start":"01:51.420 ","End":"01:53.690","Text":"If I want to take S composed with S,"},{"Start":"01:53.690 ","End":"01:59.449","Text":"I have to end off where I started."},{"Start":"01:59.449 ","End":"02:05.125","Text":"Meaning if S takes me from R^3 to R^2,"},{"Start":"02:05.125 ","End":"02:14.645","Text":"then the only other thing I can apply is something from R^2 to something whatever."},{"Start":"02:14.645 ","End":"02:23.140","Text":"I can\u0027t apply S again because S only works on R^3."},{"Start":"02:24.080 ","End":"02:26.820","Text":"Nonsense. Undefined."},{"Start":"02:26.820 ","End":"02:32.040","Text":"It says if possible, no, it\u0027s not possible"},{"Start":"02:32.040 ","End":"02:36.510","Text":"so we can\u0027t find the formula and we are done."}],"ID":10196}],"Thumbnail":null,"ID":7317}]

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1.1

1