[{"Name":"Vector Spaces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Different Vector Spaces over R","Duration":"15m 13s","ChapterTopicVideoID":25642,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.180","Text":"This clip is for more advanced students."},{"Start":"00:03.180 ","End":"00:09.480","Text":"We\u0027re going to discuss some other vector spaces beyond what we\u0027ve learned so far,"},{"Start":"00:09.480 ","End":"00:11.535","Text":"which is the R^n."},{"Start":"00:11.535 ","End":"00:14.460","Text":"For most, this is sufficient R^n,"},{"Start":"00:14.460 ","End":"00:16.470","Text":"but there are other vector spaces."},{"Start":"00:16.470 ","End":"00:20.190","Text":"I\u0027m going to introduce some of the more,"},{"Start":"00:20.190 ","End":"00:22.095","Text":"what I informally call,"},{"Start":"00:22.095 ","End":"00:24.960","Text":"famous, the more well-known,"},{"Start":"00:24.960 ","End":"00:29.625","Text":"frequently encountered, the ones that you should maybe know."},{"Start":"00:29.625 ","End":"00:34.740","Text":"But I\u0027d like to start with a review of the abstract definition of"},{"Start":"00:34.740 ","End":"00:38.580","Text":"a vector space which in brief is"},{"Start":"00:38.580 ","End":"00:42.485","Text":"a set with 2 operations that satisfies certain conditions."},{"Start":"00:42.485 ","End":"00:47.840","Text":"But more specifically, we have a set typically"},{"Start":"00:47.840 ","End":"00:53.960","Text":"called V or W and the members of the set are called vectors."},{"Start":"00:53.960 ","End":"00:56.885","Text":"There are 2 operations defined."},{"Start":"00:56.885 ","End":"01:00.620","Text":"The first is called addition of 2 vectors."},{"Start":"01:00.620 ","End":"01:10.480","Text":"We have to define, or specify how to add 2 vectors, and the sum will be also a vector."},{"Start":"01:10.480 ","End":"01:15.710","Text":"Then the other operation is multiplication of a scalar."},{"Start":"01:15.710 ","End":"01:17.270","Text":"Scalar just means number,"},{"Start":"01:17.270 ","End":"01:20.600","Text":"a real number, scalar k,"},{"Start":"01:20.600 ","End":"01:22.010","Text":"by a vector v,"},{"Start":"01:22.010 ","End":"01:26.375","Text":"and the result of this multiplication is another vector."},{"Start":"01:26.375 ","End":"01:28.100","Text":"We just call it kv,"},{"Start":"01:28.100 ","End":"01:33.120","Text":"or sometimes with a dot to emphasize the multiplication."},{"Start":"01:35.180 ","End":"01:40.770","Text":"Well, there\u0027s 8 conditions between them for each."},{"Start":"01:40.770 ","End":"01:42.950","Text":"We don\u0027t call them conditions,"},{"Start":"01:42.950 ","End":"01:44.720","Text":"we call them axioms."},{"Start":"01:44.720 ","End":"01:48.580","Text":"Let\u0027s start out with the axioms for addition."},{"Start":"01:48.580 ","End":"01:52.760","Text":"I\u0027m going more quickly, because it\u0027s a review that we have"},{"Start":"01:52.760 ","End":"01:57.935","Text":"the associative law that if we have 3 vectors, and we want to add them,"},{"Start":"01:57.935 ","End":"02:01.295","Text":"addition is only defined for 2 at a time."},{"Start":"02:01.295 ","End":"02:05.340","Text":"You could add the first 2, and then add the third to that,"},{"Start":"02:05.340 ","End":"02:08.300","Text":"or you could add the second, and the third, and the first plus that,"},{"Start":"02:08.300 ","End":"02:10.190","Text":"and you should get the same thing."},{"Start":"02:10.190 ","End":"02:13.970","Text":"Then there\u0027s also the rule that if you change the order of the addition,"},{"Start":"02:13.970 ","End":"02:16.165","Text":"you\u0027ll get the same thing."},{"Start":"02:16.165 ","End":"02:24.680","Text":"Then the axiom number 3 here is that there\u0027s a zero vector, a special 1."},{"Start":"02:24.680 ","End":"02:27.740","Text":"It has the property that if you add it to any other vector,"},{"Start":"02:27.740 ","End":"02:30.300","Text":"you just get that vector."},{"Start":"02:30.800 ","End":"02:37.850","Text":"The last axiom for addition is that every vector has an inverse vector,"},{"Start":"02:37.850 ","End":"02:43.320","Text":"which is denoted minus v in this case."},{"Start":"02:43.320 ","End":"02:46.610","Text":"If you add the vector to its inverse,"},{"Start":"02:46.610 ","End":"02:48.754","Text":"you get the zero vector."},{"Start":"02:48.754 ","End":"02:52.225","Text":"It\u0027s also sometimes called the additive inverse."},{"Start":"02:52.225 ","End":"02:59.040","Text":"Now, let\u0027s go to the axioms for the scalar multiplication."},{"Start":"02:59.360 ","End":"03:03.110","Text":"There are also 4 axioms for these."},{"Start":"03:03.110 ","End":"03:07.099","Text":"The first 2 are kinds of distributive laws."},{"Start":"03:07.099 ","End":"03:10.595","Text":"A scalar times the sum of 2 vectors,"},{"Start":"03:10.595 ","End":"03:15.975","Text":"you just multiply the scalar by each of them, and add."},{"Start":"03:15.975 ","End":"03:19.230","Text":"You could also have the sum of 2 scalars times the vector,"},{"Start":"03:19.230 ","End":"03:22.055","Text":"and you also would get the sum of"},{"Start":"03:22.055 ","End":"03:26.420","Text":"the first scalar with the vector, and the second scalar with the vector."},{"Start":"03:26.420 ","End":"03:32.630","Text":"Then there\u0027s an associative law that if you have a scalar times scalar times vector,"},{"Start":"03:32.630 ","End":"03:34.520","Text":"it doesn\u0027t matter if you take the product of"},{"Start":"03:34.520 ","End":"03:37.010","Text":"the 2 scalars first, and then multiply by the vector"},{"Start":"03:37.010 ","End":"03:43.195","Text":"or multiply the second scalar by the vector, and then the first scalar by that."},{"Start":"03:43.195 ","End":"03:46.170","Text":"Also, the scalar 1,"},{"Start":"03:46.170 ","End":"03:47.430","Text":"the number 1,"},{"Start":"03:47.430 ","End":"03:52.990","Text":"times any vector is that vector itself."},{"Start":"03:53.360 ","End":"03:56.270","Text":"That\u0027s the review. Now,"},{"Start":"03:56.270 ","End":"04:00.020","Text":"let\u0027s continue with the famous vector spaces."},{"Start":"04:00.020 ","End":"04:03.440","Text":"I\u0027ll start by reviewing what we\u0027ve already covered,"},{"Start":"04:03.440 ","End":"04:05.750","Text":"is the vector space R^n."},{"Start":"04:05.750 ","End":"04:11.360","Text":"Well, it\u0027s really many vector spaces, because n could be 1,"},{"Start":"04:11.360 ","End":"04:14.330","Text":"2, 3, 4, or 5, and so on."},{"Start":"04:14.330 ","End":"04:19.070","Text":"Just for completeness, I\u0027m including this, it\u0027s a review."},{"Start":"04:19.070 ","End":"04:24.610","Text":"The vectors are of length n,"},{"Start":"04:24.610 ","End":"04:29.780","Text":"as n real numbers in them written like this."},{"Start":"04:29.780 ","End":"04:35.660","Text":"Addition is component-wise, the first component with the first component,"},{"Start":"04:35.660 ","End":"04:37.760","Text":"second with the second, and so on,"},{"Start":"04:37.760 ","End":"04:40.555","Text":"the nth component with the nth component."},{"Start":"04:40.555 ","End":"04:45.220","Text":"Scalar multiplication, similarly multiplied by each component,"},{"Start":"04:45.220 ","End":"04:49.375","Text":"the k goes with each of these. That\u0027s a review."},{"Start":"04:49.375 ","End":"04:52.590","Text":"Now, let\u0027s go for something new."},{"Start":"04:52.590 ","End":"04:55.975","Text":"The next vector space,"},{"Start":"04:55.975 ","End":"04:59.570","Text":"although it\u0027s really a family of vector spaces"},{"Start":"04:59.570 ","End":"05:02.635","Text":"just like this was a family, because n can vary,"},{"Start":"05:02.635 ","End":"05:09.800","Text":"well, here, we have the matrices of order m by n. I don\u0027t know,"},{"Start":"05:09.800 ","End":"05:13.025","Text":"I\u0027ll tell you in a minute what the R means."},{"Start":"05:13.025 ","End":"05:18.695","Text":"But it\u0027s a family and we\u0027ll take an example where m is 2 and n is 3,"},{"Start":"05:18.695 ","End":"05:21.470","Text":"and basically just wrote down what I said."},{"Start":"05:21.470 ","End":"05:27.980","Text":"You have the matrices of order m by n. This R means that the entries of the matrix,"},{"Start":"05:27.980 ","End":"05:29.780","Text":"the elements or entries,"},{"Start":"05:29.780 ","End":"05:34.460","Text":"are real numbers, because a matrix could consist of symbols."},{"Start":"05:34.460 ","End":"05:39.450","Text":"Also, it could be other kinds of numbers, and real numbers,"},{"Start":"05:40.520 ","End":"05:44.160","Text":"complex numbers, you may have heard of them, may not."},{"Start":"05:44.160 ","End":"05:48.295","Text":"Here are a couple of examples of elements."},{"Start":"05:48.295 ","End":"05:51.565","Text":"Matrix are 2 by 3,"},{"Start":"05:51.565 ","End":"05:54.520","Text":"all the numbers here are real numbers, e,"},{"Start":"05:54.520 ","End":"05:56.980","Text":"and Pi, and whole numbers,"},{"Start":"05:56.980 ","End":"05:58.675","Text":"and negative numbers,"},{"Start":"05:58.675 ","End":"06:00.790","Text":"but in a 2 by 3."},{"Start":"06:00.790 ","End":"06:04.290","Text":"Obviously, there\u0027s infinitely many,"},{"Start":"06:04.290 ","End":"06:06.615","Text":"just give a couple of an example."},{"Start":"06:06.615 ","End":"06:12.330","Text":"The general description would be a general 2 by 3 matrix."},{"Start":"06:12.330 ","End":"06:15.895","Text":"That would be 6 real numbers that I have to give you,"},{"Start":"06:15.895 ","End":"06:17.590","Text":"call them a through f,"},{"Start":"06:17.590 ","End":"06:19.450","Text":"where all these a, b, c, d, e,"},{"Start":"06:19.450 ","End":"06:21.695","Text":"f are all real numbers."},{"Start":"06:21.695 ","End":"06:24.530","Text":"Now, we haven\u0027t finished the definition of"},{"Start":"06:24.530 ","End":"06:27.110","Text":"the vector space, because we don\u0027t just give a set,"},{"Start":"06:27.110 ","End":"06:29.855","Text":"we also have to give 2 operations,"},{"Start":"06:29.855 ","End":"06:33.215","Text":"addition, and scalar multiplication."},{"Start":"06:33.215 ","End":"06:36.170","Text":"On the 1 hand, you can look at them as matrices,"},{"Start":"06:36.170 ","End":"06:37.400","Text":"but as a vector space,"},{"Start":"06:37.400 ","End":"06:39.575","Text":"they\u0027re also considered as vectors."},{"Start":"06:39.575 ","End":"06:41.600","Text":"You could look at them from both ways."},{"Start":"06:41.600 ","End":"06:44.434","Text":"Now, since we already have"},{"Start":"06:44.434 ","End":"06:50.375","Text":"the operation of addition, and multiplication by scalar on matrices,"},{"Start":"06:50.375 ","End":"06:57.250","Text":"we just borrow those, or use these same ones as the vector operations."},{"Start":"06:57.250 ","End":"07:03.110","Text":"The addition of 2 vectors is really the addition of 2 matrices."},{"Start":"07:03.110 ","End":"07:06.510","Text":"As usual, we just take"},{"Start":"07:06.590 ","End":"07:12.800","Text":"each element with its corresponding element in the same position like this a with this a,"},{"Start":"07:12.800 ","End":"07:14.630","Text":"this b with this b, well,"},{"Start":"07:14.630 ","End":"07:18.965","Text":"have lettered them in such a way that it\u0027s obvious which 1 to add with which."},{"Start":"07:18.965 ","End":"07:22.955","Text":"Then we get another vector which is actually a matrix,"},{"Start":"07:22.955 ","End":"07:25.420","Text":"also 2 by 3."},{"Start":"07:25.420 ","End":"07:28.860","Text":"When we take a scalar,"},{"Start":"07:28.860 ","End":"07:32.930","Text":"k would be a real number, and multiply it by a matrix."},{"Start":"07:32.930 ","End":"07:39.830","Text":"We multiply each of the entries of the matrix by this scalar,"},{"Start":"07:39.830 ","End":"07:45.205","Text":"and that\u0027s what we used to do with matrices, and we\u0027ll take the same multiplication."},{"Start":"07:45.205 ","End":"07:47.635","Text":"When these are viewed as vectors,"},{"Start":"07:47.635 ","End":"07:49.905","Text":"it\u0027s really a dual viewpoint,."},{"Start":"07:49.905 ","End":"07:51.740","Text":"On the 1 hand, they\u0027re matrices,"},{"Start":"07:51.740 ","End":"07:54.500","Text":"but when we look at them as a vector space,"},{"Start":"07:54.500 ","End":"07:56.680","Text":"then they\u0027re vectors also."},{"Start":"07:56.680 ","End":"07:59.165","Text":"It\u0027s all a matter of perspective."},{"Start":"07:59.165 ","End":"08:03.155","Text":"I\u0027m going to skip the checking of the 8 axioms."},{"Start":"08:03.155 ","End":"08:08.240","Text":"You can take my word for it, or you can try it as an exercise."},{"Start":"08:08.240 ","End":"08:11.085","Text":"Let\u0027s get onto the next."},{"Start":"08:11.085 ","End":"08:14.860","Text":"Our next example is really not 1 example,"},{"Start":"08:14.860 ","End":"08:23.620","Text":"but a whole family of examples where this n is any natural number, P for polynomials."},{"Start":"08:23.620 ","End":"08:28.410","Text":"We\u0027re going to have the polynomials of degree n not"},{"Start":"08:28.410 ","End":"08:33.630","Text":"quite degree less than or equal to n. Just to write it down,"},{"Start":"08:33.630 ","End":"08:39.540","Text":"this P_n is the space of all polynomials of order less than or equal to"},{"Start":"08:39.540 ","End":"08:45.925","Text":"n. The R means that the coefficients of the polynomial are real numbers."},{"Start":"08:45.925 ","End":"08:53.080","Text":"N is a natural number or it could be 0."},{"Start":"08:53.080 ","End":"08:57.865","Text":"Polynomials of degree 0 are constants."},{"Start":"08:57.865 ","End":"09:00.880","Text":"Here we could also add 0 really."},{"Start":"09:00.880 ","End":"09:02.680","Text":"I have written a question here,"},{"Start":"09:02.680 ","End":"09:05.005","Text":"but I\u0027m going to answer it a little lighter."},{"Start":"09:05.005 ","End":"09:12.070","Text":"Why do we specify that the polynomials are of order less than, or equal to n?"},{"Start":"09:12.070 ","End":"09:16.390","Text":"Why don\u0027t we just say polynomials of degree equal to n?"},{"Start":"09:16.390 ","End":"09:18.280","Text":"In a moment I\u0027ll answer that."},{"Start":"09:18.280 ","End":"09:20.350","Text":"Let\u0027s take a specific n,"},{"Start":"09:20.350 ","End":"09:23.200","Text":"say n equals 3."},{"Start":"09:23.200 ","End":"09:27.790","Text":"We have polynomials up to degree 3."},{"Start":"09:27.790 ","End":"09:32.935","Text":"Here\u0027s an example, x cubed plus root 2_x plus 1."},{"Start":"09:32.935 ","End":"09:34.765","Text":"Here\u0027s another example,"},{"Start":"09:34.765 ","End":"09:37.690","Text":"notice that its degree is actually 2,"},{"Start":"09:37.690 ","End":"09:41.440","Text":"but it\u0027s okay, because it\u0027s less than or equal to 3."},{"Start":"09:41.440 ","End":"09:45.070","Text":"Also, a linear 1 in there its degree is 1,"},{"Start":"09:45.070 ","End":"09:48.020","Text":"but it\u0027s still less than or equal to 3,"},{"Start":"09:48.090 ","End":"09:52.570","Text":"x cubed minus Pi and 0,"},{"Start":"09:52.570 ","End":"09:54.130","Text":"0 is a tricky 1."},{"Start":"09:54.130 ","End":"09:58.150","Text":"Actually, the 0 doesn\u0027t have a defined degree,"},{"Start":"09:58.150 ","End":"10:02.035","Text":"but we allow the 0 to be included in,"},{"Start":"10:02.035 ","End":"10:07.060","Text":"say it has degree less than, or equal to anything less than or equal to 3,"},{"Start":"10:07.060 ","End":"10:08.470","Text":"it\u0027s a special 1."},{"Start":"10:08.470 ","End":"10:15.760","Text":"The general description is a_x cubed plus b_x squared plus c_x plus d,"},{"Start":"10:15.760 ","End":"10:18.950","Text":"all the coefficients are real numbers."},{"Start":"10:19.110 ","End":"10:24.490","Text":"One of the reasons now I\u0027m answering this question less than or equal to"},{"Start":"10:24.490 ","End":"10:29.355","Text":"n is I don\u0027t want to restrict the values of these."},{"Start":"10:29.355 ","End":"10:32.535","Text":"For example, if a is 0,"},{"Start":"10:32.535 ","End":"10:39.300","Text":"then we get a degree which is less than or equal to 3."},{"Start":"10:39.300 ","End":"10:43.830","Text":"Now, I need to tell you what is the sum of"},{"Start":"10:43.830 ","End":"10:46.785","Text":"2 polynomials and a scalar"},{"Start":"10:46.785 ","End":"10:50.945","Text":"times a polynomial because we have to define those 2 operations."},{"Start":"10:50.945 ","End":"10:54.130","Text":"But when we look at these polynomials,"},{"Start":"10:54.130 ","End":"10:57.670","Text":"on the 1 hand, they\u0027re vectors and the on the 1 hand polynomials,"},{"Start":"10:57.670 ","End":"11:01.840","Text":"we now have to add 2 polynomials and to multiply a constant times"},{"Start":"11:01.840 ","End":"11:07.960","Text":"a polynomial and we just use the same when viewed as vectors."},{"Start":"11:07.960 ","End":"11:09.925","Text":"For the addition,"},{"Start":"11:09.925 ","End":"11:13.180","Text":"is just like you would normally add polynomials."},{"Start":"11:13.180 ","End":"11:16.750","Text":"This plus this, you might want to collect like terms."},{"Start":"11:16.750 ","End":"11:18.010","Text":"You could leave it like this,"},{"Start":"11:18.010 ","End":"11:23.515","Text":"but usually we collect like terms and so a with a_x cubed,"},{"Start":"11:23.515 ","End":"11:25.075","Text":"b with B,"},{"Start":"11:25.075 ","End":"11:27.040","Text":"b with B,"},{"Start":"11:27.040 ","End":"11:29.185","Text":"and so on, c,"},{"Start":"11:29.185 ","End":"11:32.770","Text":"C for the x term and the constants,"},{"Start":"11:32.770 ","End":"11:35.360","Text":"so that\u0027s how we add."},{"Start":"11:35.520 ","End":"11:39.370","Text":"Multiplication by a scalar, a real number,"},{"Start":"11:39.370 ","End":"11:46.720","Text":"which is as usual as we multiply a constant times a polynomial,"},{"Start":"11:46.720 ","End":"11:52.550","Text":"we just use the distributive law to put k with each of the terms."},{"Start":"11:53.040 ","End":"11:59.710","Text":"I\u0027m going to skip the part about checking all 8 axioms for these operations,"},{"Start":"11:59.710 ","End":"12:04.090","Text":"you just try it out yourselves."},{"Start":"12:04.090 ","End":"12:10.060","Text":"I\u0027ll take my word for it that all the 8 axioms are satisfied."},{"Start":"12:10.060 ","End":"12:14.499","Text":"Now the last example of a famous vector space,"},{"Start":"12:14.499 ","End":"12:17.425","Text":"at least the last 1 for this clip."},{"Start":"12:17.425 ","End":"12:21.595","Text":"F of R, F for function."},{"Start":"12:21.595 ","End":"12:27.820","Text":"Just functions from real numbers to real numbers,"},{"Start":"12:27.820 ","End":"12:30.595","Text":"what we usually call functions."},{"Start":"12:30.595 ","End":"12:34.795","Text":"I\u0027ll explain in a moment what I wrote in brackets here."},{"Start":"12:34.795 ","End":"12:37.720","Text":"Of course there are infinitely many functions,"},{"Start":"12:37.720 ","End":"12:42.090","Text":"I just wrote 5 of them as examples,"},{"Start":"12:42.090 ","End":"12:44.370","Text":"sine x, square root of x,"},{"Start":"12:44.370 ","End":"12:47.790","Text":"natural log of x, a couple of others."},{"Start":"12:47.790 ","End":"12:52.540","Text":"Now, the reason I wrote all part of,"},{"Start":"12:52.540 ","End":"12:55.240","Text":"see if we have square root of x,"},{"Start":"12:55.240 ","End":"12:57.670","Text":"it\u0027s not strictly speaking,"},{"Start":"12:57.670 ","End":"13:00.250","Text":"defined on all of R, in fact, it\u0027s not."},{"Start":"13:00.250 ","End":"13:02.275","Text":"It\u0027s just defined for"},{"Start":"13:02.275 ","End":"13:07.420","Text":"non-negative numbers and the natural log is defined for positive numbers."},{"Start":"13:07.420 ","End":"13:11.185","Text":"This 1 here is defined when x is not equal to 0."},{"Start":"13:11.185 ","End":"13:13.330","Text":"The domain is R,"},{"Start":"13:13.330 ","End":"13:17.185","Text":"but it could also be a subset or part of it."},{"Start":"13:17.185 ","End":"13:19.930","Text":"These form a vector space."},{"Start":"13:19.930 ","End":"13:24.850","Text":"But I have to give you the operations which are the addition and"},{"Start":"13:24.850 ","End":"13:30.920","Text":"scalar multiplication as vectors."},{"Start":"13:31.650 ","End":"13:35.080","Text":"I can consider them also as functions."},{"Start":"13:35.080 ","End":"13:37.270","Text":"They\u0027re going to be both functions in"},{"Start":"13:37.270 ","End":"13:40.420","Text":"themselves and they\u0027re also going to be vectors as part of a vector space."},{"Start":"13:40.420 ","End":"13:43.750","Text":"But as functions, we already have these operations."},{"Start":"13:43.750 ","End":"13:49.555","Text":"We can add 2 functions and we can multiply a function by a number."},{"Start":"13:49.555 ","End":"13:54.280","Text":"Let\u0027s say this is 1 function and this is the other function. I can add them."},{"Start":"13:54.280 ","End":"14:03.385","Text":"I just put a plus the regular addition and simplification is optional."},{"Start":"14:03.385 ","End":"14:06.384","Text":"Like this it\u0027s already another function,"},{"Start":"14:06.384 ","End":"14:10.255","Text":"but we could collect the x squared and 2x squared and so on."},{"Start":"14:10.255 ","End":"14:13.645","Text":"Now multiplication by a scalar,"},{"Start":"14:13.645 ","End":"14:21.280","Text":"if I have scalar 4 multiply it by our function sine x plus natural log of x,"},{"Start":"14:21.280 ","End":"14:25.615","Text":"I put the dots here just to show that I\u0027m doing the skeletons vector."},{"Start":"14:25.615 ","End":"14:30.730","Text":"Well, it\u0027s just 4 times this function and I"},{"Start":"14:30.730 ","End":"14:36.070","Text":"could simplify it and use the distributive law and write it like this."},{"Start":"14:36.070 ","End":"14:39.865","Text":"But really, it\u0027s almost like a tautology,"},{"Start":"14:39.865 ","End":"14:45.040","Text":"4 times a function is 4 times the function and another example,"},{"Start":"14:45.040 ","End":"14:47.110","Text":"scalar times the vector 5,"},{"Start":"14:47.110 ","End":"14:51.560","Text":"times the function is just 5 times the function."},{"Start":"14:51.810 ","End":"15:00.475","Text":"It\u0027s just regular multiplication by a constant and regular addition of functions,"},{"Start":"15:00.475 ","End":"15:03.835","Text":"just considered in the perspective of vectors."},{"Start":"15:03.835 ","End":"15:14.360","Text":"That concludes the clip on special or famous vector spaces and so we\u0027re done."}],"ID":26467},{"Watched":false,"Name":"Vector Space over a Field","Duration":"8m 24s","ChapterTopicVideoID":25643,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.030 ","End":"00:04.120","Text":"This clip is for the more advanced students."},{"Start":"00:04.120 ","End":"00:10.910","Text":"We\u0027re still in the introduction stages of vector spaces."},{"Start":"00:10.920 ","End":"00:16.265","Text":"We\u0027re going to broaden the concept a bit because up till now,"},{"Start":"00:16.265 ","End":"00:20.160","Text":"the field, that\u0027s where the scalars come from,"},{"Start":"00:20.160 ","End":"00:21.970","Text":"has been the real numbers."},{"Start":"00:21.970 ","End":"00:25.130","Text":"Turns out we can extend this a bit."},{"Start":"00:25.130 ","End":"00:29.545","Text":"Up to now, when we talk about vector spaces,"},{"Start":"00:29.545 ","End":"00:31.945","Text":"we talked about vectors and scalars."},{"Start":"00:31.945 ","End":"00:38.620","Text":"The scalars are typically labeled k were taken from R,"},{"Start":"00:38.620 ","End":"00:40.740","Text":"which is real numbers,"},{"Start":"00:40.740 ","End":"00:43.380","Text":"and it happens to be a field."},{"Start":"00:43.380 ","End":"00:48.400","Text":"Then we would more precisely say not just that V is a vector space,"},{"Start":"00:48.400 ","End":"00:51.580","Text":"but that V is a vector space over the field"},{"Start":"00:51.580 ","End":"00:54.865","Text":"of real numbers because there are other fields too."},{"Start":"00:54.865 ","End":"00:57.500","Text":"That\u0027s the point in this clip."},{"Start":"00:57.500 ","End":"01:00.580","Text":"If you\u0027re asking me what is a field,"},{"Start":"01:00.580 ","End":"01:02.620","Text":"I suggest you review it."},{"Start":"01:02.620 ","End":"01:06.355","Text":"Perhaps I\u0027ll say a few words about what a field is."},{"Start":"01:06.355 ","End":"01:09.520","Text":"In some ways it\u0027s similar to the real numbers in"},{"Start":"01:09.520 ","End":"01:13.045","Text":"the sense that it has operations defined,"},{"Start":"01:13.045 ","End":"01:17.680","Text":"field has addition, multiplication."},{"Start":"01:17.680 ","End":"01:20.980","Text":"We have also the inverses of these,"},{"Start":"01:20.980 ","End":"01:24.145","Text":"we have also subtraction and division,"},{"Start":"01:24.145 ","End":"01:27.540","Text":"but more in terms of inverses these 2."},{"Start":"01:27.540 ","End":"01:31.425","Text":"Then there are various rules that hold,"},{"Start":"01:31.425 ","End":"01:35.670","Text":"like we have associative laws,"},{"Start":"01:35.670 ","End":"01:42.420","Text":"we have distributive laws and commutativity laws,"},{"Start":"01:42.420 ","End":"01:45.665","Text":"and we also have inverses,"},{"Start":"01:45.665 ","End":"01:48.665","Text":"in fact, is even 2 kinds of inverse."},{"Start":"01:48.665 ","End":"01:54.050","Text":"Anyway, I\u0027m not going to get into this too deeply,"},{"Start":"01:54.050 ","End":"01:57.184","Text":"but these are the concepts that go with a field."},{"Start":"01:57.184 ","End":"02:00.950","Text":"It\u0027s basically all the things you can do arithmetic on."},{"Start":"02:00.950 ","End":"02:08.195","Text":"All the rules of arithmetic apply generally to a field and not just to the real numbers."},{"Start":"02:08.195 ","End":"02:15.575","Text":"That was a brief reminder of the concept of a field."},{"Start":"02:15.575 ","End":"02:18.680","Text":"Here I just wrote down what I said earlier that,"},{"Start":"02:18.680 ","End":"02:23.560","Text":"in general replace R by a generic field called"},{"Start":"02:23.560 ","End":"02:29.400","Text":"F. When R specific cases in this course,"},{"Start":"02:29.400 ","End":"02:34.400","Text":"F is only 1 of a certain number of things that you might encounter."},{"Start":"02:34.400 ","End":"02:40.325","Text":"The completeness I\u0027ve listed the real numbers again and now here are the new ones."},{"Start":"02:40.325 ","End":"02:43.040","Text":"Complex numbers for those who have"},{"Start":"02:43.040 ","End":"02:45.500","Text":"studied them and if you\u0027re an advanced student, you may have."},{"Start":"02:45.500 ","End":"02:52.040","Text":"Remember things like a plus bi and i squared is minus 1."},{"Start":"02:52.040 ","End":"02:54.700","Text":"Anyway, complex numbers."},{"Start":"02:54.700 ","End":"02:57.590","Text":"Then there\u0027s the irrational numbers which you"},{"Start":"02:57.590 ","End":"03:00.050","Text":"sometimes learn even before the real numbers."},{"Start":"03:00.050 ","End":"03:06.110","Text":"All the quotient, m over n or p over q."},{"Start":"03:06.110 ","End":"03:11.520","Text":"Any division, the quotient of 2 integers,"},{"Start":"03:11.520 ","End":"03:18.160","Text":"but we never divide by 0."},{"Start":"03:18.800 ","End":"03:21.195","Text":"Let\u0027s see, what else?"},{"Start":"03:21.195 ","End":"03:30.795","Text":"There are actually some finite fields and these are integers modulo p,"},{"Start":"03:30.795 ","End":"03:33.615","Text":"and p is some prime number."},{"Start":"03:33.615 ","End":"03:36.030","Text":"Typically, we just go for 2, 3, 5,"},{"Start":"03:36.030 ","End":"03:40.600","Text":"or 7, mostly 3 and 5."},{"Start":"03:41.600 ","End":"03:45.395","Text":"Hopefully, you\u0027ve studied this,"},{"Start":"03:45.395 ","End":"03:51.005","Text":"these are the remainders of when you divide a number by 5."},{"Start":"03:51.005 ","End":"03:53.995","Text":"Let\u0027s say if we took a case of 5,"},{"Start":"03:53.995 ","End":"03:57.585","Text":"and then there\u0027s only 5 numbers in our field,"},{"Start":"03:57.585 ","End":"03:58.800","Text":"0, 1,"},{"Start":"03:58.800 ","End":"04:01.320","Text":"2, 3, and 4."},{"Start":"04:01.320 ","End":"04:06.200","Text":"These are the fields that we\u0027re likely to encounter if you encounter field other than"},{"Start":"04:06.200 ","End":"04:12.910","Text":"R. I\u0027ll give 4 examples of such things."},{"Start":"04:12.910 ","End":"04:19.465","Text":"Let\u0027s see, represent all of these fields."},{"Start":"04:19.465 ","End":"04:23.480","Text":"First example should remind you of R^3,"},{"Start":"04:23.480 ","End":"04:24.950","Text":"which we studied,"},{"Start":"04:24.950 ","End":"04:28.340","Text":"R^3 where vectors of 3 real numbers,"},{"Start":"04:28.340 ","End":"04:35.885","Text":"but there\u0027s also a C^3 similar kinds of vector notation with 3 components."},{"Start":"04:35.885 ","End":"04:37.970","Text":"Only the components instead of coming from"},{"Start":"04:37.970 ","End":"04:42.859","Text":"the real numbers will come from the complex numbers."},{"Start":"04:42.859 ","End":"04:49.730","Text":"I could write this out formally that C^3 is the set of all,"},{"Start":"04:49.730 ","End":"04:59.840","Text":"Z_1, Z_2, Z_3,"},{"Start":"04:59.840 ","End":"05:08.015","Text":"where Z_i belong to"},{"Start":"05:08.015 ","End":"05:14.435","Text":"field of complex numbers where i is 1, 2, or 3."},{"Start":"05:14.435 ","End":"05:16.325","Text":"Another set of triplets,"},{"Start":"05:16.325 ","End":"05:22.850","Text":"each of the 3 components is a complex number, to an example."},{"Start":"05:22.850 ","End":"05:26.240","Text":"Another example is matrices."},{"Start":"05:26.240 ","End":"05:34.450","Text":"Now we\u0027ve already discussed matrices m by n or 2 by 3 over the field R,"},{"Start":"05:34.450 ","End":"05:37.490","Text":"same thing pretty much holds instead of R,"},{"Start":"05:37.490 ","End":"05:39.200","Text":"if we take Q."},{"Start":"05:39.200 ","End":"05:42.290","Text":"It just means that we take 2 by 3 matrices,"},{"Start":"05:42.290 ","End":"05:45.560","Text":"but the entries have to be rational numbers, fractions."},{"Start":"05:45.560 ","End":"05:51.520","Text":"I can\u0027t take Pi and e and square root of 2 and things like that."},{"Start":"05:51.520 ","End":"06:00.455","Text":"It actually behaves quite similarly to the same thing over R. Another example,"},{"Start":"06:00.455 ","End":"06:05.210","Text":"polynomials of degree up to 3,"},{"Start":"06:05.210 ","End":"06:09.125","Text":"P_3 doesn\u0027t mean polynomials exactly of degree 3,"},{"Start":"06:09.125 ","End":"06:12.125","Text":"it means degree 3 or less."},{"Start":"06:12.125 ","End":"06:14.900","Text":"Just like we had it for the real numbers,"},{"Start":"06:14.900 ","End":"06:17.545","Text":"we could do it for the complex numbers."},{"Start":"06:17.545 ","End":"06:20.730","Text":"I don\u0027t mean that this is the whole list,"},{"Start":"06:20.730 ","End":"06:28.135","Text":"these are just representative example elements that might belong to this vector space."},{"Start":"06:28.135 ","End":"06:34.479","Text":"Here\u0027s a degree 3 polynomial and all the coefficients or rational numbers."},{"Start":"06:34.479 ","End":"06:35.980","Text":"Here\u0027s degree 2,"},{"Start":"06:35.980 ","End":"06:38.695","Text":"but that\u0027s okay because it\u0027s less than or equal to 3."},{"Start":"06:38.695 ","End":"06:40.835","Text":"Here\u0027s the linear polynomial,"},{"Start":"06:40.835 ","End":"06:47.800","Text":"the 0 polynomial is actually the 0 as far as vector theory goes,"},{"Start":"06:47.800 ","End":"06:50.965","Text":"the 0 for addition."},{"Start":"06:50.965 ","End":"06:57.495","Text":"Actually, 0 doesn\u0027t really have a degree undefined."},{"Start":"06:57.495 ","End":"07:00.490","Text":"Anyway, I don\u0027t want to get into that."},{"Start":"07:01.140 ","End":"07:07.760","Text":"Last example, take an example from one of these modulos,"},{"Start":"07:07.970 ","End":"07:11.635","Text":"reuse a matrix example, but this time,"},{"Start":"07:11.635 ","End":"07:15.410","Text":"2 by 2 square matrices."},{"Start":"07:15.930 ","End":"07:21.715","Text":"Looks like, you don\u0027t know that this is not real numbers or rational numbers."},{"Start":"07:21.715 ","End":"07:25.150","Text":"But just by the definition,"},{"Start":"07:25.150 ","End":"07:26.600","Text":"just by writing it this way,"},{"Start":"07:26.600 ","End":"07:31.060","Text":"it forces all the entries in the matrix and then"},{"Start":"07:31.060 ","End":"07:35.875","Text":"various matrices to be over the field Z_5,"},{"Start":"07:35.875 ","End":"07:39.490","Text":"Z_5 is a field that it only has 4 elements in it;"},{"Start":"07:39.490 ","End":"07:40.975","Text":"0, 1,"},{"Start":"07:40.975 ","End":"07:44.100","Text":"2, 3, and 4."},{"Start":"07:44.100 ","End":"07:46.935","Text":"I mentioned this over here."},{"Start":"07:46.935 ","End":"07:51.020","Text":"Additional multiplication, we do Modulo 5,"},{"Start":"07:51.020 ","End":"07:54.380","Text":"but now\u0027s not the time to get into that."},{"Start":"07:54.380 ","End":"07:57.680","Text":"The main idea was that we can generalize,"},{"Start":"07:57.680 ","End":"08:01.505","Text":"instead of just always having our scalars be real numbers,"},{"Start":"08:01.505 ","End":"08:04.374","Text":"they can come from any field."},{"Start":"08:04.374 ","End":"08:08.960","Text":"As I said, you should review what a field is."},{"Start":"08:08.960 ","End":"08:13.295","Text":"But typically we are only going to take the fields that I mentioned before,"},{"Start":"08:13.295 ","End":"08:16.220","Text":"which were real, complex,"},{"Start":"08:16.220 ","End":"08:24.510","Text":"rational, or integers modulo some prime number, and I\u0027m done."}],"ID":26468},{"Watched":false,"Name":"Exercise 1","Duration":"3m 46s","ChapterTopicVideoID":25644,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:06.270","Text":"In this exercise, we have to check if w,"},{"Start":"00:06.270 ","End":"00:07.905","Text":"to be defined in a moment,"},{"Start":"00:07.905 ","End":"00:15.750","Text":"is a subspace of the space m n of r. Now I want to remind you what this means."},{"Start":"00:15.750 ","End":"00:22.020","Text":"This is the square matrices that are n by n. If it\u0027s not square,"},{"Start":"00:22.020 ","End":"00:23.295","Text":"we would write m,"},{"Start":"00:23.295 ","End":"00:24.390","Text":"n by m,"},{"Start":"00:24.390 ","End":"00:26.850","Text":"or whatever the 2 numbers are, 3 by 2."},{"Start":"00:26.850 ","End":"00:29.205","Text":"If it\u0027s square, we just write a single number."},{"Start":"00:29.205 ","End":"00:31.100","Text":"That\u0027s a reminder. Now,"},{"Start":"00:31.100 ","End":"00:32.810","Text":"what is this subspace?"},{"Start":"00:32.810 ","End":"00:37.295","Text":"It\u0027s all the matrices A from here,"},{"Start":"00:37.295 ","End":"00:39.980","Text":"which are equal to their own transpose."},{"Start":"00:39.980 ","End":"00:43.340","Text":"Now there\u0027s another name for such matrices,"},{"Start":"00:43.340 ","End":"00:46.340","Text":"they are called the symmetric matrices,"},{"Start":"00:46.340 ","End":"00:47.870","Text":"if they\u0027re able to do your own transpose,"},{"Start":"00:47.870 ","End":"00:50.615","Text":"will give you just a simple example of a 2 by 2."},{"Start":"00:50.615 ","End":"00:52.765","Text":"That\u0027s a 1, 2,"},{"Start":"00:52.765 ","End":"00:55.490","Text":"2, 3 is symmetric."},{"Start":"00:55.490 ","End":"00:59.085","Text":"Symmetric means about this diagonal,"},{"Start":"00:59.085 ","End":"01:01.260","Text":"and if you flip it around the diagonal,"},{"Start":"01:01.260 ","End":"01:04.520","Text":"it\u0027s the same thing, or if you switch rows and columns."},{"Start":"01:04.520 ","End":"01:06.340","Text":"Now the solution;"},{"Start":"01:06.340 ","End":"01:10.385","Text":"has 3 things to check as we\u0027re going to prove that it is a subspace."},{"Start":"01:10.385 ","End":"01:15.410","Text":"First, we have to show that 0 belongs to w. When I say 0,"},{"Start":"01:15.410 ","End":"01:18.200","Text":"I mean the 0 n by n matrix,"},{"Start":"01:18.200 ","End":"01:21.140","Text":"I\u0027ll call it bolt 0."},{"Start":"01:21.140 ","End":"01:25.490","Text":"Is this symmetric? Of course, it is."},{"Start":"01:25.490 ","End":"01:29.330","Text":"The transpose of the 0 matrix is just the 0 matrix."},{"Start":"01:29.330 ","End":"01:30.860","Text":"If I flip around the diagonal,"},{"Start":"01:30.860 ","End":"01:32.840","Text":"it\u0027s all 0 and will still be all 0."},{"Start":"01:32.840 ","End":"01:36.890","Text":"That\u0027s 1. There\u0027s 2 more properties."},{"Start":"01:36.890 ","End":"01:40.250","Text":"The second one is closure under scalar multiplication,"},{"Start":"01:40.250 ","End":"01:43.370","Text":"meaning that if A is in W,"},{"Start":"01:43.370 ","End":"01:46.640","Text":"we have to show that any scalar times A is also in"},{"Start":"01:46.640 ","End":"01:50.730","Text":"W. Now I\u0027m going to reinterpret these 2,"},{"Start":"01:50.730 ","End":"01:54.660","Text":"to say that A is in W just means that A is its own transpose."},{"Start":"01:54.660 ","End":"01:56.610","Text":"Actually, it works the other way also."},{"Start":"01:56.610 ","End":"01:58.660","Text":"It should be double arrow."},{"Start":"01:58.660 ","End":"02:01.350","Text":"Similarly, kA is in W,"},{"Start":"02:01.350 ","End":"02:04.005","Text":"if kA is its own transpose."},{"Start":"02:04.005 ","End":"02:07.110","Text":"Again, [inaudible] to be pedantic."},{"Start":"02:07.110 ","End":"02:14.099","Text":"That means that we have to show that this implies this."},{"Start":"02:14.140 ","End":"02:16.280","Text":"Assume this is true."},{"Start":"02:16.280 ","End":"02:23.375","Text":"Now, kA transpose is just k times A transpose."},{"Start":"02:23.375 ","End":"02:30.830","Text":"We studied the properties of the transpose matrix that a constant scalar comes out,"},{"Start":"02:30.830 ","End":"02:33.390","Text":"but A transpose is equal to A,"},{"Start":"02:33.390 ","End":"02:37.220","Text":"so we now, shown that this is equal to this,"},{"Start":"02:37.220 ","End":"02:39.575","Text":"which is what we wanted to do,"},{"Start":"02:39.575 ","End":"02:43.580","Text":"which means that kA is in W. That\u0027s the"},{"Start":"02:43.580 ","End":"02:48.650","Text":"second out of the 3 properties we have to show for subspace."},{"Start":"02:48.650 ","End":"02:50.240","Text":"Let\u0027s get to the third."},{"Start":"02:50.240 ","End":"02:53.000","Text":"This one is called closure under addition,"},{"Start":"02:53.000 ","End":"02:56.120","Text":"that if A and B belong to the subspace,"},{"Start":"02:56.120 ","End":"02:59.150","Text":"then so does A plus B."},{"Start":"02:59.150 ","End":"03:03.095","Text":"Now to say this is the same as to say that"},{"Start":"03:03.095 ","End":"03:06.925","Text":"A is its own transpose and B is its own transpose."},{"Start":"03:06.925 ","End":"03:15.725","Text":"Again, to say that A plus B is in the subspace means that A plus B is its own transpose."},{"Start":"03:15.725 ","End":"03:18.980","Text":"Does this imply this?"},{"Start":"03:18.980 ","End":"03:21.740","Text":"Yes, it does, because if these are true,"},{"Start":"03:21.740 ","End":"03:24.790","Text":"then look A plus B transpose,"},{"Start":"03:24.790 ","End":"03:26.800","Text":"by the properties of the transpose,"},{"Start":"03:26.800 ","End":"03:30.440","Text":"is the sum of the transpose of each one separately."},{"Start":"03:30.440 ","End":"03:34.400","Text":"But, this is equal to this and this is equal to this,"},{"Start":"03:34.400 ","End":"03:36.740","Text":"and so this is just equal to A plus B."},{"Start":"03:36.740 ","End":"03:39.895","Text":"The right-hand side equal to the left-hand side."},{"Start":"03:39.895 ","End":"03:43.160","Text":"We\u0027ve shown this, which is the third property."},{"Start":"03:43.160 ","End":"03:47.790","Text":"Yes, W is a subspace and we\u0027re done."}],"ID":26469},{"Watched":false,"Name":"Exercise 2","Duration":"3m 51s","ChapterTopicVideoID":25645,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.180","Text":"In this exercise, we have to check if W is a subspace of m n of r. Remember"},{"Start":"00:06.180 ","End":"00:12.300","Text":"this means the square n by n matrices with real numbers."},{"Start":"00:12.300 ","End":"00:22.680","Text":"W is defined to be the set of matrices from here which commute with a given matrix B."},{"Start":"00:22.680 ","End":"00:24.930","Text":"If you\u0027re not sure what commute means,"},{"Start":"00:24.930 ","End":"00:27.870","Text":"it just means that when you multiply it from either side,"},{"Start":"00:27.870 ","End":"00:29.490","Text":"you get the same thing."},{"Start":"00:29.490 ","End":"00:33.150","Text":"More precisely, W is the set of all matrices A,"},{"Start":"00:33.150 ","End":"00:35.220","Text":"such that AB equals BA,"},{"Start":"00:35.220 ","End":"00:40.020","Text":"B is a fixed matrix."},{"Start":"00:40.870 ","End":"00:47.740","Text":"To show that it is, we\u0027d have to show 3 separate things."},{"Start":"00:47.740 ","End":"00:55.820","Text":"The first 1 of these 3 is that the 0 matrix just denoted it with a bold 0 is in"},{"Start":"00:55.820 ","End":"01:00.680","Text":"W. Certainly 0 commutes with B because 0"},{"Start":"01:00.680 ","End":"01:06.780","Text":"multiplied with B from the left or the right gives us 0, same thing."},{"Start":"01:08.780 ","End":"01:15.395","Text":"By definition that makes 0 belong to W. The next 1,"},{"Start":"01:15.395 ","End":"01:19.880","Text":"is to show that W is closed under multiplication by a scalar."},{"Start":"01:19.880 ","End":"01:24.260","Text":"Which just means that if you multiply something in W by a scalar"},{"Start":"01:24.260 ","End":"01:29.015","Text":"K then it should still stay in W. Let\u0027s check that."},{"Start":"01:29.015 ","End":"01:34.280","Text":"I\u0027m going to reinterpret the left and the right parts according to the definition,"},{"Start":"01:34.280 ","End":"01:38.890","Text":"A belongs to W is the same thing that AB equals BA."},{"Start":"01:38.890 ","End":"01:44.240","Text":"KA belongs to W is the same as saying that KA multiplied by B,"},{"Start":"01:44.240 ","End":"01:47.015","Text":"is the same as B multiplied by K A."},{"Start":"01:47.015 ","End":"01:50.240","Text":"We have to show that this implies this."},{"Start":"01:50.240 ","End":"01:55.215","Text":"It\u0027s just a little bit of matrix algebra."},{"Start":"01:55.215 ","End":"01:57.305","Text":"Suppose that this is true,"},{"Start":"01:57.305 ","End":"02:01.685","Text":"then I can multiply both sides by K,"},{"Start":"02:01.685 ","End":"02:05.585","Text":"but by the rules of matrix multiplication,"},{"Start":"02:05.585 ","End":"02:08.705","Text":"a constant K can be slipped inside,"},{"Start":"02:08.705 ","End":"02:10.550","Text":"either in front of the A, well,"},{"Start":"02:10.550 ","End":"02:16.250","Text":"in front of the first part or in front of the second part."},{"Start":"02:16.250 ","End":"02:19.010","Text":"In each case it lands with the A."},{"Start":"02:19.010 ","End":"02:22.475","Text":"This is exactly what we wanted to show."},{"Start":"02:22.475 ","End":"02:25.610","Text":"This means that K A is in W,"},{"Start":"02:25.610 ","End":"02:30.170","Text":"so that\u0027s the second of 3 parts."},{"Start":"02:30.170 ","End":"02:34.765","Text":"The final part is what we call closure under addition."},{"Start":"02:34.765 ","End":"02:38.760","Text":"If we have 2 matrices in W,"},{"Start":"02:38.760 ","End":"02:45.530","Text":"I called them A and C because B is reserved for our special fixed matrix."},{"Start":"02:45.530 ","End":"02:53.265","Text":"If 2 matrices A and C and W we have to show that there sum A plus C is also in W. I mean,"},{"Start":"02:53.265 ","End":"02:58.560","Text":"reinterpreting these statements by the definition of W,"},{"Start":"02:58.560 ","End":"03:03.695","Text":"this 1 says that A commutes with B and C commutes with B."},{"Start":"03:03.695 ","End":"03:08.255","Text":"This part says that A plus C commutes with B like so."},{"Start":"03:08.255 ","End":"03:12.025","Text":"We have to show that this implies this."},{"Start":"03:12.025 ","End":"03:17.030","Text":"Let\u0027s rewrite this as 2 separate equations,"},{"Start":"03:17.030 ","End":"03:19.460","Text":"1 above the other like a system."},{"Start":"03:19.460 ","End":"03:24.630","Text":"Next, we\u0027re going to do an addition at the left-hand side,"},{"Start":"03:24.630 ","End":"03:27.745","Text":"at the right-hand side of an equation."},{"Start":"03:27.745 ","End":"03:31.090","Text":"Next we use the distributive law that\u0027s actually 2 parts,"},{"Start":"03:31.090 ","End":"03:34.580","Text":"1 is distributed from the right and from the left and"},{"Start":"03:34.580 ","End":"03:38.330","Text":"so we get that this equals this and this equals this."},{"Start":"03:38.330 ","End":"03:41.345","Text":"But that\u0027s exactly what we were trying to show,"},{"Start":"03:41.345 ","End":"03:47.360","Text":"which means that A plus C is in W so we\u0027ve shown the third part and so yes,"},{"Start":"03:47.360 ","End":"03:51.630","Text":"indeed, W is a subspace and we\u0027re d1."}],"ID":26470},{"Watched":false,"Name":"Exercise 3","Duration":"2m 15s","ChapterTopicVideoID":25646,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"In this exercise,"},{"Start":"00:01.920 ","End":"00:06.915","Text":"we have to check if W is a subspace of M_n of R,"},{"Start":"00:06.915 ","End":"00:10.020","Text":"means square n by n matrices of real numbers,"},{"Start":"00:10.020 ","End":"00:15.150","Text":"where W is the set of matrices whose determinant is 0,"},{"Start":"00:15.150 ","End":"00:17.100","Text":"or we could write it like this,"},{"Start":"00:17.100 ","End":"00:23.190","Text":"that have all A such that determinant of A is 0."},{"Start":"00:23.190 ","End":"00:25.410","Text":"Write det A,"},{"Start":"00:25.410 ","End":"00:28.890","Text":"you could also write it as A inside bars,"},{"Start":"00:28.890 ","End":"00:31.185","Text":"I prefer the det."},{"Start":"00:31.185 ","End":"00:36.975","Text":"Let\u0027s see, we have to show 3 things if we\u0027re going to prove it."},{"Start":"00:36.975 ","End":"00:41.415","Text":"But to disprove it, we just have to show that 1 of the 3 things."},{"Start":"00:41.415 ","End":"00:47.460","Text":"If we started off trying to prove it yet 0 actually is in the space"},{"Start":"00:47.460 ","End":"00:53.730","Text":"W. But when we get to the last part of closure under addition, it fails."},{"Start":"00:53.730 ","End":"01:00.750","Text":"Now, all I have to do is find 1 counterexample of not being closed under addition."},{"Start":"01:00.750 ","End":"01:02.775","Text":"Let\u0027s take n equals 2,"},{"Start":"01:02.775 ","End":"01:04.980","Text":"2 by 2 matrices."},{"Start":"01:04.980 ","End":"01:08.685","Text":"Wasn\u0027t hard to find examples just play with the numbers."},{"Start":"01:08.685 ","End":"01:12.585","Text":"Make sure the product of the 2 diagonals is equal like this one,"},{"Start":"01:12.585 ","End":"01:14.880","Text":"1 times 6 is 3 times 2."},{"Start":"01:14.880 ","End":"01:17.010","Text":"The determinant is 0,"},{"Start":"01:17.010 ","End":"01:18.535","Text":"and here\u0027s another one,"},{"Start":"01:18.535 ","End":"01:21.155","Text":"B, its determinant is 0."},{"Start":"01:21.155 ","End":"01:25.485","Text":"But if we add them and figure out A plus B,"},{"Start":"01:25.485 ","End":"01:27.570","Text":"then 1 plus 1 is 2,"},{"Start":"01:27.570 ","End":"01:29.115","Text":"2 plus 1 is 3, and so on."},{"Start":"01:29.115 ","End":"01:30.810","Text":"We\u0027ll get 2, 3, 4, 7."},{"Start":"01:30.810 ","End":"01:37.635","Text":"But the determinant of this is not 0 because 2 times 7 is not the same as 4 times 3."},{"Start":"01:37.635 ","End":"01:40.095","Text":"So that\u0027s a counterexample."},{"Start":"01:40.095 ","End":"01:44.310","Text":"I might just add that also the second condition of"},{"Start":"01:44.310 ","End":"01:48.780","Text":"multiplication by a scalar also doesn\u0027t work because actually we"},{"Start":"01:48.780 ","End":"01:52.830","Text":"have a formula that the determinant of k times a"},{"Start":"01:52.830 ","End":"01:57.585","Text":"matrix is not k times the determinant of the matrix,"},{"Start":"01:57.585 ","End":"02:02.670","Text":"but k to the power of n times the determinant of A."},{"Start":"02:02.670 ","End":"02:09.855","Text":"I\u0027m just pointing out it would have failed also on the second of the 3 conditions."},{"Start":"02:09.855 ","End":"02:15.610","Text":"Counterexample, not a subspace, we\u0027re done."}],"ID":26471},{"Watched":false,"Name":"Exercise 4","Duration":"1m 54s","ChapterTopicVideoID":25647,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.675","Text":"In this exercise, we have to check if W is a subspace"},{"Start":"00:04.675 ","End":"00:10.660","Text":"of the set of square n by n matrices of real numbers,"},{"Start":"00:10.660 ","End":"00:17.035","Text":"where W is the set of matrices equal to their own square."},{"Start":"00:17.035 ","End":"00:19.405","Text":"That\u0027s probably not clear."},{"Start":"00:19.405 ","End":"00:26.695","Text":"More precisely, it\u0027s the set of all A from here such that A squared equals A."},{"Start":"00:26.695 ","End":"00:29.455","Text":"If you\u0027re still not sure what A squared means,"},{"Start":"00:29.455 ","End":"00:34.880","Text":"A squared is just A times A for matrices."},{"Start":"00:35.270 ","End":"00:39.195","Text":"Now, actually, I\u0027m going to disprove it,"},{"Start":"00:39.195 ","End":"00:42.795","Text":"there\u0027s no good reason why this should be a subspace."},{"Start":"00:42.795 ","End":"00:46.370","Text":"In fact, the first of the 3 conditions does hold."},{"Start":"00:46.370 ","End":"00:48.770","Text":"The 0 matrix does belong,"},{"Start":"00:48.770 ","End":"00:50.210","Text":"0 squared is 0,"},{"Start":"00:50.210 ","End":"00:53.375","Text":"but the other 2 don\u0027t apply."},{"Start":"00:53.375 ","End":"00:55.580","Text":"Let\u0027s just give a counter-example,"},{"Start":"00:55.580 ","End":"00:57.970","Text":"play around with the numbers."},{"Start":"00:57.970 ","End":"01:01.890","Text":"I found a pretty simple counterexample."},{"Start":"01:01.890 ","End":"01:03.195","Text":"Let\u0027s take n is 2."},{"Start":"01:03.195 ","End":"01:06.180","Text":"We having square 2 by 2 matrices,"},{"Start":"01:06.180 ","End":"01:09.724","Text":"let\u0027s just take the identity matrix in both cases."},{"Start":"01:09.724 ","End":"01:17.835","Text":"Certainly the identity matrix squared is itself and take A and B both to be this."},{"Start":"01:17.835 ","End":"01:20.100","Text":"A squared is A and B squared is B."},{"Start":"01:20.100 ","End":"01:23.590","Text":"Now how about the addition?"},{"Start":"01:24.170 ","End":"01:28.025","Text":"A plus B would be this matrix."},{"Start":"01:28.025 ","End":"01:32.800","Text":"Now this would not satisfy or doesn\u0027t satisfy that condition."},{"Start":"01:32.800 ","End":"01:37.515","Text":"Because A plus B squared is just this thing, times itself."},{"Start":"01:37.515 ","End":"01:39.740","Text":"If you do this multiplication,"},{"Start":"01:39.740 ","End":"01:41.430","Text":"we get 4, 0, 0, 0,"},{"Start":"01:41.430 ","End":"01:45.800","Text":"4, which is not the same as A plus B that we were supposed to get."},{"Start":"01:45.800 ","End":"01:48.950","Text":"If A plus B, we\u0027re in W."},{"Start":"01:48.950 ","End":"01:54.480","Text":"This counterexample shows that it\u0027s not a subspace and we\u0027re done."}],"ID":26472},{"Watched":false,"Name":"Exercise 5","Duration":"1m 36s","ChapterTopicVideoID":25648,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.960","Text":"In this exercise, we have to check if W is a subspace of this,"},{"Start":"00:06.960 ","End":"00:10.695","Text":"which is the square n by n matrices of real numbers,"},{"Start":"00:10.695 ","End":"00:16.170","Text":"where W is the set of upper triangular matrices."},{"Start":"00:16.170 ","End":"00:20.140","Text":"Now you probably might have forgotten what that means."},{"Start":"00:20.140 ","End":"00:24.125","Text":"I\u0027ll just illustrate it in the case where n is 3."},{"Start":"00:24.125 ","End":"00:30.770","Text":"Upper triangular means that we have something, anything here, here,"},{"Start":"00:30.770 ","End":"00:33.695","Text":"here, from the diagonal upwards,"},{"Start":"00:33.695 ","End":"00:37.835","Text":"but below the diagonal, we have 0s."},{"Start":"00:37.835 ","End":"00:43.745","Text":"There\u0027s like a 0 part everywhere below the diagonal."},{"Start":"00:43.745 ","End":"00:47.495","Text":"Now is this a subspace?"},{"Start":"00:47.495 ","End":"00:49.910","Text":"Turns out the answer is yes."},{"Start":"00:49.910 ","End":"00:51.230","Text":"Let\u0027s check the 3 conditions."},{"Start":"00:51.230 ","End":"00:57.805","Text":"The 0 matrix is certainly upper triangular because the 0 matrix has 0s here."},{"Start":"00:57.805 ","End":"01:01.660","Text":"In general, not just for n equals 3,"},{"Start":"01:01.660 ","End":"01:04.595","Text":"multiplying by a scalar."},{"Start":"01:04.595 ","End":"01:07.875","Text":"Well, if you take this and multiply it by some k,"},{"Start":"01:07.875 ","End":"01:09.280","Text":"wherever you had 0s,"},{"Start":"01:09.280 ","End":"01:13.730","Text":"you\u0027re still going to get 0s, so yes."},{"Start":"01:14.320 ","End":"01:17.990","Text":"If I take 2 matrices with this shape,"},{"Start":"01:17.990 ","End":"01:24.300","Text":"both upper triangular, then they\u0027re both going to have a 0 in this triangle."},{"Start":"01:24.300 ","End":"01:28.909","Text":"if you add them, it\u0027s still going to be 0 in this area below the diagonal."},{"Start":"01:28.909 ","End":"01:31.205","Text":"Yes, the sum also."},{"Start":"01:31.205 ","End":"01:36.690","Text":"The answer is yes, it is a subspace and we are done."}],"ID":26473},{"Watched":false,"Name":"Exercise 6","Duration":"3m 18s","ChapterTopicVideoID":25649,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"In this exercise, we have to check if W,"},{"Start":"00:04.080 ","End":"00:05.730","Text":"to be defined in a moment,"},{"Start":"00:05.730 ","End":"00:08.310","Text":"is a subspace of this,"},{"Start":"00:08.310 ","End":"00:12.570","Text":"which is the square n by n matrices of real numbers."},{"Start":"00:12.570 ","End":"00:21.445","Text":"W is the subset of matrices whose product with a given matrix B is 0."},{"Start":"00:21.445 ","End":"00:27.315","Text":"More precisely, its set of all A such that A times B is 0."},{"Start":"00:27.315 ","End":"00:29.535","Text":"This is a bit imprecise product,"},{"Start":"00:29.535 ","End":"00:33.180","Text":"I should say product from the left or from the right and this means that"},{"Start":"00:33.180 ","End":"00:35.220","Text":"our matrix is on the left and"},{"Start":"00:35.220 ","End":"00:39.060","Text":"the given matrix B is on the right, that this product is 0."},{"Start":"00:39.060 ","End":"00:46.205","Text":"I\u0027m going to show that it is a subspace so we have to show 3 parts."},{"Start":"00:46.205 ","End":"00:49.850","Text":"Firstly, we have to check is the 0 matrix in"},{"Start":"00:49.850 ","End":"00:58.010","Text":"W. It certainly is because it satisfies the condition that it times B is 0."},{"Start":"00:58.010 ","End":"01:01.910","Text":"0 times B is 0, so that 0 is in W. Then"},{"Start":"01:01.910 ","End":"01:08.810","Text":"the next part is what we call closure under scalar multiplications,"},{"Start":"01:08.810 ","End":"01:10.115","Text":"just put a check mark here,"},{"Start":"01:10.115 ","End":"01:11.810","Text":"since we\u0027ve done that part."},{"Start":"01:11.810 ","End":"01:18.170","Text":"This part means that if a matrix is in W,"},{"Start":"01:18.170 ","End":"01:22.730","Text":"then the constant scalar times it is also there."},{"Start":"01:22.730 ","End":"01:27.660","Text":"I can re-interpret these 2 things from"},{"Start":"01:27.660 ","End":"01:34.110","Text":"the definition of W. This part says that A times B is 0,"},{"Start":"01:34.110 ","End":"01:37.505","Text":"and this says that kA times B is 0."},{"Start":"01:37.505 ","End":"01:40.040","Text":"Does this imply this?"},{"Start":"01:40.040 ","End":"01:43.040","Text":"Well, yes it does, because if this is 0,"},{"Start":"01:43.040 ","End":"01:48.320","Text":"certainly I can multiply both sides by k, I mean,"},{"Start":"01:48.320 ","End":"01:52.340","Text":"so k times 0 is 0."},{"Start":"01:52.340 ","End":"01:57.800","Text":"I can just put a brackets around the kA because it\u0027s one of the properties of matrices,"},{"Start":"01:57.800 ","End":"02:01.635","Text":"so kA times B is 0."},{"Start":"02:01.635 ","End":"02:05.535","Text":"That\u0027s equivalent to saying that kA is in W,"},{"Start":"02:05.535 ","End":"02:07.950","Text":"so that\u0027s the second part."},{"Start":"02:07.950 ","End":"02:13.300","Text":"The third part is the part we call closure under addition."},{"Start":"02:13.300 ","End":"02:18.940","Text":"What we have to do is show that if we have 2 matrices in W,"},{"Start":"02:18.940 ","End":"02:23.740","Text":"that their sum is also in W. I used the letters A and C because B is"},{"Start":"02:23.740 ","End":"02:29.015","Text":"already reserved for our special matrix from the definition."},{"Start":"02:29.015 ","End":"02:31.890","Text":"Let\u0027s see if this is true."},{"Start":"02:31.890 ","End":"02:40.320","Text":"Reinterpreting this part says that A times B is 0 and that C times B is 0."},{"Start":"02:40.320 ","End":"02:45.455","Text":"This part interpret as A plus C times B is 0."},{"Start":"02:45.455 ","End":"02:48.950","Text":"We want to know, does this imply this?"},{"Start":"02:48.950 ","End":"02:51.290","Text":"I\u0027ll just write them one above the other."},{"Start":"02:51.290 ","End":"02:53.360","Text":"It\u0027s just easier to see."},{"Start":"02:53.360 ","End":"03:00.085","Text":"We have these 2 equalities, let\u0027s add them."},{"Start":"03:00.085 ","End":"03:04.830","Text":"We get this and now use a distributive law to take B out,"},{"Start":"03:04.830 ","End":"03:08.345","Text":"because B has to stay on the right and we get this."},{"Start":"03:08.345 ","End":"03:14.450","Text":"But that just means that A plus C is also in W like we wanted so that\u0027s the 3 parts,"},{"Start":"03:14.450 ","End":"03:18.810","Text":"so yes, it\u0027s a subspace and we\u0027re done."}],"ID":26474},{"Watched":false,"Name":"Exercise 7","Duration":"2m 43s","ChapterTopicVideoID":25650,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.950","Text":"In this exercise, we have the space of N by N square matrices with real numbers."},{"Start":"00:07.950 ","End":"00:11.445","Text":"We have to check if W is a subspace,"},{"Start":"00:11.445 ","End":"00:17.970","Text":"where W is defined to be the set of those matrices from here which have a trace(0)."},{"Start":"00:17.970 ","End":"00:23.039","Text":"More formally, we can write it like this. Trace(A) is 0."},{"Start":"00:23.039 ","End":"00:27.900","Text":"You should go and review the section on the trace of a matrix but"},{"Start":"00:27.900 ","End":"00:33.585","Text":"the trace basically just means that some of the elements along the main diagonal,"},{"Start":"00:33.585 ","End":"00:37.035","Text":"and there are certain properties it has."},{"Start":"00:37.035 ","End":"00:41.789","Text":"Does the 0 matrix belong to W?"},{"Start":"00:41.789 ","End":"00:45.950","Text":"Well, certainly, the trace of the 0 matrix"},{"Start":"00:45.950 ","End":"00:50.645","Text":"is 0 because the sum of the elements on the diagonal will only elements is 0."},{"Start":"00:50.645 ","End":"00:55.160","Text":"It\u0027s 0, which makes it in W. The next property we have to"},{"Start":"00:55.160 ","End":"00:59.750","Text":"show called the closure under scalar multiplication."},{"Start":"00:59.750 ","End":"01:01.355","Text":"If A is inside,"},{"Start":"01:01.355 ","End":"01:05.315","Text":"then k times A should also be in the subspace."},{"Start":"01:05.315 ","End":"01:09.470","Text":"Re-interpreting each of these to say that i is in"},{"Start":"01:09.470 ","End":"01:14.440","Text":"W is the same thing to say that the trace(A) is 0 and the other"},{"Start":"01:14.440 ","End":"01:18.720","Text":"interprets as the trace (kA) is 0"},{"Start":"01:18.720 ","End":"01:25.240","Text":"but there\u0027s a property if traced that says that trace of a constant times a matrix,"},{"Start":"01:25.240 ","End":"01:28.000","Text":"you can just take that constant outside."},{"Start":"01:28.000 ","End":"01:32.650","Text":"Where we\u0027re heading is that we have to show that this implies this."},{"Start":"01:32.650 ","End":"01:37.375","Text":"If this is true, then trace(kA) is K times trace(A),"},{"Start":"01:37.375 ","End":"01:39.295","Text":"but trace(A) is 0."},{"Start":"01:39.295 ","End":"01:40.930","Text":"This is k times 0."},{"Start":"01:40.930 ","End":"01:43.135","Text":"This is indeed 0,"},{"Start":"01:43.135 ","End":"01:51.180","Text":"which means that kA really is in W. That\u0027s one part, second part."},{"Start":"01:51.180 ","End":"01:55.840","Text":"Now the third part is called the closure under addition."},{"Start":"01:55.840 ","End":"01:59.410","Text":"But we have to answer the question if A and B are in W,"},{"Start":"01:59.410 ","End":"02:02.810","Text":"is A plus B also in W?"},{"Start":"02:02.810 ","End":"02:06.930","Text":"Reinterpreting from the definition of W. To say that A and B"},{"Start":"02:06.930 ","End":"02:10.935","Text":"are in W means a trace(A) is 0 and trace(B) is 0."},{"Start":"02:10.935 ","End":"02:15.860","Text":"To say that this is in W is the same thing as to say that the trace of A plus B is 0."},{"Start":"02:15.860 ","End":"02:18.560","Text":"Does this imply this?"},{"Start":"02:18.560 ","End":"02:22.550","Text":"The answer is yes because one of the properties of the trace"},{"Start":"02:22.550 ","End":"02:27.530","Text":"is that the trace of a sum of 2 matrices is the sum of the traces."},{"Start":"02:27.530 ","End":"02:30.830","Text":"Trace(A) plus B is this plus this,"},{"Start":"02:30.830 ","End":"02:32.480","Text":"but each of these is 0."},{"Start":"02:32.480 ","End":"02:38.030","Text":"We do get 0, which means that A plus B is indeed in W,"},{"Start":"02:38.030 ","End":"02:39.875","Text":"and that\u0027s the third one."},{"Start":"02:39.875 ","End":"02:43.800","Text":"Yes, it\u0027s a subspace. We\u0027re done."}],"ID":26475},{"Watched":false,"Name":"Exercise 8","Duration":"3m 56s","ChapterTopicVideoID":25651,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.050","Text":"In this exercise, we start off with the space of"},{"Start":"00:04.050 ","End":"00:08.025","Text":"the square n by n matrices of real numbers."},{"Start":"00:08.025 ","End":"00:09.570","Text":"That\u0027s how it\u0027s denoted."},{"Start":"00:09.570 ","End":"00:14.670","Text":"We have to check if W is a subspace where W is the set of"},{"Start":"00:14.670 ","End":"00:20.670","Text":"matrices with this unusual property that the sum of each row is 0."},{"Start":"00:20.670 ","End":"00:23.745","Text":"If you sum any of the n rows,"},{"Start":"00:23.745 ","End":"00:26.865","Text":"you will get 0 in all cases."},{"Start":"00:26.865 ","End":"00:34.945","Text":"It turns out this is a subspace so we have to show 3 parts, 3 axioms."},{"Start":"00:34.945 ","End":"00:37.415","Text":"But to make it simpler,"},{"Start":"00:37.415 ","End":"00:39.260","Text":"let\u0027s take n equals 3,"},{"Start":"00:39.260 ","End":"00:41.450","Text":"but the proof will be totally general,"},{"Start":"00:41.450 ","End":"00:42.560","Text":"won\u0027t rely on 3."},{"Start":"00:42.560 ","End":"00:45.760","Text":"I just don\u0027t want to put dot everywhere."},{"Start":"00:45.760 ","End":"00:47.615","Text":"As you go along, you\u0027ll see,"},{"Start":"00:47.615 ","End":"00:50.930","Text":"follow this and see that 3 is nothing special."},{"Start":"00:50.930 ","End":"00:56.090","Text":"I could do this for any n. The 3 properties, the 3 axioms."},{"Start":"00:56.090 ","End":"00:57.925","Text":"First of all, it\u0027s 0,"},{"Start":"00:57.925 ","End":"01:01.320","Text":"0 matrix is just 3 by 3,"},{"Start":"01:01.320 ","End":"01:04.280","Text":"were all zeros, each row is 0, 0,"},{"Start":"01:04.280 ","End":"01:09.020","Text":"0, and the sum of each row is 0 plus 0 plus 0, which is 0."},{"Start":"01:09.020 ","End":"01:18.035","Text":"So 0 matrix, easy to see is in W, has the property."},{"Start":"01:18.035 ","End":"01:24.980","Text":"Next, we need to show that it\u0027s closed on the scalar multiplication."},{"Start":"01:24.980 ","End":"01:29.675","Text":"I\u0027m doing this a little bit informally in hand-waving style."},{"Start":"01:29.675 ","End":"01:32.000","Text":"I think it\u0027s easier that way."},{"Start":"01:32.000 ","End":"01:35.120","Text":"Let\u0027s suppose that we have some matrix, I don\u0027t know, X,"},{"Start":"01:35.120 ","End":"01:39.125","Text":"call it, where each row adds up to 0."},{"Start":"01:39.125 ","End":"01:43.475","Text":"If I multiply that matrix by k,"},{"Start":"01:43.475 ","End":"01:45.875","Text":"the sum of each row will still be 0."},{"Start":"01:45.875 ","End":"01:49.055","Text":"I mean, well, if 1 of the rows is,"},{"Start":"01:49.055 ","End":"01:51.260","Text":"let\u0027s say a, b,"},{"Start":"01:51.260 ","End":"01:55.330","Text":"c, then a plus b plus c is 0."},{"Start":"01:55.330 ","End":"01:57.600","Text":"In the matrix kX,"},{"Start":"01:57.600 ","End":"01:59.535","Text":"that same row, I mean,"},{"Start":"01:59.535 ","End":"02:00.950","Text":"if we took the second row here,"},{"Start":"02:00.950 ","End":"02:02.825","Text":"we take the second row here or whatever,"},{"Start":"02:02.825 ","End":"02:05.705","Text":"that would just be ka, kb, kc."},{"Start":"02:05.705 ","End":"02:08.150","Text":"What is this sum?"},{"Start":"02:08.150 ","End":"02:10.940","Text":"Well, this sum by the distributive law,"},{"Start":"02:10.940 ","End":"02:16.640","Text":"we take k outside the brackets and this is 0 so it\u0027s still 0."},{"Start":"02:16.640 ","End":"02:18.935","Text":"That\u0027s 1 and 2."},{"Start":"02:18.935 ","End":"02:20.360","Text":"Now for 3,"},{"Start":"02:20.360 ","End":"02:22.415","Text":"the closure under addition,"},{"Start":"02:22.415 ","End":"02:24.875","Text":"let\u0027s suppose that we have 2 matrices,"},{"Start":"02:24.875 ","End":"02:28.010","Text":"X and Y, each in the subspace,"},{"Start":"02:28.010 ","End":"02:31.460","Text":"meaning that they have zero-sum rows,"},{"Start":"02:31.460 ","End":"02:32.825","Text":"all of the rows."},{"Start":"02:32.825 ","End":"02:37.625","Text":"Let\u0027s see if the property remains true for X plus Y."},{"Start":"02:37.625 ","End":"02:40.085","Text":"Let\u0027s take a specific row,"},{"Start":"02:40.085 ","End":"02:42.335","Text":"say the second row."},{"Start":"02:42.335 ","End":"02:43.820","Text":"I mean, I could have said,"},{"Start":"02:43.820 ","End":"02:47.914","Text":"I guess the ith throw just any definite row number."},{"Start":"02:47.914 ","End":"02:51.370","Text":"Let\u0027s look at it in X and in Y."},{"Start":"02:51.370 ","End":"02:55.500","Text":"That row in X will be let\u0027s say a, b,"},{"Start":"02:55.500 ","End":"02:59.280","Text":"c, and in Y it\u0027ll be capital A, B,"},{"Start":"02:59.280 ","End":"03:04.770","Text":"C. If we add X plus Y,"},{"Start":"03:04.770 ","End":"03:07.830","Text":"will be a plus A, b plus B,"},{"Start":"03:07.830 ","End":"03:12.155","Text":"c plus C. But because of the property,"},{"Start":"03:12.155 ","End":"03:16.070","Text":"a plus b plus c is 0 because X is in the subspace and A"},{"Start":"03:16.070 ","End":"03:20.690","Text":"plus B plus C capital is 0 because Y is in the subset."},{"Start":"03:20.690 ","End":"03:24.390","Text":"We just rewrite the order of the addition."},{"Start":"03:24.560 ","End":"03:33.500","Text":"The sum of these 3 is equal just by rearranging the order to a plus b plus c lowercase,"},{"Start":"03:33.500 ","End":"03:36.210","Text":"plus uppercase A, B,"},{"Start":"03:36.210 ","End":"03:38.715","Text":"C. This is 0 and this is 0,"},{"Start":"03:38.715 ","End":"03:40.380","Text":"so this is 0."},{"Start":"03:40.380 ","End":"03:43.110","Text":"That row in the sum is also 0,"},{"Start":"03:43.110 ","End":"03:44.150","Text":"and that\u0027s for any row,"},{"Start":"03:44.150 ","End":"03:47.570","Text":"not just the second or the ith row."},{"Start":"03:47.570 ","End":"03:56.880","Text":"That\u0027s it. That\u0027s the third part and we\u0027re done. That\u0027s it for the clip."}],"ID":26476},{"Watched":false,"Name":"Exercise 9","Duration":"5m 25s","ChapterTopicVideoID":25652,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.640","Text":"In this exercise, we have to check if W is a subspace of P_n of R,"},{"Start":"00:06.640 ","End":"00:13.890","Text":"this is the polynomials of degree up to and including n with real coefficients and"},{"Start":"00:13.890 ","End":"00:21.735","Text":"W is defined as the set of polynomials which have 4 as a root."},{"Start":"00:21.735 ","End":"00:24.205","Text":"Let me rephrase that."},{"Start":"00:24.205 ","End":"00:29.720","Text":"It means the set of all polynomials p of x such that p of 4 is 0,"},{"Start":"00:29.720 ","End":"00:33.870","Text":"4 is a root means when you substitute 4 you get 0."},{"Start":"00:34.330 ","End":"00:38.990","Text":"We\u0027re going to show that it is a subspace and you have to prove 3 things."},{"Start":"00:38.990 ","End":"00:43.855","Text":"The first is that 0 is in the subspace,"},{"Start":"00:43.855 ","End":"00:46.820","Text":"0 means the 0 polynomial p of x,"},{"Start":"00:46.820 ","End":"00:51.800","Text":"which is always 0 for every x, is that in W?"},{"Start":"00:51.800 ","End":"00:58.175","Text":"This exercise involves P_n of R and I want to remind you what this is."},{"Start":"00:58.175 ","End":"01:08.580","Text":"This is the vector space of all polynomials with real coefficients of degree not n,"},{"Start":"01:08.580 ","End":"01:14.270","Text":"but less than or equal to n. We want to check if W is"},{"Start":"01:14.270 ","End":"01:22.090","Text":"a subspace where w is the subset of all polynomials which have 4 as a root."},{"Start":"01:22.090 ","End":"01:25.310","Text":"To have 4 as a root means when you substitute them, you get zeros."},{"Start":"01:25.310 ","End":"01:32.605","Text":"I could rewrite it more formally as W is the set of all p of x such that p of 4 is 0."},{"Start":"01:32.605 ","End":"01:34.415","Text":"I\u0027ll even give you an example."},{"Start":"01:34.415 ","End":"01:41.490","Text":"Suppose I take p of x equals x squared minus 16,"},{"Start":"01:41.490 ","End":"01:48.965","Text":"then p would be in W because if I substitute 4 I get 0."},{"Start":"01:48.965 ","End":"01:51.290","Text":"But if I took another polynomial, I don\u0027t know,"},{"Start":"01:51.290 ","End":"01:56.565","Text":"q of x is just x cubed minus 5,"},{"Start":"01:56.565 ","End":"02:03.179","Text":"then that\u0027s not in W because when I substitute 4 I most certainly do not get 0."},{"Start":"02:03.179 ","End":"02:04.910","Text":"Now that you\u0027ve got the idea,"},{"Start":"02:04.910 ","End":"02:08.900","Text":"we\u0027re going to show that it is a subspace so we have to show 3 things."},{"Start":"02:08.900 ","End":"02:14.975","Text":"The first thing is that the 0 polynomial has to be in the subspace."},{"Start":"02:14.975 ","End":"02:21.620","Text":"The 0 polynomial means the polynomial which is 0 for every x."},{"Start":"02:22.070 ","End":"02:27.560","Text":"Of course this is going to be in W. Because if p of x is 0 for every x,"},{"Start":"02:27.560 ","End":"02:33.230","Text":"then in particular p of 4 is 0 and so this 0 polynomial really is in"},{"Start":"02:33.230 ","End":"02:41.945","Text":"W. The second thing we have to show is what we call closure under scalar multiplication."},{"Start":"02:41.945 ","End":"02:44.120","Text":"That if p is in W,"},{"Start":"02:44.120 ","End":"02:49.400","Text":"then kp is also in W. Let\u0027s check this."},{"Start":"02:49.400 ","End":"02:54.050","Text":"Let me reinterpret both sides of this implication to say that"},{"Start":"02:54.050 ","End":"02:58.934","Text":"p is in W means that p of 4 is 0."},{"Start":"02:58.934 ","End":"03:05.959","Text":"To say that kp is in W is equivalent to saying that kp of 4 is 0."},{"Start":"03:05.959 ","End":"03:12.275","Text":"We want to know, does this imply this? Well, yes."},{"Start":"03:12.275 ","End":"03:15.395","Text":"Because if p is in W,"},{"Start":"03:15.395 ","End":"03:18.470","Text":"then it means that p of 4 is 0,"},{"Start":"03:18.470 ","End":"03:20.960","Text":"like we said and if p of 4 is 0,"},{"Start":"03:20.960 ","End":"03:26.660","Text":"I can multiply both sides by k and get k times p of 4 is 0."},{"Start":"03:26.660 ","End":"03:32.510","Text":"Now the k can go inside we can take the polynomial kp."},{"Start":"03:32.510 ","End":"03:36.725","Text":"This is how k times a polynomial is defined in fact."},{"Start":"03:36.725 ","End":"03:43.370","Text":"The polynomial kp of 4 is also 0 and hence it means that"},{"Start":"03:43.370 ","End":"03:52.620","Text":"kp is really in W. That was the second of 3 things we have to show."},{"Start":"03:52.620 ","End":"03:56.250","Text":"The last 1 is called closure under addition."},{"Start":"03:56.250 ","End":"04:00.230","Text":"That was the question mark,"},{"Start":"04:00.230 ","End":"04:04.730","Text":"we want to show this that if p and q are both in W,"},{"Start":"04:04.730 ","End":"04:13.160","Text":"then the sum of the polynomials p plus q is also in W. I\u0027m going to reinterpret"},{"Start":"04:13.160 ","End":"04:18.260","Text":"these statements here to say that p and q are in"},{"Start":"04:18.260 ","End":"04:24.710","Text":"W is the same as saying that p of 4 is 0 and q of 4 is 0."},{"Start":"04:24.710 ","End":"04:28.085","Text":"To say that p plus q is in W,"},{"Start":"04:28.085 ","End":"04:33.660","Text":"is to say that p plus q of 4 is 0."},{"Start":"04:34.250 ","End":"04:39.119","Text":"Our question is, does this imply this?"},{"Start":"04:39.119 ","End":"04:45.250","Text":"Supposing that these are both true,"},{"Start":"04:45.670 ","End":"04:50.150","Text":"by definition of the sum of polynomials,"},{"Start":"04:50.150 ","End":"04:53.270","Text":"we take the sum separately."},{"Start":"04:53.270 ","End":"05:01.450","Text":"This would be p of 4 plus q of 4,"},{"Start":"05:01.450 ","End":"05:08.240","Text":"but these already we\u0027ve taken to be 0 so it\u0027s 0 plus 0, which is 0."},{"Start":"05:08.240 ","End":"05:16.280","Text":"Which means that p plus q really does belong to W. Perhaps we should write it"},{"Start":"05:16.280 ","End":"05:25.060","Text":"with a symbolic x and so that\u0027s the third part proven and we\u0027re done."}],"ID":26477},{"Watched":false,"Name":"Exercise 10","Duration":"3m 5s","ChapterTopicVideoID":25653,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.830 ","End":"00:02.880","Text":"Before we start the exercise,"},{"Start":"00:02.880 ","End":"00:05.670","Text":"just want to remind you what P_n of R means."},{"Start":"00:05.670 ","End":"00:12.270","Text":"It\u0027s a set of all polynomials p with real coefficients with"},{"Start":"00:12.270 ","End":"00:20.730","Text":"degree less than or equal to n. Not just equal to n,"},{"Start":"00:20.730 ","End":"00:24.720","Text":"but less than or equal to n. In this case,"},{"Start":"00:24.720 ","End":"00:26.940","Text":"we want to check if W is a subspace,"},{"Start":"00:26.940 ","End":"00:33.315","Text":"where W consists of all the polynomials with degree less than or equal to 4,"},{"Start":"00:33.315 ","End":"00:35.805","Text":"that\u0027s a specific number."},{"Start":"00:35.805 ","End":"00:42.325","Text":"In formal terms, W is a set of all p of x that belong to here,"},{"Start":"00:42.325 ","End":"00:46.480","Text":"such that the degree of p is less than or equal to 4."},{"Start":"00:46.970 ","End":"00:50.700","Text":"I\u0027m going to show that W is a subspace,"},{"Start":"00:50.700 ","End":"00:54.725","Text":"and so we have to show that it satisfies 3 axioms."},{"Start":"00:54.725 ","End":"01:00.650","Text":"The first is that W has to contain the 0 polynomial."},{"Start":"01:00.650 ","End":"01:06.560","Text":"Now, there is some controversy about the degree of the 0 polynomial,"},{"Start":"01:06.560 ","End":"01:10.310","Text":"but everyone agrees that it\u0027s less than or equal to 4."},{"Start":"01:10.310 ","End":"01:13.310","Text":"In this course, we take it as minus infinity,"},{"Start":"01:13.310 ","End":"01:15.300","Text":"but it\u0027s not universally accepted."},{"Start":"01:15.300 ","End":"01:19.859","Text":"In any event, it\u0027s always in the subspace."},{"Start":"01:20.180 ","End":"01:24.480","Text":"The second part is closure and the scalar multiplication."},{"Start":"01:24.480 ","End":"01:27.620","Text":"We have to show that if a polynomial is in W,"},{"Start":"01:27.620 ","End":"01:30.320","Text":"then if you multiply it by a scalar,"},{"Start":"01:30.320 ","End":"01:33.440","Text":"it\u0027s still in W. To be precise,"},{"Start":"01:33.440 ","End":"01:35.420","Text":"we have to distinguish 2 cases."},{"Start":"01:35.420 ","End":"01:37.310","Text":"If it\u0027s a non 0 scalar,"},{"Start":"01:37.310 ","End":"01:39.110","Text":"then it makes no difference to the degree,"},{"Start":"01:39.110 ","End":"01:42.575","Text":"the powers of x are not changed, just the coefficients."},{"Start":"01:42.575 ","End":"01:45.650","Text":"If we multiply it by 0,"},{"Start":"01:45.650 ","End":"01:48.320","Text":"then it gives us the 0 polynomial,"},{"Start":"01:48.320 ","End":"01:51.290","Text":"which already from part 1 is in there."},{"Start":"01:51.290 ","End":"01:55.235","Text":"Either case, multiplying by a scalar 0 or non 0,"},{"Start":"01:55.235 ","End":"02:00.200","Text":"will leave a polynomial in the subspace."},{"Start":"02:00.200 ","End":"02:02.795","Text":"The degree will remain less than or equal to 4."},{"Start":"02:02.795 ","End":"02:05.255","Text":"The third part."},{"Start":"02:05.255 ","End":"02:10.130","Text":"That\u0027s to show that it\u0027s closed under addition,"},{"Start":"02:10.130 ","End":"02:13.820","Text":"which means that the sum of 2 polynomials of degree less than or equal to"},{"Start":"02:13.820 ","End":"02:18.930","Text":"4 has to also be less than or equal to 4."},{"Start":"02:18.940 ","End":"02:21.170","Text":"I think that\u0027s fairly clear."},{"Start":"02:21.170 ","End":"02:25.810","Text":"I mean, if it only contains powers of x to the 4 or less,"},{"Start":"02:25.810 ","End":"02:31.175","Text":"when you add them, it\u0027s still only going to have powers of x to the 4 or less."},{"Start":"02:31.175 ","End":"02:33.870","Text":"That\u0027s part 3 also."},{"Start":"02:33.870 ","End":"02:35.600","Text":"Yes, it is a subspace."},{"Start":"02:35.600 ","End":"02:38.155","Text":"I\u0027d just like to make an observation."},{"Start":"02:38.155 ","End":"02:42.290","Text":"Not essential and you can just quit now."},{"Start":"02:42.290 ","End":"02:48.590","Text":"We\u0027ve shown that if n is bigger than 4 at any rate,"},{"Start":"02:48.590 ","End":"02:52.470","Text":"then p_4 of R,"},{"Start":"02:52.470 ","End":"02:57.035","Text":"polynomials of degree less than or equal to 4 is actually a subspace of"},{"Start":"02:57.035 ","End":"03:03.290","Text":"the polynomials of degree n. Just an observation,"},{"Start":"03:03.290 ","End":"03:05.940","Text":"that\u0027s all. We\u0027re done."}],"ID":26478},{"Watched":false,"Name":"Exercise 11","Duration":"1m 40s","ChapterTopicVideoID":25654,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.980","Text":"Here we have another exercise involving P_n of R. Just to remind you,"},{"Start":"00:04.980 ","End":"00:12.870","Text":"it\u0027s the polynomials of degree less than or equal to n with real number coefficients."},{"Start":"00:12.870 ","End":"00:17.130","Text":"We have to check if W is a subspace where W"},{"Start":"00:17.130 ","End":"00:21.510","Text":"consists of all the polynomials with integer coefficients."},{"Start":"00:21.510 ","End":"00:27.250","Text":"Integer is a whole number but could be positive or negative or 0."},{"Start":"00:27.250 ","End":"00:31.430","Text":"Actually, I\u0027m going to show that W is not a subspace."},{"Start":"00:31.430 ","End":"00:33.710","Text":"It doesn\u0027t satisfy all 3 axioms."},{"Start":"00:33.710 ","End":"00:35.930","Text":"I just have to show you 1 axiom."},{"Start":"00:35.930 ","End":"00:37.865","Text":"We\u0027ll take the second,"},{"Start":"00:37.865 ","End":"00:42.200","Text":"which is closure under scalar multiplication."},{"Start":"00:42.200 ","End":"00:44.660","Text":"We would normally ask ourselves,"},{"Start":"00:44.660 ","End":"00:46.520","Text":"if p is in W,"},{"Start":"00:46.520 ","End":"00:49.040","Text":"is KP also in W?"},{"Start":"00:49.040 ","End":"00:52.640","Text":"The answer is no or at least not always."},{"Start":"00:52.640 ","End":"00:55.450","Text":"Just need 1 counterexample."},{"Start":"00:55.450 ","End":"00:57.510","Text":"Here\u0027s our counterexample."},{"Start":"00:57.510 ","End":"01:05.655","Text":"Let\u0027s take p of x to be 1 plus 2x plus 3x cubed,"},{"Start":"01:05.655 ","End":"01:09.710","Text":"assume that n is bigger than 3 or bigger or equal to 3."},{"Start":"01:09.710 ","End":"01:16.069","Text":"Certainly, the coefficients are integers 1, 2, and 3."},{"Start":"01:16.069 ","End":"01:20.535","Text":"But if I multiply by a scalar like"},{"Start":"01:20.535 ","End":"01:28.260","Text":"0.5 and then I\u0027ll get this polynomial which does not have integer coefficients."},{"Start":"01:28.260 ","End":"01:31.455","Text":"One of them is integer but the others are not."},{"Start":"01:31.455 ","End":"01:37.175","Text":"It\u0027s not in W. We do not have closure under scalar multiplication,"},{"Start":"01:37.175 ","End":"01:40.830","Text":"and this is not a subspace. We\u0027re done."}],"ID":26479},{"Watched":false,"Name":"Exercise 12","Duration":"2m 44s","ChapterTopicVideoID":25655,"CourseChapterTopicPlaylistID":246315,"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 to check if W is"},{"Start":"00:04.530 ","End":"00:10.655","Text":"a subspace of the space of polynomials of degree less than or equal to n,"},{"Start":"00:10.655 ","End":"00:15.810","Text":"where W is defined to be the set of polynomials which have"},{"Start":"00:15.810 ","End":"00:21.045","Text":"only even powers of x among the terms."},{"Start":"00:21.045 ","End":"00:23.700","Text":"I\u0027ll just give you an example."},{"Start":"00:23.700 ","End":"00:35.490","Text":"Say 3x^4 minus x^6 plus x squared minus 5,"},{"Start":"00:35.490 ","End":"00:37.380","Text":"all even powers of x,"},{"Start":"00:37.380 ","End":"00:41.410","Text":"even the constant, it\u0027s like to x^0."},{"Start":"00:41.470 ","End":"00:45.830","Text":"An example of something not in W."},{"Start":"00:45.830 ","End":"00:48.714","Text":"Let me just write that this is in W,"},{"Start":"00:48.714 ","End":"00:54.250","Text":"not in W would be x cubed minus x^4."},{"Start":"00:54.250 ","End":"00:58.670","Text":"That would not be in W because it has an odd power."},{"Start":"00:58.670 ","End":"01:01.845","Text":"This 1 is even, but if it had even 1 odd power,"},{"Start":"01:01.845 ","End":"01:05.930","Text":"then it\u0027s not in W. I\u0027m going to show"},{"Start":"01:05.930 ","End":"01:10.685","Text":"that it is a subspace that we have to show it satisfies 3 axioms."},{"Start":"01:10.685 ","End":"01:14.270","Text":"The 0 polynomial, it\u0027s a bit debatable."},{"Start":"01:14.270 ","End":"01:15.500","Text":"It\u0027s not immediately clear."},{"Start":"01:15.500 ","End":"01:18.050","Text":"I can consider it as having only even powers of x."},{"Start":"01:18.050 ","End":"01:20.695","Text":"It doesn\u0027t really have any terms."},{"Start":"01:20.695 ","End":"01:23.955","Text":"Well, it certainly has no terms with odd degree."},{"Start":"01:23.955 ","End":"01:24.800","Text":"Let\u0027s put it that way."},{"Start":"01:24.800 ","End":"01:28.310","Text":"There\u0027s no x or x^3 or x^5 or anything like that."},{"Start":"01:28.310 ","End":"01:30.280","Text":"It only has even powers."},{"Start":"01:30.280 ","End":"01:32.905","Text":"We\u0027ll take it by definition."},{"Start":"01:32.905 ","End":"01:38.105","Text":"Next, we check closure under multiplication by a scalar."},{"Start":"01:38.105 ","End":"01:41.210","Text":"To be precise, we have to distinguish 2 cases."},{"Start":"01:41.210 ","End":"01:43.550","Text":"If it\u0027s a non-0 scalar,"},{"Start":"01:43.550 ","End":"01:46.325","Text":"then the degrees of all the terms doesn\u0027t change,"},{"Start":"01:46.325 ","End":"01:48.920","Text":"it\u0027s just the coefficient changes."},{"Start":"01:48.920 ","End":"01:54.380","Text":"On the other hand, if you multiply the polynomial by the scalar 0,"},{"Start":"01:54.380 ","End":"01:56.750","Text":"it just gives the 0 polynomial,"},{"Start":"01:56.750 ","End":"02:05.200","Text":"and we\u0027ve already agreed that that only has even powers or terms of even degree."},{"Start":"02:06.010 ","End":"02:08.495","Text":"That\u0027s okay."},{"Start":"02:08.495 ","End":"02:10.670","Text":"Now on part 3,"},{"Start":"02:10.670 ","End":"02:13.279","Text":"which is called closure under addition,"},{"Start":"02:13.279 ","End":"02:23.525","Text":"I have to show that if I add 2 polynomial which only have even degrees,"},{"Start":"02:23.525 ","End":"02:27.525","Text":"then it also has only even degrees."},{"Start":"02:27.525 ","End":"02:35.840","Text":"Certainly, there\u0027s nowhere odd degrees can suddenly appear if I add only even degrees."},{"Start":"02:35.840 ","End":"02:38.550","Text":"That\u0027s also fairly clear."},{"Start":"02:38.550 ","End":"02:41.850","Text":"So yes, W is a subspace."},{"Start":"02:41.850 ","End":"02:44.290","Text":"We are done."}],"ID":26480},{"Watched":false,"Name":"Exercise 13","Duration":"52s","ChapterTopicVideoID":25656,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.240","Text":"In this exercise, we again have the space of polynomials of degree"},{"Start":"00:06.240 ","End":"00:13.680","Text":"less than or equal to n. We take w to be the subset,"},{"Start":"00:13.680 ","End":"00:20.760","Text":"which is all the polynomials with degree between 4 and 7 inclusive."},{"Start":"00:20.760 ","End":"00:24.090","Text":"We want to know if W is a subspace,"},{"Start":"00:24.090 ","End":"00:27.320","Text":"and I\u0027m claiming that it\u0027s not,"},{"Start":"00:27.320 ","End":"00:29.720","Text":"it doesn\u0027t satisfy all 3 axioms."},{"Start":"00:29.720 ","End":"00:37.895","Text":"In fact, it already fails on the first axiom that the 0 polynomial is not in there."},{"Start":"00:37.895 ","End":"00:41.630","Text":"There is some debate on the degree of the 0 polynomial,"},{"Start":"00:41.630 ","End":"00:43.880","Text":"and we take it as minus infinity."},{"Start":"00:43.880 ","End":"00:48.690","Text":"But in any event, no 1 claims that it\u0027s between 4 and 7."},{"Start":"00:48.790 ","End":"00:52.860","Text":"No not a subspace and we\u0027re done."}],"ID":26481},{"Watched":false,"Name":"Exercise 14","Duration":"47s","ChapterTopicVideoID":25657,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.700","Text":"In this exercise, we want to check if W is a subspace."},{"Start":"00:05.700 ","End":"00:10.535","Text":"This is the set of all polynomials with degree less than or equal to n,"},{"Start":"00:10.535 ","End":"00:16.880","Text":"and W, those polynomials which satisfy that p of 0 is 1."},{"Start":"00:16.880 ","End":"00:19.320","Text":"In other words, the value of the polynomial,"},{"Start":"00:19.320 ","End":"00:21.075","Text":"if you substitute 0 is 1."},{"Start":"00:21.075 ","End":"00:23.220","Text":"Is this a subspace?"},{"Start":"00:23.220 ","End":"00:26.080","Text":"I\u0027m claiming no."},{"Start":"00:26.450 ","End":"00:29.190","Text":"We have to satisfy 3 axioms,"},{"Start":"00:29.190 ","End":"00:30.620","Text":"we\u0027ve already failed on the first."},{"Start":"00:30.620 ","End":"00:35.690","Text":"The 0 polynomial does not satisfy p of 0 is 1."},{"Start":"00:35.690 ","End":"00:37.590","Text":"If p is a 0 polynomial,"},{"Start":"00:37.590 ","End":"00:42.194","Text":"then p of everything is 0, so p of 0 would be 0."},{"Start":"00:42.194 ","End":"00:48.120","Text":"No, and W is not a subspace, and we\u0027re done."}],"ID":26482},{"Watched":false,"Name":"Exercise 15","Duration":"3m 20s","ChapterTopicVideoID":25658,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.355","Text":"In this exercise, we have to check if W is a subspace of F of R,"},{"Start":"00:05.355 ","End":"00:08.025","Text":"the functions defined on the real numbers,"},{"Start":"00:08.025 ","End":"00:10.980","Text":"where W consists of all even functions."},{"Start":"00:10.980 ","End":"00:15.870","Text":"In other words, W is the set of all f such that f of"},{"Start":"00:15.870 ","End":"00:22.395","Text":"x equals f of minus x for all x in R,"},{"Start":"00:22.395 ","End":"00:24.660","Text":"that\u0027s what an even function means."},{"Start":"00:24.660 ","End":"00:30.345","Text":"For example, if I took f of x is equal to cosine x,"},{"Start":"00:30.345 ","End":"00:36.390","Text":"that would be an even function because cosine of minus x is the same as cosine of x."},{"Start":"00:36.390 ","End":"00:39.720","Text":"Or another example, g of x is x squared,"},{"Start":"00:39.720 ","End":"00:44.485","Text":"another even function because if I put in x or minus x, it\u0027s the same."},{"Start":"00:44.485 ","End":"00:47.955","Text":"Let\u0027s prove that it is a subspace,"},{"Start":"00:47.955 ","End":"00:51.555","Text":"and then there are 3 axioms to check first all that"},{"Start":"00:51.555 ","End":"00:57.540","Text":"0 function is in W. 0 function is even,"},{"Start":"00:57.540 ","End":"00:59.675","Text":"If I could call it z of x,"},{"Start":"00:59.675 ","End":"01:00.830","Text":"I could call it f of x,"},{"Start":"01:00.830 ","End":"01:03.260","Text":"or call it z for 0, z of x,"},{"Start":"01:03.260 ","End":"01:05.880","Text":"is 0 for all x."},{"Start":"01:07.280 ","End":"01:12.190","Text":"Z of minus x is also 0 for any x,"},{"Start":"01:12.190 ","End":"01:15.290","Text":"so it doesn\u0027t matter what you plug in,"},{"Start":"01:15.290 ","End":"01:19.770","Text":"you always get the same thing, so that\u0027s even."},{"Start":"01:19.810 ","End":"01:26.835","Text":"The second part is to show closure under scalar multiplication,"},{"Start":"01:26.835 ","End":"01:28.485","Text":"so if I have f,"},{"Start":"01:28.485 ","End":"01:30.090","Text":"that\u0027s an even function,"},{"Start":"01:30.090 ","End":"01:31.500","Text":"f is in w,"},{"Start":"01:31.500 ","End":"01:37.845","Text":"I have to show that kf is also in w,"},{"Start":"01:37.845 ","End":"01:40.445","Text":"and I\u0027ll just give the essential step."},{"Start":"01:40.445 ","End":"01:44.315","Text":"The essential step is that kf,"},{"Start":"01:44.315 ","End":"01:47.465","Text":"the function kf of minus x.,"},{"Start":"01:47.465 ","End":"01:50.675","Text":"I\u0027ve show that it\u0027s the same as the function kf at x,"},{"Start":"01:50.675 ","End":"01:55.715","Text":"kf of minus x is like k times f of minus x,"},{"Start":"01:55.715 ","End":"01:58.040","Text":"and because f is even,"},{"Start":"01:58.040 ","End":"02:01.640","Text":"f of minus x is the same as f of x,"},{"Start":"02:01.640 ","End":"02:06.620","Text":"and k times f of x is the same as kf of x."},{"Start":"02:06.620 ","End":"02:08.585","Text":"That\u0027s written out in full,"},{"Start":"02:08.585 ","End":"02:12.170","Text":"but you could just hand wave and say,"},{"Start":"02:12.170 ","End":"02:14.180","Text":"if a function\u0027s the same at x,"},{"Start":"02:14.180 ","End":"02:17.885","Text":"and minus x always then multiplying it by a constant,"},{"Start":"02:17.885 ","End":"02:20.165","Text":"still going to keep the equality."},{"Start":"02:20.165 ","End":"02:23.120","Text":"Now let\u0027s get to the third part,"},{"Start":"02:23.120 ","End":"02:26.400","Text":"that if we have 2 even functions,"},{"Start":"02:26.400 ","End":"02:30.470","Text":"if f and g even functions,"},{"Start":"02:30.470 ","End":"02:34.505","Text":"we have to show that f plus g is also even."},{"Start":"02:34.505 ","End":"02:39.130","Text":"It\u0027s fairly intuitive that if f is the same that x, and minus x,"},{"Start":"02:39.130 ","End":"02:41.360","Text":"and g is the same at x and minus x,"},{"Start":"02:41.360 ","End":"02:43.265","Text":"and then add them, it\u0027s going to be the same."},{"Start":"02:43.265 ","End":"02:45.665","Text":"I\u0027ll just write down the critical step,"},{"Start":"02:45.665 ","End":"02:49.845","Text":"f plus g minus x,"},{"Start":"02:49.845 ","End":"02:55.800","Text":"the sum of 2 functions is defined just by adding them for each x,"},{"Start":"02:55.800 ","End":"03:00.440","Text":"so it\u0027s f of minus x plus g of minus x."},{"Start":"03:00.440 ","End":"03:03.370","Text":"Now f of minus x is the same as f of x,"},{"Start":"03:03.370 ","End":"03:06.450","Text":"and g of minus x is the same as g of x,"},{"Start":"03:06.450 ","End":"03:11.535","Text":"and this sum is the same as the sum of the functions at x."},{"Start":"03:11.535 ","End":"03:12.870","Text":"If you look at the first, and last,"},{"Start":"03:12.870 ","End":"03:15.285","Text":"you see f plus g is also even,"},{"Start":"03:15.285 ","End":"03:18.300","Text":"and so we\u0027re done with this exercise,"},{"Start":"03:18.300 ","End":"03:20.470","Text":"the answer is yes."}],"ID":26483},{"Watched":false,"Name":"Exercise 16","Duration":"2m 46s","ChapterTopicVideoID":25659,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.915","Text":"In this exercise, you want to check if W,"},{"Start":"00:03.915 ","End":"00:05.490","Text":"to be defined in a moment,"},{"Start":"00:05.490 ","End":"00:07.380","Text":"is a subspace of F of R,"},{"Start":"00:07.380 ","End":"00:11.340","Text":"the set of real space of real functions."},{"Start":"00:11.340 ","End":"00:14.865","Text":"Well, W consists of all the bounded functions."},{"Start":"00:14.865 ","End":"00:21.510","Text":"Bounded means that there is some number M"},{"Start":"00:21.510 ","End":"00:29.760","Text":"such that f of x in absolute value is less than or equal to M for all x."},{"Start":"00:29.760 ","End":"00:33.780","Text":"Let\u0027s see. I claim that yes,"},{"Start":"00:33.780 ","End":"00:35.325","Text":"it is a subspace,"},{"Start":"00:35.325 ","End":"00:37.529","Text":"and we have to check the 3 axioms."},{"Start":"00:37.529 ","End":"00:45.840","Text":"The first is that 0 function has to be in W. Is the 0 function bounded?"},{"Start":"00:45.840 ","End":"00:48.545","Text":"It certainly is, it\u0027s always 0."},{"Start":"00:48.545 ","End":"00:53.105","Text":"It\u0027s certainly less than or equal to any M you want to take."},{"Start":"00:53.105 ","End":"00:55.070","Text":"You can even take M equals,"},{"Start":"00:55.070 ","End":"00:58.110","Text":"I don\u0027t know, 1/10."},{"Start":"00:58.120 ","End":"01:00.875","Text":"Even a small positive number,"},{"Start":"01:00.875 ","End":"01:03.110","Text":"absolute value of f of x is always 0."},{"Start":"01:03.110 ","End":"01:04.940","Text":"Yeah, 0 function is bounded,"},{"Start":"01:04.940 ","End":"01:07.445","Text":"any constant function is bounded."},{"Start":"01:07.445 ","End":"01:12.380","Text":"Next 1 is closure under scalar multiplication."},{"Start":"01:12.380 ","End":"01:16.250","Text":"If I have a bounded function and I multiply it by some scalar,"},{"Start":"01:16.250 ","End":"01:20.465","Text":"by some constant, is it still bounded? Sure it is."},{"Start":"01:20.465 ","End":"01:26.870","Text":"If the absolute value of f is less than or equal to M for all x,"},{"Start":"01:26.870 ","End":"01:30.900","Text":"then if I take the function kf, k times f,"},{"Start":"01:30.900 ","End":"01:34.280","Text":"that\u0027s going to be less than or equal to k,"},{"Start":"01:34.280 ","End":"01:38.405","Text":"and absolute value, because of course k could be negative."},{"Start":"01:38.405 ","End":"01:42.500","Text":"It\u0027s the absolute value of k times the old bound."},{"Start":"01:42.500 ","End":"01:44.765","Text":"If I multiply by 10,"},{"Start":"01:44.765 ","End":"01:49.280","Text":"then if I was less than or equal to 50,"},{"Start":"01:49.280 ","End":"01:54.210","Text":"I\u0027d now be less than or equal to 500 if I multiplied by 10."},{"Start":"01:54.210 ","End":"01:55.740","Text":"This is my new M,"},{"Start":"01:55.740 ","End":"01:56.970","Text":"I called it M prime."},{"Start":"01:56.970 ","End":"02:00.110","Text":"This is kf is also bounded,"},{"Start":"02:00.110 ","End":"02:03.150","Text":"but a different bound possibly."},{"Start":"02:03.560 ","End":"02:10.060","Text":"The third axiom is that it has to be closed under addition."},{"Start":"02:10.060 ","End":"02:14.455","Text":"In other words, if I take 2 bounded functions and add them,"},{"Start":"02:14.455 ","End":"02:16.420","Text":"is the result also bounded?"},{"Start":"02:16.420 ","End":"02:18.470","Text":"The answer is yes."},{"Start":"02:18.470 ","End":"02:21.900","Text":"Like if I have 2 functions, f and g,"},{"Start":"02:21.900 ","End":"02:24.810","Text":"each 1 of the different bound,"},{"Start":"02:24.810 ","End":"02:31.230","Text":"f would be less than or equal to M_1 for all x and g will be less than or equal to M_2."},{"Start":"02:31.230 ","End":"02:33.210","Text":"If I add f plus g,"},{"Start":"02:33.210 ","End":"02:38.670","Text":"there\u0027s no way it could be greater than the sum of these 2 M_1 plus M_2,"},{"Start":"02:38.670 ","End":"02:41.850","Text":"and that\u0027s like my new M for this f plus g."},{"Start":"02:41.850 ","End":"02:47.330","Text":"The answer is yes, it is a subspace and we\u0027re done."}],"ID":26484},{"Watched":false,"Name":"Exercise 17","Duration":"1m 17s","ChapterTopicVideoID":25660,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:04.650","Text":"In this exercise, we have the space of real functions"},{"Start":"00:04.650 ","End":"00:10.650","Text":"and we have a subset W consisting of all the continuous functions."},{"Start":"00:10.650 ","End":"00:14.175","Text":"We want to know if W is a subspace."},{"Start":"00:14.175 ","End":"00:21.025","Text":"Well, I hope you remember some of your calculus on continuous functions."},{"Start":"00:21.025 ","End":"00:25.575","Text":"Let me just say that I\u0027m going to show that W is a subspace."},{"Start":"00:25.575 ","End":"00:28.260","Text":"We have to check the 3 axioms."},{"Start":"00:28.260 ","End":"00:30.990","Text":"The 0 function is continuous."},{"Start":"00:30.990 ","End":"00:32.940","Text":"Certainly, it\u0027s an elementary function,"},{"Start":"00:32.940 ","End":"00:34.740","Text":"it\u0027s even a constant function."},{"Start":"00:34.740 ","End":"00:38.095","Text":"There\u0027s no gaps in the graph."},{"Start":"00:38.095 ","End":"00:40.475","Text":"Definitely continuous."},{"Start":"00:40.475 ","End":"00:42.770","Text":"If I take a continuous function,"},{"Start":"00:42.770 ","End":"00:45.950","Text":"multiply it by a scalar or a constant,"},{"Start":"00:45.950 ","End":"00:47.120","Text":"is it still continuous?"},{"Start":"00:47.120 ","End":"00:49.970","Text":"Sure, it is. There\u0027s any number of ways to see this."},{"Start":"00:49.970 ","End":"00:51.590","Text":"I could even just say that the product of"},{"Start":"00:51.590 ","End":"00:57.610","Text":"2 times a constant function will be continuous."},{"Start":"00:57.610 ","End":"01:00.360","Text":"Yes, it\u0027s still continuous."},{"Start":"01:00.360 ","End":"01:04.830","Text":"There\u0027s not going to be any gaps if I scale it with a scalar."},{"Start":"01:05.300 ","End":"01:11.925","Text":"We also learned that the sum of 2 continuous functions is continuous."},{"Start":"01:11.925 ","End":"01:16.149","Text":"3 axioms were satisfied, so it really is a subspace."},{"Start":"01:16.149 ","End":"01:18.070","Text":"We\u0027re done."}],"ID":26485},{"Watched":false,"Name":"Exercise 18","Duration":"1m 19s","ChapterTopicVideoID":25661,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"For this exercise, I suggest you review"},{"Start":"00:03.150 ","End":"00:07.710","Text":"your Calculus 1 and the definition of a differentiable function,"},{"Start":"00:07.710 ","End":"00:09.600","Text":"or rather its properties."},{"Start":"00:09.600 ","End":"00:14.220","Text":"In any event, here we have the space of real functions and"},{"Start":"00:14.220 ","End":"00:18.660","Text":"we have a subset W consisting of all the differentiable functions,"},{"Start":"00:18.660 ","End":"00:22.125","Text":"and we want to know if these form a subspace."},{"Start":"00:22.125 ","End":"00:24.450","Text":"The answer is yes."},{"Start":"00:24.450 ","End":"00:28.750","Text":"We have to show that the 3 axioms are satisfied."},{"Start":"00:28.880 ","End":"00:31.799","Text":"The 0 function, is it differentiable?"},{"Start":"00:31.799 ","End":"00:33.790","Text":"Sure it is. It\u0027s an elementary function."},{"Start":"00:33.790 ","End":"00:34.920","Text":"It\u0027s a constant function."},{"Start":"00:34.920 ","End":"00:36.290","Text":"I can tell you what the derivative is."},{"Start":"00:36.290 ","End":"00:39.420","Text":"It\u0027s 0. Certainly, yes."},{"Start":"00:39.490 ","End":"00:46.130","Text":"As for closure under multiplication by a scalar and closure under addition,"},{"Start":"00:46.130 ","End":"00:51.970","Text":"that was also learnt in the section in calculus on differentiable functions."},{"Start":"00:51.970 ","End":"00:56.690","Text":"There might have been some theorem saying that if f and g are differentiable,"},{"Start":"00:56.690 ","End":"01:01.275","Text":"then so is af plus bg,"},{"Start":"01:01.275 ","End":"01:03.120","Text":"or maybe in a different form maybe it"},{"Start":"01:03.120 ","End":"01:07.430","Text":"was written that multiplying by a constant or in fact,"},{"Start":"01:07.430 ","End":"01:10.670","Text":"multiplying any 2 differentiable functions is differentiable,"},{"Start":"01:10.670 ","End":"01:13.610","Text":"and adding 2 will still keep it differentiable."},{"Start":"01:13.610 ","End":"01:16.610","Text":"For many reasons, the answers to all these is yes,"},{"Start":"01:16.610 ","End":"01:20.070","Text":"and so it is a subspace, and we are done"}],"ID":26486},{"Watched":false,"Name":"Exercise 19","Duration":"1m 1s","ChapterTopicVideoID":25662,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.400","Text":"In this exercise, we want to check if W is a subspace of the space of real functions,"},{"Start":"00:08.400 ","End":"00:12.645","Text":"where W consists of all constant functions."},{"Start":"00:12.645 ","End":"00:15.375","Text":"The answer is surely yes."},{"Start":"00:15.375 ","End":"00:18.825","Text":"It is a subspace let\u0027s look at the 3 axioms."},{"Start":"00:18.825 ","End":"00:21.060","Text":"Is the 0 functioning there?"},{"Start":"00:21.060 ","End":"00:22.920","Text":"Is the 0 function a constant function?"},{"Start":"00:22.920 ","End":"00:25.455","Text":"Yeah, sure, it\u0027s constantly 0."},{"Start":"00:25.455 ","End":"00:28.680","Text":"Closure on the multiplication by a scalar."},{"Start":"00:28.680 ","End":"00:30.130","Text":"Yeah, definitely."},{"Start":"00:30.130 ","End":"00:32.330","Text":"If I have a function that\u0027s constant,"},{"Start":"00:32.330 ","End":"00:35.195","Text":"let\u0027s say it\u0027s equal to constantly 5,"},{"Start":"00:35.195 ","End":"00:37.190","Text":"and I multiply it by say,"},{"Start":"00:37.190 ","End":"00:40.760","Text":"3, then it\u0027s going to be constantly 15."},{"Start":"00:40.760 ","End":"00:43.670","Text":"Yeah, no big deal here."},{"Start":"00:43.670 ","End":"00:45.920","Text":"Similarly closure under addition,"},{"Start":"00:45.920 ","End":"00:51.250","Text":"if 1 function is constantly 3 and the other function is constantly 8,"},{"Start":"00:51.250 ","End":"00:54.050","Text":"then the sum of them is going to be constantly 11."},{"Start":"00:54.050 ","End":"00:57.185","Text":"For example, I think it\u0027s fairly obvious."},{"Start":"00:57.185 ","End":"00:58.790","Text":"The answer is yes,"},{"Start":"00:58.790 ","End":"01:01.860","Text":"it is a subspace and we\u0027re done."}],"ID":26487},{"Watched":false,"Name":"Exercise 20","Duration":"1m 3s","ChapterTopicVideoID":25663,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.500 ","End":"00:04.380","Text":"In this exercise, we have the space of"},{"Start":"00:04.380 ","End":"00:09.810","Text":"real functions and we want to know if W is a subspace and here is"},{"Start":"00:09.810 ","End":"00:14.730","Text":"the definition of W. It\u0027s the set of all functions f such"},{"Start":"00:14.730 ","End":"00:20.700","Text":"that the integral from 0 to 1 of that function is 4."},{"Start":"00:20.700 ","End":"00:23.430","Text":"We\u0027re assuming that f is integrable,"},{"Start":"00:23.430 ","End":"00:24.720","Text":"or I could have just said,"},{"Start":"00:24.720 ","End":"00:30.615","Text":"all f such that f is integrable and the integral from 0 to 1 is 4."},{"Start":"00:30.615 ","End":"00:32.475","Text":"Is this a subspace?"},{"Start":"00:32.475 ","End":"00:34.890","Text":"The answer is no."},{"Start":"00:34.890 ","End":"00:38.865","Text":"In fact, it fails even the first axiom."},{"Start":"00:38.865 ","End":"00:44.750","Text":"The 0 function is not in W because it doesn\u0027t satisfy this."},{"Start":"00:44.750 ","End":"00:46.639","Text":"If f is the 0 function,"},{"Start":"00:46.639 ","End":"00:49.985","Text":"then what we have here is the integral from 0 to 1 of"},{"Start":"00:49.985 ","End":"00:56.835","Text":"0 dx and that integral is 0 which is not 4."},{"Start":"00:56.835 ","End":"00:59.165","Text":"Anyway, we\u0027re not a subspace."},{"Start":"00:59.165 ","End":"01:01.459","Text":"We failed at least 1 of the axioms,"},{"Start":"01:01.459 ","End":"01:03.660","Text":"and then we\u0027re done."}],"ID":26488},{"Watched":false,"Name":"Exercise 21","Duration":"48s","ChapterTopicVideoID":25637,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.175","Text":"In this exercise, we have to check if W is a subspace of this,"},{"Start":"00:05.175 ","End":"00:10.155","Text":"the set of the space of real functions."},{"Start":"00:10.155 ","End":"00:17.880","Text":"W, it\u0027s a set of all functions that have a derivative of 0 everywhere."},{"Start":"00:17.880 ","End":"00:21.090","Text":"Well, assuming that there are differentiable."},{"Start":"00:21.090 ","End":"00:25.620","Text":"It turns out the answer is"},{"Start":"00:25.620 ","End":"00:29.810","Text":"yes because to say that it has"},{"Start":"00:29.810 ","End":"00:34.460","Text":"a derivative of 0 is to say that the function is a constant function."},{"Start":"00:34.460 ","End":"00:37.970","Text":"We already had a previous exercise, I forget what number it"},{"Start":"00:37.970 ","End":"00:42.350","Text":"was that the constant functions are a subspace."},{"Start":"00:42.350 ","End":"00:44.600","Text":"That\u0027s equivalent. Derivative 0,"},{"Start":"00:44.600 ","End":"00:49.320","Text":"constant function, same thing. Okay, that\u0027s it."}],"ID":26462},{"Watched":false,"Name":"Exercise 22","Duration":"33s","ChapterTopicVideoID":25638,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"In this exercise, you want to check"},{"Start":"00:01.740 ","End":"00:06.390","Text":"if W is a subspace of the space of real functions."},{"Start":"00:06.390 ","End":"00:10.125","Text":"Our W is defined as those functions f,"},{"Start":"00:10.125 ","End":"00:14.920","Text":"such that the derivative is constantly 1."},{"Start":"00:15.290 ","End":"00:17.340","Text":"The answer is no,"},{"Start":"00:17.340 ","End":"00:20.940","Text":"it\u0027s not a subspace because it fails the first axiom."},{"Start":"00:20.940 ","End":"00:23.820","Text":"The 0 function doesn\u0027t satisfy this,"},{"Start":"00:23.820 ","End":"00:28.080","Text":"because the derivative of the 0 function is 0 everywhere,"},{"Start":"00:28.080 ","End":"00:30.180","Text":"certainly not 1."},{"Start":"00:30.180 ","End":"00:32.120","Text":"So that\u0027s it."},{"Start":"00:32.120 ","End":"00:34.510","Text":"We\u0027re done"}],"ID":26463},{"Watched":false,"Name":"Exercise 23","Duration":"4m 2s","ChapterTopicVideoID":25639,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.709","Text":"In this exercise, you want to check if W is a subspace of F of R"},{"Start":"00:04.709 ","End":"00:10.260","Text":"the set of real-valued functions on the real numbers."},{"Start":"00:10.260 ","End":"00:12.990","Text":"W consists of those functions"},{"Start":"00:12.990 ","End":"00:20.475","Text":"which have the property that f of x plus 1 equals f of x always."},{"Start":"00:20.475 ","End":"00:22.530","Text":"This would mean, for example,"},{"Start":"00:22.530 ","End":"00:26.020","Text":"that f of 5 equals f of 4,"},{"Start":"00:26.210 ","End":"00:30.540","Text":"f of 9 1/2 is equal to f of 8 1/2, and so on."},{"Start":"00:30.540 ","End":"00:34.320","Text":"Each time if I increase x by 1,"},{"Start":"00:34.320 ","End":"00:38.110","Text":"the value of the function doesn\u0027t change."},{"Start":"00:38.800 ","End":"00:42.500","Text":"I could give you some good examples of functions that"},{"Start":"00:42.500 ","End":"00:46.870","Text":"satisfy this from trigonometry, I won\u0027t."},{"Start":"00:46.870 ","End":"00:50.600","Text":"Just take a simple example of a constant function."},{"Start":"00:50.600 ","End":"00:52.445","Text":"If f is a constant function,"},{"Start":"00:52.445 ","End":"00:56.990","Text":"then certainly will have the property that if I increase x by 1,"},{"Start":"00:56.990 ","End":"01:00.680","Text":"the function will have the same value because it\u0027s the same value everywhere."},{"Start":"01:00.680 ","End":"01:05.809","Text":"Let\u0027s just leave it abstract and let\u0027s check the 3 axioms."},{"Start":"01:05.809 ","End":"01:07.910","Text":"I\u0027m going to show that it is a subspace."},{"Start":"01:07.910 ","End":"01:11.480","Text":"We have to check the 3 axioms."},{"Start":"01:11.480 ","End":"01:15.440","Text":"The first 1, does W contain 0."},{"Start":"01:15.440 ","End":"01:20.840","Text":"In this case, 0 is the 0 function and 0 is the function,"},{"Start":"01:20.840 ","End":"01:22.895","Text":"call it F or Z,"},{"Start":"01:22.895 ","End":"01:26.980","Text":"which is 0 for all x."},{"Start":"01:26.980 ","End":"01:30.110","Text":"Certainly, it has the property we already mentioned."},{"Start":"01:30.110 ","End":"01:34.640","Text":"The constant functions have that property because the 0 function of x plus"},{"Start":"01:34.640 ","End":"01:40.110","Text":"1 is 0 and also at x is 0."},{"Start":"01:40.110 ","End":"01:44.445","Text":"Yes, the 0 function is in our subspace."},{"Start":"01:44.445 ","End":"01:54.590","Text":"The next property axiom is closure under multiplication by a scalar."},{"Start":"01:54.590 ","End":"01:56.660","Text":"If f has the property,"},{"Start":"01:56.660 ","End":"01:59.195","Text":"does kf have the property?"},{"Start":"01:59.195 ","End":"02:02.450","Text":"Let\u0027s translate that into different words to say that f is"},{"Start":"02:02.450 ","End":"02:07.340","Text":"inside means that f of x plus 1 equals f of x, for all x,"},{"Start":"02:07.340 ","End":"02:09.800","Text":"I\u0027m not going to keep writing for all x for all x,"},{"Start":"02:09.800 ","End":"02:15.950","Text":"kf means that kf of x plus 1 and at x is the same value."},{"Start":"02:15.950 ","End":"02:17.990","Text":"Now does this imply this?"},{"Start":"02:17.990 ","End":"02:21.095","Text":"Well, let\u0027s take this as being true and prove this."},{"Start":"02:21.095 ","End":"02:25.915","Text":"We have the property that f of x plus 1 is f of x for all x."},{"Start":"02:25.915 ","End":"02:30.170","Text":"Can certainly multiply the equality by a constant k."},{"Start":"02:30.170 ","End":"02:32.320","Text":"So this would be true."},{"Start":"02:32.320 ","End":"02:37.529","Text":"Then I could say that k times f of x plus 1 is just the function kf"},{"Start":"02:37.529 ","End":"02:41.790","Text":"and x plus 1 and similarly for k, f at x."},{"Start":"02:41.790 ","End":"02:45.574","Text":"Indeed kf has the property and it\u0027s in W,"},{"Start":"02:45.574 ","End":"02:47.275","Text":"That\u0027s 2 out of 3."},{"Start":"02:47.275 ","End":"02:49.620","Text":"Now we have to go for the third 1,"},{"Start":"02:49.620 ","End":"02:52.060","Text":"which as you remember is closure."},{"Start":"02:52.060 ","End":"02:57.505","Text":"Under addition, if f has the property and g has the property,"},{"Start":"02:57.505 ","End":"03:03.070","Text":"does the function f plus g also have that property?"},{"Start":"03:03.070 ","End":"03:08.089","Text":"If I translate all that to say that f and g belong"},{"Start":"03:08.089 ","End":"03:11.390","Text":"means that f of x plus 1 is f of x for all x"},{"Start":"03:11.390 ","End":"03:14.420","Text":"and g of x plus 1 is g of x for all x."},{"Start":"03:14.420 ","End":"03:19.520","Text":"To say that f plus g belongs is to say that f plus g at x plus 1,"},{"Start":"03:19.520 ","End":"03:21.935","Text":"then at x are the same for all x."},{"Start":"03:21.935 ","End":"03:25.805","Text":"We want to know if this implies this."},{"Start":"03:25.805 ","End":"03:30.230","Text":"More convenient is to rewrite this in vertical form."},{"Start":"03:30.230 ","End":"03:32.315","Text":"Now we have these 2."},{"Start":"03:32.315 ","End":"03:36.985","Text":"Certainly, we can add equalities or equations."},{"Start":"03:36.985 ","End":"03:40.725","Text":"This plus this is equal to this plus this."},{"Start":"03:40.725 ","End":"03:42.880","Text":"That\u0027s what I\u0027ve written here."},{"Start":"03:42.880 ","End":"03:45.530","Text":"By definition of f plus g."},{"Start":"03:45.530 ","End":"03:49.860","Text":"This is just f plus g at x plus 1,"},{"Start":"03:49.860 ","End":"03:52.485","Text":"and this is f plus g at x."},{"Start":"03:52.485 ","End":"03:54.059","Text":"These are equal"},{"Start":"03:54.059 ","End":"03:58.520","Text":"and so indeed the function f plus g also belongs to W,"},{"Start":"03:58.520 ","End":"04:02.460","Text":"and that was the third part and so we are done."}],"ID":26464},{"Watched":false,"Name":"Exercise 24","Duration":"7m 20s","ChapterTopicVideoID":25640,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:07.030","Text":"In this exercise, we have to check if W is a subspace of this,"},{"Start":"00:07.030 ","End":"00:09.720","Text":"and I\u0027ll explain in a moment what this is."},{"Start":"00:09.720 ","End":"00:16.230","Text":"Where W is defined as the set of all triplets of complex numbers,"},{"Start":"00:16.230 ","End":"00:18.135","Text":"z_1, z_2, z_3,"},{"Start":"00:18.135 ","End":"00:21.315","Text":"which satisfy 2 equations."},{"Start":"00:21.315 ","End":"00:25.770","Text":"The second one has to be the conjugate of the first one,"},{"Start":"00:25.770 ","End":"00:31.710","Text":"and the third one is the sum of the first one and its conjugate."},{"Start":"00:31.710 ","End":"00:35.550","Text":"Now, what is C^3 of R?"},{"Start":"00:35.550 ","End":"00:39.525","Text":"C^3 over R. Normally,"},{"Start":"00:39.525 ","End":"00:41.415","Text":"if you say C^3,"},{"Start":"00:41.415 ","End":"00:50.715","Text":"it\u0027s the space of 3d complex numbers known as 3d triplets of complex numbers."},{"Start":"00:50.715 ","End":"00:56.810","Text":"The field is automatically assumed to be the complex numbers."},{"Start":"00:56.810 ","End":"00:58.820","Text":"But when we write this,"},{"Start":"00:58.820 ","End":"01:02.340","Text":"then we\u0027re not taking the complex numbers,"},{"Start":"01:02.340 ","End":"01:06.130","Text":"we\u0027re taking it over the field of real numbers."},{"Start":"01:06.130 ","End":"01:08.700","Text":"It turns out that this is a vector space."},{"Start":"01:08.700 ","End":"01:13.620","Text":"Of course, you can multiply a real number by complex numbers."},{"Start":"01:13.630 ","End":"01:17.990","Text":"That\u0027s the difference that this is C^3,"},{"Start":"01:17.990 ","End":"01:22.985","Text":"but the scalars come from real numbers."},{"Start":"01:22.985 ","End":"01:30.080","Text":"I would also like to explain this unusual condition for W. I\u0027ll give an example."},{"Start":"01:30.080 ","End":"01:40.425","Text":"Let\u0027s take as an example the vector 1 plus i,"},{"Start":"01:40.425 ","End":"01:46.270","Text":"1 minus i, 2."},{"Start":"01:46.270 ","End":"01:54.980","Text":"Then this will be a vector in W because the second one is the conjugate of the first one,"},{"Start":"01:54.980 ","End":"01:56.615","Text":"the plus becomes a minus."},{"Start":"01:56.615 ","End":"01:59.269","Text":"If I add this to its conjugate,"},{"Start":"01:59.269 ","End":"02:00.650","Text":"which is like this plus this,"},{"Start":"02:00.650 ","End":"02:04.070","Text":"I really get the third one, which is 2."},{"Start":"02:04.070 ","End":"02:06.140","Text":"That explains these conditions."},{"Start":"02:06.140 ","End":"02:11.075","Text":"Of course, most vectors will not satisfy this."},{"Start":"02:11.075 ","End":"02:14.570","Text":"Let me say right away that we don\u0027t have to check."},{"Start":"02:14.570 ","End":"02:16.430","Text":"I\u0027m going to show that actually,"},{"Start":"02:16.430 ","End":"02:18.889","Text":"it is that the answer is yes,"},{"Start":"02:18.889 ","End":"02:20.840","Text":"it is a subspace."},{"Start":"02:20.840 ","End":"02:23.210","Text":"As usual, we have 3 things to show,"},{"Start":"02:23.210 ","End":"02:25.075","Text":"there are 3 axioms."},{"Start":"02:25.075 ","End":"02:28.790","Text":"The first one is to show that the 0 vector is in W."},{"Start":"02:28.790 ","End":"02:35.010","Text":"The 0 vector is just of course 0,0,0, 3 complex numbers."},{"Start":"02:35.320 ","End":"02:38.869","Text":"It certainly satisfies these conditions."},{"Start":"02:38.869 ","End":"02:43.400","Text":"The second is the conjugate of the first,"},{"Start":"02:43.400 ","End":"02:44.795","Text":"well, they\u0027re all 0."},{"Start":"02:44.795 ","End":"02:47.900","Text":"The third is the sum of the first and"},{"Start":"02:47.900 ","End":"02:50.930","Text":"its conjugate because the conjugate of 0 is also 0,"},{"Start":"02:50.930 ","End":"02:53.195","Text":"so, yeah, the 0 vectors in W,"},{"Start":"02:53.195 ","End":"02:54.880","Text":"that was the first one."},{"Start":"02:54.880 ","End":"03:02.970","Text":"Now let\u0027s move on to the second axiom closure under multiplication by a scalar."},{"Start":"03:03.190 ","End":"03:07.520","Text":"We have to show that if a vector u is in the subspace,"},{"Start":"03:07.520 ","End":"03:11.150","Text":"then so is a scalar times ku,"},{"Start":"03:11.150 ","End":"03:13.505","Text":"but this is important."},{"Start":"03:13.505 ","End":"03:17.855","Text":"You have to take the scalar from the field of real numbers,"},{"Start":"03:17.855 ","End":"03:19.930","Text":"k is a real number."},{"Start":"03:19.930 ","End":"03:24.285","Text":"Let\u0027s give the components of u names a, b, c,"},{"Start":"03:24.285 ","End":"03:29.710","Text":"and then ku as usual we multiply component-wise will be this."},{"Start":"03:29.710 ","End":"03:33.920","Text":"Here I just copied the condition because it\u0027s scrolled off screen."},{"Start":"03:33.920 ","End":"03:39.300","Text":"The condition for being in W. In our case,"},{"Start":"03:39.300 ","End":"03:41.955","Text":"we took a, b, c as the components."},{"Start":"03:41.955 ","End":"03:47.450","Text":"The equivalent of this is that the second is the conjugate of the first,"},{"Start":"03:47.450 ","End":"03:50.940","Text":"b is conjugate of a,"},{"Start":"03:50.940 ","End":"03:53.310","Text":"and c is a plus its conjugate."},{"Start":"03:53.310 ","End":"03:55.555","Text":"Similarly, for ku"},{"Start":"03:55.555 ","End":"04:00.840","Text":"the second component is the conjugate of the first and the third, so on."},{"Start":"04:00.840 ","End":"04:06.390","Text":"How do I show that this implies this?"},{"Start":"04:06.390 ","End":"04:10.305","Text":"I start from here, just copied it."},{"Start":"04:10.305 ","End":"04:13.790","Text":"There are 2 equalities here separated by a comma."},{"Start":"04:13.790 ","End":"04:15.500","Text":"Each of these 2 equalities,"},{"Start":"04:15.500 ","End":"04:18.260","Text":"I\u0027ll multiply by k on both sides."},{"Start":"04:18.260 ","End":"04:24.780","Text":"Certainly, I can do that, so kb is ka conjugate, and so on."},{"Start":"04:26.690 ","End":"04:29.615","Text":"Here\u0027s the delicate bit."},{"Start":"04:29.615 ","End":"04:32.645","Text":"Because k is a real number,"},{"Start":"04:32.645 ","End":"04:35.720","Text":"k is the same as its own conjugate."},{"Start":"04:35.720 ","End":"04:42.155","Text":"Because really k could be written as k plus 0i,"},{"Start":"04:42.155 ","End":"04:46.950","Text":"and the conjugate is k minus 0i."},{"Start":"04:46.950 ","End":"04:49.230","Text":"But these 2 are the same things."},{"Start":"04:49.230 ","End":"04:50.900","Text":"For a real number,"},{"Start":"04:50.900 ","End":"04:52.669","Text":"it\u0027s equal to its own conjugate."},{"Start":"04:52.669 ","End":"04:59.140","Text":"I can put a bar over these k\u0027s here and here."},{"Start":"04:59.140 ","End":"05:02.510","Text":"Now, the property of the conjugate says the conjugate"},{"Start":"05:02.510 ","End":"05:05.315","Text":"of the product is the product of conjugates."},{"Start":"05:05.315 ","End":"05:08.315","Text":"I can extend this bar to a single bar,"},{"Start":"05:08.315 ","End":"05:10.745","Text":"and likewise for this bar."},{"Start":"05:10.745 ","End":"05:14.215","Text":"But this is precisely this condition."},{"Start":"05:14.215 ","End":"05:20.870","Text":"That gives us the ku is also in W. That\u0027s the second of the 3 axioms."},{"Start":"05:20.870 ","End":"05:26.935","Text":"Now let\u0027s go on to the third which is closure under addition."},{"Start":"05:26.935 ","End":"05:31.110","Text":"We have to show that if 2 vectors are in W,"},{"Start":"05:31.110 ","End":"05:33.300","Text":"so is their sum."},{"Start":"05:33.300 ","End":"05:35.370","Text":"Once again, I copied this,"},{"Start":"05:35.370 ","End":"05:37.605","Text":"I know I\u0027m going to need it."},{"Start":"05:37.605 ","End":"05:41.280","Text":"Let\u0027s name the components of the vectors let u,"},{"Start":"05:41.280 ","End":"05:43.080","Text":"be a, b, c,"},{"Start":"05:43.080 ","End":"05:45.900","Text":"and v is going to be A, B, C,"},{"Start":"05:45.900 ","End":"05:50.925","Text":"so u plus v addition is component-wise, will be this."},{"Start":"05:50.925 ","End":"05:54.165","Text":"Now if I interpret these 2 conditions,"},{"Start":"05:54.165 ","End":"05:57.220","Text":"u is in W gives me this."},{"Start":"05:57.220 ","End":"06:00.850","Text":"Because instead of Z_1, Z_2, Z_3, I have a, b, c,"},{"Start":"06:00.850 ","End":"06:05.230","Text":"and that v is in W gives me the same thing with capital letters."},{"Start":"06:05.230 ","End":"06:08.230","Text":"To say that u plus v is in W is"},{"Start":"06:08.230 ","End":"06:11.754","Text":"we take Z_1, Z_2, Z_3 to be these,"},{"Start":"06:11.754 ","End":"06:13.990","Text":"and that gives us this condition."},{"Start":"06:13.990 ","End":"06:16.480","Text":"That\u0027s a reinterpretation of these 2."},{"Start":"06:16.480 ","End":"06:18.955","Text":"Now I have to show that this implies this."},{"Start":"06:18.955 ","End":"06:23.035","Text":"I\u0027m going to show the right-hand side here, imply this."},{"Start":"06:23.035 ","End":"06:26.290","Text":"Here what I did is I just rewrote this long line."},{"Start":"06:26.290 ","End":"06:30.145","Text":"I broke it up into 2 separate lines,"},{"Start":"06:30.145 ","End":"06:33.675","Text":"but each line still has 2 equations in it."},{"Start":"06:33.675 ","End":"06:38.615","Text":"Now if add the first equation on the top to the first equation on the bottom,"},{"Start":"06:38.615 ","End":"06:40.525","Text":"then I\u0027ll get this."},{"Start":"06:40.525 ","End":"06:44.570","Text":"If I add the second equation on the top to the second equation on the bottom,"},{"Start":"06:44.570 ","End":"06:47.405","Text":"then I will get this."},{"Start":"06:47.405 ","End":"06:51.635","Text":"If I look here, not just yet,"},{"Start":"06:51.635 ","End":"06:55.960","Text":"I\u0027m going to use again the property of the conjugate for sums,"},{"Start":"06:55.960 ","End":"07:04.070","Text":"so I can extend the bar here all the way over a plus A and similarly here."},{"Start":"07:04.070 ","End":"07:08.845","Text":"Now, this is equivalent to what\u0027s on the right here."},{"Start":"07:08.845 ","End":"07:13.820","Text":"This gives us that u plus v is also in W. That\u0027s what we wanted."},{"Start":"07:13.820 ","End":"07:16.730","Text":"That\u0027s the third of the 3 axioms,"},{"Start":"07:16.730 ","End":"07:20.250","Text":"and so we are done."}],"ID":26465},{"Watched":false,"Name":"Exercise 25","Duration":"2m 29s","ChapterTopicVideoID":25641,"CourseChapterTopicPlaylistID":246315,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.710","Text":"In this exercise, I have to check if W,"},{"Start":"00:04.710 ","End":"00:09.510","Text":"which is the set of all triples of complex numbers, z_1, z_2, z_3,"},{"Start":"00:09.510 ","End":"00:12.345","Text":"which satisfy these 2 equations,"},{"Start":"00:12.345 ","End":"00:16.650","Text":"is a subset of the vector space C^3."},{"Start":"00:16.650 ","End":"00:20.600","Text":"Now, normally I wouldn\u0027t write over the complex field C,"},{"Start":"00:20.600 ","End":"00:22.050","Text":"because it\u0027s understood,"},{"Start":"00:22.050 ","End":"00:25.845","Text":"but If you look at the previous exercise,"},{"Start":"00:25.845 ","End":"00:30.180","Text":"we had the field, R, real numbers."},{"Start":"00:30.180 ","End":"00:33.845","Text":"This is in contrast to the previous exercise."},{"Start":"00:33.845 ","End":"00:37.505","Text":"There, we showed that it was indeed a subspace."},{"Start":"00:37.505 ","End":"00:41.705","Text":"I can tell you right now that in this case,"},{"Start":"00:41.705 ","End":"00:43.940","Text":"the answer is going to be,"},{"Start":"00:43.940 ","End":"00:48.545","Text":"no, that it is not a subspace."},{"Start":"00:48.545 ","End":"00:52.730","Text":"You might want to go look back at the previous exercise there."},{"Start":"00:52.730 ","End":"00:58.695","Text":"The delicate part was closure under multiplication by scalars."},{"Start":"00:58.695 ","End":"01:01.950","Text":"We use the fact that the scalar was a real number."},{"Start":"01:01.950 ","End":"01:10.880","Text":"In fact, here this violates that closure by scalar multiplication."},{"Start":"01:10.880 ","End":"01:14.020","Text":"The easiest thing and probably the best thing to do"},{"Start":"01:14.020 ","End":"01:15.670","Text":"is to give a counterexample."},{"Start":"01:15.670 ","End":"01:18.270","Text":"All I have to do is give 1 example"},{"Start":"01:18.270 ","End":"01:24.735","Text":"which doesn\u0027t satisfy multiplication by scalar and we\u0027re done."},{"Start":"01:24.735 ","End":"01:26.860","Text":"Here\u0027s the example."},{"Start":"01:26.860 ","End":"01:31.510","Text":"Here\u0027s a vector which is in the subspace W."},{"Start":"01:31.510 ","End":"01:33.340","Text":"Let\u0027s just verify that,"},{"Start":"01:33.340 ","End":"01:34.615","Text":"here are the conditions."},{"Start":"01:34.615 ","End":"01:37.120","Text":"The second is the conjugate of the first."},{"Start":"01:37.120 ","End":"01:40.480","Text":"Yes, minus i is the conjugate of i."},{"Start":"01:40.480 ","End":"01:43.525","Text":"This is 0 plus i, 0 minus i."},{"Start":"01:43.525 ","End":"01:46.295","Text":"The last 1 is the sum of those,"},{"Start":"01:46.295 ","End":"01:49.335","Text":"this plus this is this."},{"Start":"01:49.335 ","End":"01:53.335","Text":"We\u0027re okay, this is in the subspace W,"},{"Start":"01:53.335 ","End":"01:56.945","Text":"but if I choose the scalar k,"},{"Start":"01:56.945 ","End":"01:58.710","Text":"not as a real number,"},{"Start":"01:58.710 ","End":"02:00.465","Text":"let\u0027s take it as i."},{"Start":"02:00.465 ","End":"02:02.990","Text":"That\u0027s our K and multiply it."},{"Start":"02:02.990 ","End":"02:05.370","Text":"I get i times i is i squared,"},{"Start":"02:05.370 ","End":"02:06.720","Text":"which is minus 1."},{"Start":"02:06.720 ","End":"02:09.755","Text":"Minus i squared is 1 item, i times 0 is 0."},{"Start":"02:09.755 ","End":"02:12.850","Text":"This is not in W because,"},{"Start":"02:12.850 ","End":"02:16.505","Text":"for example, z_2 is not z_1 bar."},{"Start":"02:16.505 ","End":"02:19.005","Text":"The conjugate of minus 1 is minus 1."},{"Start":"02:19.005 ","End":"02:21.605","Text":"It\u0027s a real number, it\u0027s not equal to 1."},{"Start":"02:21.605 ","End":"02:23.810","Text":"Here we have a counterexample."},{"Start":"02:23.810 ","End":"02:27.410","Text":"So W is not a subspace in this case."},{"Start":"02:27.410 ","End":"02:29.910","Text":"We\u0027re done."}],"ID":26466}],"Thumbnail":null,"ID":246315},{"Name":"Linear Combination, Dependence and Span","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Linear Combinations","Duration":"20m 16s","ChapterTopicVideoID":25904,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.505","Text":"Continuing with linear combinations, and this is for the more advanced students."},{"Start":"00:05.505 ","End":"00:06.870","Text":"When I say advanced,"},{"Start":"00:06.870 ","End":"00:13.395","Text":"I mean that you\u0027ve studied some of the more famous vector spaces beyond just R^n."},{"Start":"00:13.395 ","End":"00:17.670","Text":"Basic students just covered R^n."},{"Start":"00:17.670 ","End":"00:22.350","Text":"I\u0027m assuming you know about the space of matrices of size m by"},{"Start":"00:22.350 ","End":"00:26.190","Text":"n, and the space of polynomials of degree"},{"Start":"00:26.190 ","End":"00:35.140","Text":"up to n. We\u0027ll start with this 1, and we\u0027ll take 3 matrices."},{"Start":"00:35.140 ","End":"00:39.580","Text":"I guess we\u0027re taking m, and n to be 2."},{"Start":"00:39.580 ","End":"00:42.770","Text":"Yeah, that\u0027s right. 2 by 2 matrices,"},{"Start":"00:42.770 ","End":"00:44.510","Text":"but they\u0027re also considered as vectors."},{"Start":"00:44.510 ","End":"00:46.490","Text":"These are 3 matrices,"},{"Start":"00:46.490 ","End":"00:52.465","Text":"but we\u0027re also viewing them as 3 vectors in this vector space."},{"Start":"00:52.465 ","End":"01:01.765","Text":"I happen to note that the following equality holds that C is twice A plus 4 times B."},{"Start":"01:01.765 ","End":"01:04.310","Text":"When I say I happen to know it all, because I cooked"},{"Start":"01:04.310 ","End":"01:07.460","Text":"the numbers up that way so that it would work out."},{"Start":"01:07.460 ","End":"01:12.050","Text":"But it does bring the question that if I didn\u0027t give you the 2 and the 4,"},{"Start":"01:12.050 ","End":"01:18.080","Text":"how would you find numbers that C is a combination of A and B."},{"Start":"01:18.080 ","End":"01:20.680","Text":"They\u0027re going to show you systematically,"},{"Start":"01:20.680 ","End":"01:24.595","Text":"there\u0027s actually 2 main methods."},{"Start":"01:24.595 ","End":"01:27.565","Text":"First, I\u0027ll phrase it as an exercise."},{"Start":"01:27.565 ","End":"01:33.550","Text":"We\u0027re in the vector space of 2 by 2 matrices over the reals, and we have A, B,"},{"Start":"01:33.550 ","End":"01:35.620","Text":"and C as above,"},{"Start":"01:35.620 ","End":"01:39.775","Text":"and we want to show that C is a linear combination of A and B."},{"Start":"01:39.775 ","End":"01:43.165","Text":"But we pretend we haven\u0027t seen the 2 and the 4."},{"Start":"01:43.165 ","End":"01:44.650","Text":"Now, like I said,"},{"Start":"01:44.650 ","End":"01:47.035","Text":"there\u0027s going to be 2 different solutions."},{"Start":"01:47.035 ","End":"01:51.670","Text":"1 of them is more straight forward, naive saying, okay,"},{"Start":"01:51.670 ","End":"01:53.905","Text":"we don\u0027t know it\u0027s a 2 and a 4 here,"},{"Start":"01:53.905 ","End":"01:55.585","Text":"so let\u0027s just put,"},{"Start":"01:55.585 ","End":"02:00.020","Text":"say an x and a y in place of those numbers."},{"Start":"02:00.020 ","End":"02:02.480","Text":"Here we are. Yeah, first solution,"},{"Start":"02:02.480 ","End":"02:06.215","Text":"we copy this, but we don\u0027t know the 2 and the 4. We have x and y."},{"Start":"02:06.215 ","End":"02:12.035","Text":"Now we get an equation, and we can multiply out."},{"Start":"02:12.035 ","End":"02:17.250","Text":"Remember when we multiply a scalar times, this is a vector,"},{"Start":"02:17.250 ","End":"02:18.465","Text":"but it\u0027s also a matrix,"},{"Start":"02:18.465 ","End":"02:24.620","Text":"we just multiply each entry here, and similarly here."},{"Start":"02:24.620 ","End":"02:26.720","Text":"If we do that,"},{"Start":"02:26.720 ","End":"02:29.760","Text":"then x times minus 1,"},{"Start":"02:29.760 ","End":"02:35.734","Text":"2.5 times x, everything gets an x here, and here we just multiply everything by y."},{"Start":"02:35.734 ","End":"02:38.150","Text":"That\u0027s the scalar multiplication."},{"Start":"02:38.150 ","End":"02:39.605","Text":"Now the addition,"},{"Start":"02:39.605 ","End":"02:41.750","Text":"so minus x plus y,"},{"Start":"02:41.750 ","End":"02:45.350","Text":"we just do each position,"},{"Start":"02:45.350 ","End":"02:47.660","Text":"first row, first column with first row,"},{"Start":"02:47.660 ","End":"02:51.505","Text":"first column, and so on for the whole 4 of them."},{"Start":"02:51.505 ","End":"02:54.800","Text":"Minus x plus y is minus x plus y,"},{"Start":"02:54.800 ","End":"02:59.060","Text":"2.5x minus 4y is here, and similarly for the other 2."},{"Start":"02:59.060 ","End":"03:02.515","Text":"Now we have an equation in"},{"Start":"03:02.515 ","End":"03:07.279","Text":"matrices, and that means that each of the corresponding positions,"},{"Start":"03:07.279 ","End":"03:10.505","Text":"the entries is equal."},{"Start":"03:10.505 ","End":"03:12.050","Text":"We get 4 equations."},{"Start":"03:12.050 ","End":"03:15.125","Text":"This equals this, and this equals this."},{"Start":"03:15.125 ","End":"03:17.545","Text":"A system of linear equations,"},{"Start":"03:17.545 ","End":"03:20.970","Text":"4 equations in 2 unknowns."},{"Start":"03:20.970 ","End":"03:24.615","Text":"Let\u0027s see if it has any solution."},{"Start":"03:24.615 ","End":"03:27.185","Text":"We, as usual,"},{"Start":"03:27.185 ","End":"03:34.774","Text":"we\u0027ll solve it using matrices, and this is the augmented matrix we get from this system."},{"Start":"03:34.774 ","End":"03:37.070","Text":"See here just the coefficients minus 1,"},{"Start":"03:37.070 ","End":"03:40.825","Text":"1, and 2, we just lose the x and the y."},{"Start":"03:40.825 ","End":"03:44.930","Text":"Numbers on the right go on this side of the partition."},{"Start":"03:44.930 ","End":"03:47.075","Text":"Anyway, you know the deal."},{"Start":"03:47.075 ","End":"03:52.590","Text":"We want to bring this 1 now to row echelon form."},{"Start":"03:52.670 ","End":"03:57.440","Text":"This entry, I thought a 1 to minus 1, just as good."},{"Start":"03:57.440 ","End":"04:03.035","Text":"We\u0027re going to add, or subtract multiples of the top row from the other rows."},{"Start":"04:03.035 ","End":"04:07.835","Text":"Or perhaps, maybe we should get rid of fractions."},{"Start":"04:07.835 ","End":"04:12.380","Text":"I\u0027ll go for the getting rid of fractions versus double this, and double this,"},{"Start":"04:12.380 ","End":"04:16.500","Text":"and we won\u0027t have these 1/2s here, and this looks better."},{"Start":"04:16.500 ","End":"04:20.440","Text":"Now we\u0027ll add or subtract multiples of the top row from the other 3."},{"Start":"04:20.440 ","End":"04:24.380","Text":"Specifically, we\u0027ll add 5 times this row to this row,"},{"Start":"04:24.380 ","End":"04:26.165","Text":"and similarly for the others,"},{"Start":"04:26.165 ","End":"04:28.715","Text":"and we get this matrix."},{"Start":"04:28.715 ","End":"04:32.450","Text":"Now look, there\u0027s a lot of stuff that will cancel."},{"Start":"04:32.450 ","End":"04:35.870","Text":"This divides by 3 or minus 3."},{"Start":"04:35.870 ","End":"04:37.940","Text":"This I can divide by minus 7,"},{"Start":"04:37.940 ","End":"04:40.400","Text":"this I can divide by 5."},{"Start":"04:40.400 ","End":"04:46.335","Text":"This is the notation for the row operations and we end up with this."},{"Start":"04:46.335 ","End":"04:50.040","Text":"Now look, the last 3 rows are the same."},{"Start":"04:50.040 ","End":"04:55.010","Text":"As I subtract the second row from the third and the fourth row,"},{"Start":"04:55.010 ","End":"04:59.505","Text":"that will just zero those out and we end up with this."},{"Start":"04:59.505 ","End":"05:02.850","Text":"Of course, we ignore 0 rows as if they\u0027re not there."},{"Start":"05:02.850 ","End":"05:08.734","Text":"Now we have this augmented matrix,"},{"Start":"05:08.734 ","End":"05:15.575","Text":"and now it\u0027s already in echelon form like so,"},{"Start":"05:15.575 ","End":"05:21.205","Text":"it\u0027s time to go back to the system of equations form."},{"Start":"05:21.205 ","End":"05:23.580","Text":"Remember the variables were x and y,"},{"Start":"05:23.580 ","End":"05:25.760","Text":"so we just put the x and y back in."},{"Start":"05:25.760 ","End":"05:30.815","Text":"Minus 1x plus y equals 2 and y equals 4."},{"Start":"05:30.815 ","End":"05:34.130","Text":"Back substitution, but it\u0027s a trivial case."},{"Start":"05:34.130 ","End":"05:35.960","Text":"We already have y here."},{"Start":"05:35.960 ","End":"05:38.810","Text":"If I put 4 in here,"},{"Start":"05:38.810 ","End":"05:41.600","Text":"I\u0027ve got minus x plus 4 is 2."},{"Start":"05:41.600 ","End":"05:44.845","Text":"Minus x is minus 2 and x is 2."},{"Start":"05:44.845 ","End":"05:48.460","Text":"Yeah, this is a check is what we expected to get because remember,"},{"Start":"05:48.460 ","End":"05:50.195","Text":"I had the numbers cooked up at first,"},{"Start":"05:50.195 ","End":"05:52.360","Text":"I started out with 2 and 4."},{"Start":"05:52.360 ","End":"05:53.950","Text":"In this 1,"},{"Start":"05:53.950 ","End":"05:55.760","Text":"we just remember what x and y mean."},{"Start":"05:55.760 ","End":"05:58.460","Text":"There were the numbers that you put here."},{"Start":"05:58.460 ","End":"06:02.090","Text":"If we do that, we get this,"},{"Start":"06:02.090 ","End":"06:07.955","Text":"which shows that indeed C is a linear combination of A and B."},{"Start":"06:07.955 ","End":"06:10.430","Text":"Now, that was just Method 1."},{"Start":"06:10.430 ","End":"06:17.575","Text":"Let\u0027s go to the second solution method and we need a reminder of what the problem was."},{"Start":"06:17.575 ","End":"06:19.380","Text":"We had A,"},{"Start":"06:19.380 ","End":"06:21.720","Text":"B, and C like so,"},{"Start":"06:21.720 ","End":"06:25.430","Text":"2 by 2 matrices also considered as vectors,"},{"Start":"06:25.430 ","End":"06:31.385","Text":"and we wanted to show that C is a linear combination of A and B."},{"Start":"06:31.385 ","End":"06:41.000","Text":"Now, the trick this time is to really show how the matrices can be viewed as vectors."},{"Start":"06:41.000 ","End":"06:47.630","Text":"In general, 2 by 2 matrix is just for numbers and we decide on"},{"Start":"06:47.630 ","End":"06:54.995","Text":"a particular order and I put a little snake or zigzag or whatever to show you,"},{"Start":"06:54.995 ","End":"06:57.890","Text":"we\u0027ll take them in the order a, b, c,"},{"Start":"06:57.890 ","End":"07:01.700","Text":"d. This would be like lexicographic order."},{"Start":"07:01.700 ","End":"07:04.370","Text":"The order you\u0027d be reading in,"},{"Start":"07:04.370 ","End":"07:09.760","Text":"you go from left to right and the rows go from top to bottom."},{"Start":"07:09.760 ","End":"07:13.550","Text":"We\u0027ll associate this with the vector a,"},{"Start":"07:13.550 ","End":"07:16.340","Text":"b, c, d, written out this way."},{"Start":"07:16.340 ","End":"07:19.480","Text":"Now, you don\u0027t have to do it in this order,"},{"Start":"07:19.480 ","End":"07:21.665","Text":"you could choose any other order."},{"Start":"07:21.665 ","End":"07:23.690","Text":"But once you\u0027ve chosen the order,"},{"Start":"07:23.690 ","End":"07:26.780","Text":"you have to be consistent that for all the matrices,"},{"Start":"07:26.780 ","End":"07:29.620","Text":"you\u0027re going to use that same order."},{"Start":"07:29.620 ","End":"07:37.290","Text":"In our case, we\u0027ll stick with this from top-left to bottom-right."},{"Start":"07:38.740 ","End":"07:42.320","Text":"Vector A will become minus 1,"},{"Start":"07:42.320 ","End":"07:45.030","Text":"2.5, minus 4,"},{"Start":"07:45.030 ","End":"07:46.965","Text":"4, which is this."},{"Start":"07:46.965 ","End":"07:49.140","Text":"B is 1 minus 4,"},{"Start":"07:49.140 ","End":"07:51.690","Text":"0.5, 1, which is this."},{"Start":"07:51.690 ","End":"07:53.160","Text":"C is this,"},{"Start":"07:53.160 ","End":"08:00.150","Text":"which is here and we write them in a augmented matrix with the names A,"},{"Start":"08:00.150 ","End":"08:03.340","Text":"B, C and this is what they are."},{"Start":"08:03.340 ","End":"08:09.119","Text":"But being consistent, always starting off the same snake."},{"Start":"08:12.290 ","End":"08:18.210","Text":"I forgot to mention the 1 that we want as a linear combination of the other 2,"},{"Start":"08:18.210 ","End":"08:20.750","Text":"that\u0027s the 1 we put on the last row."},{"Start":"08:20.750 ","End":"08:22.930","Text":"Other than that, not important."},{"Start":"08:22.930 ","End":"08:24.930","Text":"Yes, C is at the bottom."},{"Start":"08:24.930 ","End":"08:31.865","Text":"Yeah. Now we start trying to do row operations on this with the idea of getting"},{"Start":"08:31.865 ","End":"08:39.170","Text":"all 0s on the bottom here to the left of the C. May not be possible,"},{"Start":"08:39.170 ","End":"08:43.040","Text":"in which case, C won\u0027t be a linear combination of A and B,"},{"Start":"08:43.040 ","End":"08:46.250","Text":"but we happen to know that it will be possible."},{"Start":"08:46.250 ","End":"08:49.505","Text":"Let\u0027s start with the row operations."},{"Start":"08:49.505 ","End":"08:55.975","Text":"We\u0027re going to add the top row to the second row that will zero this 1 out."},{"Start":"08:55.975 ","End":"09:00.980","Text":"If I add twice this to this, that will make that 0."},{"Start":"09:00.980 ","End":"09:05.670","Text":"After we do that, this is what we get."},{"Start":"09:05.670 ","End":"09:06.690","Text":"You can check the numbers,"},{"Start":"09:06.690 ","End":"09:10.160","Text":"and notice that the letters A,"},{"Start":"09:10.160 ","End":"09:12.845","Text":"B, C, they stay."},{"Start":"09:12.845 ","End":"09:15.815","Text":"Meaning if I add this to this,"},{"Start":"09:15.815 ","End":"09:20.640","Text":"then I get here B plus A and here C plus twice A,"},{"Start":"09:20.640 ","End":"09:24.850","Text":"they stay as the letters."},{"Start":"09:24.850 ","End":"09:28.330","Text":"What should we do next?"},{"Start":"09:29.460 ","End":"09:35.320","Text":"Choice of 2 things. You can either get rid of the fractions here,"},{"Start":"09:35.320 ","End":"09:36.955","Text":"but this time I won\u0027t bother."},{"Start":"09:36.955 ","End":"09:40.585","Text":"I noticed that 6 is 4 times 1.5."},{"Start":"09:40.585 ","End":"09:44.935","Text":"If I subtract 4 times this from this I\u0027ll get a 0 here."},{"Start":"09:44.935 ","End":"09:49.405","Text":"Let\u0027s subtract 4 times the second row from the third row."},{"Start":"09:49.405 ","End":"09:51.925","Text":"This is the notation."},{"Start":"09:51.925 ","End":"09:57.624","Text":"We got lucky because not only did we end up with a 0 here as planned,"},{"Start":"09:57.624 ","End":"09:59.935","Text":"but we also got 0 here and here,"},{"Start":"09:59.935 ","End":"10:04.720","Text":"because you see 14 is 4 times 3.5 and 20 is 4 times 5."},{"Start":"10:04.720 ","End":"10:06.715","Text":"At this point we say,"},{"Start":"10:06.715 ","End":"10:10.195","Text":"good and we look at what\u0027s written here."},{"Start":"10:10.195 ","End":"10:18.130","Text":"Now we write that this is equal to 0 so this is what we get."},{"Start":"10:18.130 ","End":"10:23.320","Text":"Now it\u0027s easy to extract C that\u0027s why we put the C on the last row,"},{"Start":"10:23.320 ","End":"10:24.610","Text":"it works out nicely."},{"Start":"10:24.610 ","End":"10:28.375","Text":"Just bring everything else to the other side and simplify."},{"Start":"10:28.375 ","End":"10:32.470","Text":"We get C equals 2A plus 4B,"},{"Start":"10:32.470 ","End":"10:39.170","Text":"the 2 and the 4, just like we got in the first solution so we know we\u0027re all right."},{"Start":"10:39.560 ","End":"10:44.849","Text":"That concludes this particular exercise,"},{"Start":"10:44.849 ","End":"10:46.860","Text":"and let\u0027s do another one now."},{"Start":"10:46.860 ","End":"10:51.280","Text":"We\u0027ll go to another vector space."},{"Start":"10:51.690 ","End":"10:55.585","Text":"This time we go into the world of polynomials,"},{"Start":"10:55.585 ","End":"10:58.570","Text":"we\u0027ll consider 3 polynomials."},{"Start":"10:58.570 ","End":"11:05.935","Text":"Notice that they\u0027re all up to third degree."},{"Start":"11:05.935 ","End":"11:11.035","Text":"These all live in the space, P_3 over R,"},{"Start":"11:11.035 ","End":"11:17.050","Text":"which means polynomials up to and including degree 3 could have degree less."},{"Start":"11:17.050 ","End":"11:22.850","Text":"These happen to all be degree exactly 3. and that\u0027s a vector space."},{"Start":"11:26.250 ","End":"11:28.660","Text":"We don\u0027t really notice I mean,"},{"Start":"11:28.660 ","End":"11:33.115","Text":"I\u0027m just claiming that this polynomial, which I called,"},{"Start":"11:33.115 ","End":"11:38.980","Text":"r, is twice polynomial p plus 4 times polynomial q,"},{"Start":"11:38.980 ","End":"11:40.330","Text":"I cooked it up that way."},{"Start":"11:40.330 ","End":"11:41.740","Text":"I wouldn\u0027t expect you to see it."},{"Start":"11:41.740 ","End":"11:45.040","Text":"In fact, that\u0027s going to be the subject of this exercise,"},{"Start":"11:45.040 ","End":"11:47.830","Text":"is how would we find the 2 and the 4?"},{"Start":"11:47.830 ","End":"11:51.640","Text":"We\u0027ll just phrase it as an exercise."},{"Start":"11:51.640 ","End":"11:57.010","Text":"The exercise says in the vector space P_3 of R let p,"},{"Start":"11:57.010 ","End":"12:01.005","Text":"q and r be polynomials as above."},{"Start":"12:01.005 ","End":"12:06.255","Text":"We have to show that r is a linear combination of p and q,"},{"Start":"12:06.255 ","End":"12:10.535","Text":"we haven\u0027t seen the 2 and the 4 we want to find them."},{"Start":"12:10.535 ","End":"12:13.210","Text":"Another in the previous example,"},{"Start":"12:13.210 ","End":"12:17.800","Text":"we\u0027re going to do it with 2 different ways."},{"Start":"12:17.800 ","End":"12:21.145","Text":"Solution one will be the more naive one."},{"Start":"12:21.145 ","End":"12:26.110","Text":"We just say, we want some numbers here and here,"},{"Start":"12:26.110 ","End":"12:33.070","Text":"but we don\u0027t know the 2 and the 4 so I won\u0027t use x and y because x is taken,"},{"Start":"12:33.070 ","End":"12:35.635","Text":"we\u0027ll call it a and b."},{"Start":"12:35.635 ","End":"12:40.945","Text":"This is just this but with a and b in place of the 2 and the 4."},{"Start":"12:40.945 ","End":"12:42.640","Text":"Hopefully when we solve this,"},{"Start":"12:42.640 ","End":"12:46.160","Text":"we\u0027ll get that a is 2 and b is 4."},{"Start":"12:47.250 ","End":"12:50.890","Text":"How do we go about this?"},{"Start":"12:50.890 ","End":"12:53.680","Text":"I did 2 operations in one,"},{"Start":"12:53.680 ","End":"12:55.960","Text":"I did the scalar multiplication,"},{"Start":"12:55.960 ","End":"12:59.635","Text":"well 2 of them and the addition all in one."},{"Start":"12:59.635 ","End":"13:02.440","Text":"Scalar multiplication means you put a here,"},{"Start":"13:02.440 ","End":"13:03.895","Text":"here, here, and here,"},{"Start":"13:03.895 ","End":"13:05.320","Text":"and the b we multiply here,"},{"Start":"13:05.320 ","End":"13:14.080","Text":"here and then we just add up the corresponding 1 to the same exponent."},{"Start":"13:14.080 ","End":"13:19.405","Text":"Like here we get a times minus 1 x cubed and here b x cubed,"},{"Start":"13:19.405 ","End":"13:23.350","Text":"so minus a plus bx cubed and then so many x squared,"},{"Start":"13:23.350 ","End":"13:28.180","Text":"so many x and a constant and I\u0027ve used colors to help us."},{"Start":"13:28.180 ","End":"13:32.440","Text":"We just compare the ones with the same colors,"},{"Start":"13:32.440 ","End":"13:36.025","Text":"but the ones with the same degree."},{"Start":"13:36.025 ","End":"13:41.680","Text":"I also switched left and right so we have minus a plus b equals 2,"},{"Start":"13:41.680 ","End":"13:44.215","Text":"and similarly for the other 3."},{"Start":"13:44.215 ","End":"13:50.290","Text":"That\u0027s 4 equations in 2 unknowns, a and b."},{"Start":"13:50.290 ","End":"13:53.530","Text":"As usual, we solved it with a matrix."},{"Start":"13:53.530 ","End":"13:55.975","Text":"This is the matrix,"},{"Start":"13:55.975 ","End":"13:58.240","Text":"notice minus 1, 1,"},{"Start":"13:58.240 ","End":"14:05.540","Text":"2 and similarly for the others and we want to bring this to row echelon form."},{"Start":"14:05.880 ","End":"14:08.410","Text":"I\u0027m going to multiply second,"},{"Start":"14:08.410 ","End":"14:12.805","Text":"and third row by 2 to get rid of these fractions, these 1/2."},{"Start":"14:12.805 ","End":"14:17.100","Text":"This is the row notation and this is what we get."},{"Start":"14:17.100 ","End":"14:19.980","Text":"Double these 2 rows, we get this."},{"Start":"14:19.980 ","End":"14:24.570","Text":"At this point, you want to add multiples of the first row to"},{"Start":"14:24.570 ","End":"14:32.560","Text":"the other 3 rows as specifically 5 times this to this and as written here."},{"Start":"14:32.560 ","End":"14:37.480","Text":"After we do these operations, we get this."},{"Start":"14:37.480 ","End":"14:46.165","Text":"Now I noticed that this row is divisible by 3 or minus 3 similarly here, 7 and 5."},{"Start":"14:46.165 ","End":"14:51.355","Text":"I\u0027ll do these operations to make the numbers smaller."},{"Start":"14:51.355 ","End":"14:56.235","Text":"After I\u0027ve done that, we get this."},{"Start":"14:56.235 ","End":"14:57.930","Text":"If it looks familiar,"},{"Start":"14:57.930 ","End":"14:59.310","Text":"because it is familiar,"},{"Start":"14:59.310 ","End":"15:04.660","Text":"because I used the same numbers from the previous exercise."},{"Start":"15:04.660 ","End":"15:08.050","Text":"Anyway, the last 3 rows are equal,"},{"Start":"15:08.050 ","End":"15:12.730","Text":"so I subtract the second from the third and the fourth and then"},{"Start":"15:12.730 ","End":"15:17.680","Text":"we end up with 2 rows of 0s which I just eliminate."},{"Start":"15:17.680 ","End":"15:24.805","Text":"What\u0027s left is echelon form on the restricted matrix anyway."},{"Start":"15:24.805 ","End":"15:33.385","Text":"Now we want to go back from matrix to system of linear equation and this is what we get."},{"Start":"15:33.385 ","End":"15:38.110","Text":"I remember we were using a and b not x and y, it doesn\u0027t matter."},{"Start":"15:38.110 ","End":"15:40.480","Text":"Once again, we get b equals 4."},{"Start":"15:40.480 ","End":"15:41.680","Text":"If you plug it in here,"},{"Start":"15:41.680 ","End":"15:43.930","Text":"we see that a is 2,"},{"Start":"15:43.930 ","End":"15:48.490","Text":"which we already knew because I gave you a spoiler and now"},{"Start":"15:48.490 ","End":"15:52.570","Text":"we just put these back in here."},{"Start":"15:52.570 ","End":"15:54.940","Text":"This is what we had before with the a and the b,"},{"Start":"15:54.940 ","End":"15:59.800","Text":"replace a by 2, replace b by 4, we get this."},{"Start":"15:59.800 ","End":"16:05.380","Text":"This demonstrates that r is indeed a linear combination of p and q,"},{"Start":"16:05.380 ","End":"16:07.705","Text":"because there are numbers here and here."},{"Start":"16:07.705 ","End":"16:10.390","Text":"That\u0027s this times p times q is"},{"Start":"16:10.390 ","End":"16:18.190","Text":"r. I said that I\u0027d solve it with 2 different methods."},{"Start":"16:18.190 ","End":"16:20.845","Text":"Let\u0027s jump to the second."},{"Start":"16:20.845 ","End":"16:23.410","Text":"I\u0027ve reminded you of the problem."},{"Start":"16:23.410 ","End":"16:24.985","Text":"We had these polynomials,"},{"Start":"16:24.985 ","End":"16:26.425","Text":"p, q, r,"},{"Start":"16:26.425 ","End":"16:28.435","Text":"and we wanted to show that r,"},{"Start":"16:28.435 ","End":"16:31.600","Text":"is the linear combination of p and q."},{"Start":"16:31.600 ","End":"16:38.575","Text":"In this method, I look upon these not as polynomials but as vectors."},{"Start":"16:38.575 ","End":"16:48.070","Text":"There are 4 numbers here so we just can look at them as like elements in R^4,"},{"Start":"16:48.070 ","End":"16:52.959","Text":"I mean polynomials of degree up to 3 have 4 constants"},{"Start":"16:52.959 ","End":"16:57.550","Text":"in them like the general polynomial would be ax cubed,"},{"Start":"16:57.550 ","End":"17:01.270","Text":"bx squared, cx plus d and I just look at the coefficients a,"},{"Start":"17:01.270 ","End":"17:05.080","Text":"b, c, d and work with those."},{"Start":"17:05.080 ","End":"17:08.635","Text":"They really do look like vectors now."},{"Start":"17:08.635 ","End":"17:11.140","Text":"Let\u0027s take all of these."},{"Start":"17:11.140 ","End":"17:13.825","Text":"I build an augmented matrix."},{"Start":"17:13.825 ","End":"17:19.330","Text":"I put on the right the letters, the names p,"},{"Start":"17:19.330 ","End":"17:24.550","Text":"q and r. The last one is the one that I want as a combination of the other 2,"},{"Start":"17:24.550 ","End":"17:28.210","Text":"since r is the one I\u0027m looking for, it goes last."},{"Start":"17:28.210 ","End":"17:30.625","Text":"The p and q could have been in any order."},{"Start":"17:30.625 ","End":"17:33.970","Text":"Now look the numbers, let\u0027s take r,"},{"Start":"17:33.970 ","End":"17:38.170","Text":"for example, I see that the a, b, c,"},{"Start":"17:38.170 ","End":"17:42.070","Text":"d are, 2 minus 11 minus 6,"},{"Start":"17:42.070 ","End":"17:47.229","Text":"12 so here I put 2 minus 11 minus 6, 12."},{"Start":"17:47.229 ","End":"17:57.025","Text":"It\u0027s like I was looking at r as if it was the vector 2 minus 11 minus 6, 12."},{"Start":"17:57.025 ","End":"17:59.920","Text":"Similarly p minus 1,"},{"Start":"17:59.920 ","End":"18:04.130","Text":"2.5 minus 4, 4 and q."},{"Start":"18:05.070 ","End":"18:12.130","Text":"As before, the idea is to do row operations with the aim of trying to"},{"Start":"18:12.130 ","End":"18:19.970","Text":"get zeros in the bottom row of the restricted matrix."},{"Start":"18:22.970 ","End":"18:28.480","Text":"Let\u0027s see, could have gotten rid of fractions."},{"Start":"18:28.480 ","End":"18:29.530","Text":"I don\u0027t think I\u0027ll bother."},{"Start":"18:29.530 ","End":"18:36.695","Text":"We\u0027ll just add the top row to the second row and twice this to this,"},{"Start":"18:36.695 ","End":"18:40.620","Text":"it\u0027s what I just said in row notation."},{"Start":"18:40.620 ","End":"18:45.420","Text":"Check the computations, but this brings us 0s here and here."},{"Start":"18:45.420 ","End":"18:49.755","Text":"Now we\u0027re still not all 0s here."},{"Start":"18:49.755 ","End":"18:53.580","Text":"We could get this one to be 0 because 6 is 4 times"},{"Start":"18:53.580 ","End":"18:58.840","Text":"1.5 so I could subtract 4 times this row from this row."},{"Start":"18:58.840 ","End":"19:01.940","Text":"This is the row operation for that."},{"Start":"19:01.940 ","End":"19:05.360","Text":"If we do that, then we get the 0 here"},{"Start":"19:05.360 ","End":"19:08.600","Text":"but we\u0027re also very lucky that we got 0s here and here,"},{"Start":"19:08.600 ","End":"19:10.860","Text":"or perhaps it wasn\u0027t luck."},{"Start":"19:10.860 ","End":"19:13.120","Text":"It may be I cooked it up that way."},{"Start":"19:13.120 ","End":"19:18.350","Text":"Anyway, notice that we also worked on the right-hand side that when we"},{"Start":"19:18.350 ","End":"19:25.585","Text":"subtracted 4 times this row from this row, for example."},{"Start":"19:25.585 ","End":"19:28.920","Text":"Then we got this,"},{"Start":"19:28.920 ","End":"19:31.620","Text":"what was here, minus 4 times this,"},{"Start":"19:31.620 ","End":"19:35.270","Text":"and of course earlier back when I did this plus this,"},{"Start":"19:35.270 ","End":"19:39.810","Text":"I got p plus q here and when I took twice this plus this,"},{"Start":"19:39.810 ","End":"19:43.114","Text":"we work on all the rows with these row operations."},{"Start":"19:43.114 ","End":"19:45.140","Text":"Anyway, now we have all 0s."},{"Start":"19:45.140 ","End":"19:47.325","Text":"This is the expression here."},{"Start":"19:47.325 ","End":"19:49.310","Text":"Because we put our last r,"},{"Start":"19:49.310 ","End":"19:53.375","Text":"it\u0027s easy to separate from the rest."},{"Start":"19:53.375 ","End":"19:59.590","Text":"We let this equal 0 and then we isolate the r and we get that"},{"Start":"19:59.590 ","End":"20:05.725","Text":"r is 2p plus 4 q and here the 2 and the 4 that we knew we should get."},{"Start":"20:05.725 ","End":"20:11.930","Text":"I think we\u0027re not only done with this exercise,"},{"Start":"20:11.930 ","End":"20:17.220","Text":"but with this whole clip on linear combinations."}],"ID":26714},{"Watched":false,"Name":"Linear Independence - set of vectors","Duration":"11m 25s","ChapterTopicVideoID":25905,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.850","Text":"We\u0027re heading towards a more formal definition of what it means for a set of vectors to"},{"Start":"00:05.850 ","End":"00:12.540","Text":"be linearly independent or linearly dependent but I\u0027m going to start off light,"},{"Start":"00:12.540 ","End":"00:16.410","Text":"so to speak, with a preface like introduction."},{"Start":"00:16.410 ","End":"00:21.780","Text":"Let\u0027s return to our familiar set of vectors called the set A."},{"Start":"00:21.780 ","End":"00:26.250","Text":"The vectors u, v and w are as described here."},{"Start":"00:26.250 ","End":"00:29.940","Text":"Let\u0027s write an equation where a,"},{"Start":"00:29.940 ","End":"00:31.950","Text":"b and c are the unknowns."},{"Start":"00:31.950 ","End":"00:39.375","Text":"We want to write a times u plus b times v plus c times w equals 0."},{"Start":"00:39.375 ","End":"00:41.040","Text":"Look at it as an equation in a,"},{"Start":"00:41.040 ","End":"00:43.320","Text":"b and c. U,"},{"Start":"00:43.320 ","End":"00:44.790","Text":"v and w are not unknown,"},{"Start":"00:44.790 ","End":"00:45.915","Text":"they\u0027re the ones here."},{"Start":"00:45.915 ","End":"00:47.235","Text":"Let\u0027s replace them."},{"Start":"00:47.235 ","End":"00:50.370","Text":"I called it expanding, it\u0027s substituting."},{"Start":"00:50.370 ","End":"00:57.775","Text":"A times this plus b times this plus c times this is vector 0 which is 0,0,0."},{"Start":"00:57.775 ","End":"01:01.630","Text":"Now, there\u0027s 1 solution to this equation which"},{"Start":"01:01.630 ","End":"01:05.200","Text":"is obvious and it\u0027s called the trivial solution."},{"Start":"01:05.200 ","End":"01:09.870","Text":"The file at a equals 0 and b equals 0 and c equals 0,"},{"Start":"01:09.870 ","End":"01:13.840","Text":"certainly this will be true without any computations."},{"Start":"01:13.840 ","End":"01:17.920","Text":"This solution and that\u0027s why we call it the trivial solution."},{"Start":"01:17.920 ","End":"01:20.380","Text":"It\u0027s obvious and it\u0027s always there."},{"Start":"01:20.380 ","End":"01:23.944","Text":"Whenever we have a linear combination of vectors,"},{"Start":"01:23.944 ","End":"01:27.940","Text":"z is 0 then we can always choose coefficients"},{"Start":"01:27.940 ","End":"01:32.135","Text":"0 and be guaranteed that at least it has the trivial solution."},{"Start":"01:32.135 ","End":"01:33.995","Text":"But the interesting question is,"},{"Start":"01:33.995 ","End":"01:38.005","Text":"does it have any other solutions besides the trivial solution?"},{"Start":"01:38.005 ","End":"01:39.890","Text":"The question I asked first is,"},{"Start":"01:39.890 ","End":"01:43.190","Text":"does it have only the trivial solution or maybe some others?"},{"Start":"01:43.190 ","End":"01:49.105","Text":"If it has gotten on non-trivial solution means not all 3 0."},{"Start":"01:49.105 ","End":"01:52.580","Text":"What can we conclude if it has a non-trivial solution?"},{"Start":"01:52.580 ","End":"01:54.350","Text":"Just say for instance,"},{"Start":"01:54.350 ","End":"01:57.159","Text":"that we solve and we get a,"},{"Start":"01:57.159 ","End":"01:58.890","Text":"b and c equal to 1,"},{"Start":"01:58.890 ","End":"02:02.360","Text":"2 and 3 respectively which is actually not a solution."},{"Start":"02:02.360 ","End":"02:04.580","Text":"But I\u0027m just saying, for instance,"},{"Start":"02:04.580 ","End":"02:13.640","Text":"suppose that we had that then what would we conclude about the vectors u, v and w?"},{"Start":"02:13.640 ","End":"02:18.380","Text":"Turns out that the answer is that they are linearly dependent."},{"Start":"02:18.380 ","End":"02:24.700","Text":"Whenever you find a non-trivial solution then the 3 vectors u,"},{"Start":"02:24.700 ","End":"02:28.625","Text":"v, and w are linearly dependent."},{"Start":"02:28.625 ","End":"02:31.790","Text":"I\u0027ll give you a brief explanation of why this is so that"},{"Start":"02:31.790 ","End":"02:35.435","Text":"they are linearly dependent in such a case."},{"Start":"02:35.435 ","End":"02:40.560","Text":"We chose 1, 2, 3 but I mean any nontrivial case."},{"Start":"02:40.560 ","End":"02:42.255","Text":"One of them is not 0."},{"Start":"02:42.255 ","End":"02:44.450","Text":"Let\u0027s say b is not 0."},{"Start":"02:44.450 ","End":"02:48.830","Text":"In this case, all 3 are not 0 but if b was not 0,"},{"Start":"02:48.830 ","End":"02:51.220","Text":"I circled is not a 0."},{"Start":"02:51.220 ","End":"02:53.690","Text":"Then we can get one of the vectors,"},{"Start":"02:53.690 ","End":"02:55.400","Text":"this one in terms of the others."},{"Start":"02:55.400 ","End":"03:00.260","Text":"We just bring this part and this part to the other side of the equation."},{"Start":"03:00.260 ","End":"03:03.040","Text":"Since b is not 0, we divide by it."},{"Start":"03:03.040 ","End":"03:08.855","Text":"We can isolate and get this 1 as a linear combination of the other 2."},{"Start":"03:08.855 ","End":"03:11.525","Text":"This will always be the case with a non-trivial solution."},{"Start":"03:11.525 ","End":"03:14.840","Text":"You pick one of the coefficients that isn\u0027t 0 then you can"},{"Start":"03:14.840 ","End":"03:19.355","Text":"isolate that vector in terms of the others."},{"Start":"03:19.355 ","End":"03:21.925","Text":"That\u0027s the reason."},{"Start":"03:21.925 ","End":"03:25.520","Text":"This next paragraph, if I just summarize it,"},{"Start":"03:25.520 ","End":"03:27.995","Text":"just says that the opposite is also true."},{"Start":"03:27.995 ","End":"03:33.590","Text":"If the only solution to this equation is the trivial solution,"},{"Start":"03:33.590 ","End":"03:35.120","Text":"the trivial solution is always there."},{"Start":"03:35.120 ","End":"03:37.670","Text":"But if it\u0027s the only one the 0,0,0"},{"Start":"03:37.670 ","End":"03:42.580","Text":"solution then they\u0027re going to be independent linearly."},{"Start":"03:42.580 ","End":"03:44.880","Text":"I won\u0027t even read out what this says."},{"Start":"03:44.880 ","End":"03:50.705","Text":"Basically if we find a solution other than the trivial one then they are dependent."},{"Start":"03:50.705 ","End":"03:53.870","Text":"If we can show that that\u0027s the only solution,"},{"Start":"03:53.870 ","End":"03:56.710","Text":"the trivial one then they\u0027re independent."},{"Start":"03:56.710 ","End":"04:04.070","Text":"Now let\u0027s actually check if this equation has a non-trivial solution where not all of a,"},{"Start":"04:04.070 ","End":"04:05.660","Text":"b and c are 0 or not."},{"Start":"04:05.660 ","End":"04:10.845","Text":"I mean, this was just a made-up hypothetical instance."},{"Start":"04:10.845 ","End":"04:14.170","Text":"Let\u0027s solve this equation."},{"Start":"04:14.380 ","End":"04:18.485","Text":"Here it is, we\u0027re going to actually try and solve it."},{"Start":"04:18.485 ","End":"04:24.470","Text":"First we do the scalar multiplications and end up with this."},{"Start":"04:24.470 ","End":"04:28.070","Text":"Now we\u0027re going to add component wise."},{"Start":"04:28.070 ","End":"04:34.484","Text":"Like the first component will be a plus 4b plus 7c and we\u0027ll compare that to 0."},{"Start":"04:34.484 ","End":"04:39.920","Text":"I\u0027ve skip the step but you can see how we end up with these 3 equations."},{"Start":"04:39.920 ","End":"04:42.710","Text":"I showed you how we got the first one."},{"Start":"04:42.970 ","End":"04:48.890","Text":"We want to see if this has non-trivial solutions. You know what?"},{"Start":"04:48.890 ","End":"04:52.385","Text":"I don\u0027t want to break the flow with tedious computations."},{"Start":"04:52.385 ","End":"04:54.785","Text":"Let me just tell you."},{"Start":"04:54.785 ","End":"04:58.970","Text":"I\u0027ll show you that there is a non-trivial solution."},{"Start":"04:58.970 ","End":"05:01.550","Text":"I\u0027ll leave the computation\u0027s to the end."},{"Start":"05:01.550 ","End":"05:06.585","Text":"I owe you to show you how I got this."},{"Start":"05:06.585 ","End":"05:09.260","Text":"In our particular case,"},{"Start":"05:09.260 ","End":"05:13.390","Text":"we do have a non-trivial solution there."},{"Start":"05:13.390 ","End":"05:19.910","Text":"Here. I just copied the equation again and remember that this was u,"},{"Start":"05:19.910 ","End":"05:24.470","Text":"this was v, This was w. We do have actual numbers here."},{"Start":"05:24.470 ","End":"05:26.264","Text":"I put a as 1,"},{"Start":"05:26.264 ","End":"05:30.440","Text":"b is minus 2 and c is 1."},{"Start":"05:30.440 ","End":"05:36.270","Text":"We actually have found a nontrivial solution for"},{"Start":"05:36.270 ","End":"05:40.740","Text":"the equation but we wanted it in the form"},{"Start":"05:40.740 ","End":"05:47.100","Text":"like au plus bv plus cw equals 0."},{"Start":"05:47.100 ","End":"05:54.625","Text":"Here it is. Let\u0027s get back to our formal definition of linear dependence."},{"Start":"05:54.625 ","End":"05:59.390","Text":"I don\u0027t want to start off with a general situation with n vectors."},{"Start":"05:59.390 ","End":"06:06.120","Text":"Let\u0027s start off with 3 vectors and then I\u0027ll generalize to any number of vectors."},{"Start":"06:06.280 ","End":"06:13.340","Text":"For 3 vectors and pretty much restating what we had before that given 3 vectors, u,"},{"Start":"06:13.340 ","End":"06:20.100","Text":"v, w if this equation has only the trivial solution."},{"Start":"06:20.100 ","End":"06:23.865","Text":"Then the vectors are linearly independent."},{"Start":"06:23.865 ","End":"06:27.505","Text":"There we did dependent first and then independent this way."},{"Start":"06:27.505 ","End":"06:30.265","Text":"This is the only solution."},{"Start":"06:30.265 ","End":"06:32.410","Text":"Trivial solution is always a solution."},{"Start":"06:32.410 ","End":"06:36.080","Text":"The only question is whether it\u0027s the only one or whether there were others."},{"Start":"06:36.210 ","End":"06:40.450","Text":"Here\u0027s the other side of the coin if the equation has"},{"Start":"06:40.450 ","End":"06:44.340","Text":"a solution which is not all 0 it could"},{"Start":"06:44.340 ","End":"06:47.460","Text":"be that just one of them is non-zero or 2 of them are"},{"Start":"06:47.460 ","End":"06:52.170","Text":"non-zero or even 3 of them non-zero but not all 3 of them are 0,"},{"Start":"06:52.170 ","End":"06:53.770","Text":"at least one is non-zero."},{"Start":"06:53.770 ","End":"06:59.920","Text":"Then it\u0027s not trivial the solution and then the vectors turn out to be dependent."},{"Start":"06:59.920 ","End":"07:01.600","Text":"That\u0027s for 3 vectors."},{"Start":"07:01.600 ","End":"07:05.420","Text":"Now I want to generalize a bit,"},{"Start":"07:05.420 ","End":"07:09.565","Text":"maybe not a bit is generalized to any number of vectors."},{"Start":"07:09.565 ","End":"07:15.230","Text":"This time we take n vectors and we need to use subscripts,"},{"Start":"07:15.230 ","End":"07:18.200","Text":"say u_1, u_2 up to u_n."},{"Start":"07:18.200 ","End":"07:20.975","Text":"Here\u0027s the equation that we\u0027re interested in."},{"Start":"07:20.975 ","End":"07:23.120","Text":"Put a coefficient in front of each one."},{"Start":"07:23.120 ","End":"07:24.740","Text":"Take a linear combination."},{"Start":"07:24.740 ","End":"07:28.100","Text":"If this equation and the equation is in a_1 through"},{"Start":"07:28.100 ","End":"07:33.595","Text":"a_n has only the trivial solution where all the a\u0027s are 0"},{"Start":"07:33.595 ","End":"07:37.810","Text":"then the vectors u_1 through u_n are"},{"Start":"07:37.810 ","End":"07:43.205","Text":"linearly independent or the set containing these vectors is linearly independent."},{"Start":"07:43.205 ","End":"07:45.710","Text":"On the other side,"},{"Start":"07:45.710 ","End":"07:50.150","Text":"if the equation has a solution where not all of them are"},{"Start":"07:50.150 ","End":"07:53.990","Text":"0 could be some of them 0 as long as not all of"},{"Start":"07:53.990 ","End":"07:58.460","Text":"them are 0 then this is a non-trivial solution and the vectors are linearly"},{"Start":"07:58.460 ","End":"08:04.630","Text":"dependent or the set of vectors is linearly dependent."},{"Start":"08:04.630 ","End":"08:07.695","Text":"That\u0027s the formal definition."},{"Start":"08:07.695 ","End":"08:11.500","Text":"It\u0027s equivalent to the one we had before where 1"},{"Start":"08:11.500 ","End":"08:16.910","Text":"is linear combination of some of the others because like I said before,"},{"Start":"08:16.910 ","End":"08:19.805","Text":"as soon as you have one of these which is non-zero,"},{"Start":"08:19.805 ","End":"08:23.900","Text":"you can extract it as a combination of the others by putting everything to"},{"Start":"08:23.900 ","End":"08:28.750","Text":"the other side and dividing by the non-zero coefficient."},{"Start":"08:28.750 ","End":"08:32.000","Text":"We\u0027re not quite finished because I have this debt."},{"Start":"08:32.000 ","End":"08:35.375","Text":"I didn\u0027t solve this for you."},{"Start":"08:35.375 ","End":"08:40.290","Text":"I wanted to find a non-trivial solution where not all of a,"},{"Start":"08:40.290 ","End":"08:41.580","Text":"b, c are 0."},{"Start":"08:41.580 ","End":"08:43.835","Text":"I remember I produced out of a hat,"},{"Start":"08:43.835 ","End":"08:49.295","Text":"rotated here faintly a solution and non-trivial."},{"Start":"08:49.295 ","End":"08:54.420","Text":"Let\u0027s see how I could go about this."},{"Start":"08:54.420 ","End":"08:58.390","Text":"We convert the equation to matrix form."},{"Start":"08:58.390 ","End":"09:01.090","Text":"Now, I have to tell you that when it\u0027s homogeneous,"},{"Start":"09:01.090 ","End":"09:03.310","Text":"meaning the right-hand sides are all 0,"},{"Start":"09:03.310 ","End":"09:06.970","Text":"it\u0027s not customary to use the augmented matrix,"},{"Start":"09:06.970 ","End":"09:11.920","Text":"the separator and all zeros because these things just stay all 0 and it\u0027s just,"},{"Start":"09:11.920 ","End":"09:15.055","Text":"the waste is dragging these with you."},{"Start":"09:15.055 ","End":"09:17.800","Text":"But in this case we\u0027ll leave them in."},{"Start":"09:17.800 ","End":"09:23.790","Text":"It doesn\u0027t hurt. This is the matrix form, the augmented."},{"Start":"09:23.790 ","End":"09:27.015","Text":"Now we do row operations."},{"Start":"09:27.015 ","End":"09:30.810","Text":"Subtract twice the first from the second and 3 times the"},{"Start":"09:30.810 ","End":"09:34.000","Text":"first from the third. I\u0027m going to leave you to check it."},{"Start":"09:34.000 ","End":"09:37.870","Text":"This is what we get and now we subtract"},{"Start":"09:37.870 ","End":"09:44.820","Text":"twice this row from this row we get this and all these are zeros,"},{"Start":"09:44.820 ","End":"09:46.885","Text":"so we just basically throw that out."},{"Start":"09:46.885 ","End":"09:53.195","Text":"Now we want to go back from the matrix to the equation form."},{"Start":"09:53.195 ","End":"09:55.539","Text":"This is what we get."},{"Start":"09:55.539 ","End":"10:00.970","Text":"I put boxes around the a and the b because the leading entries are"},{"Start":"10:00.970 ","End":"10:06.850","Text":"the ones that become dependent but the other variable c is free,"},{"Start":"10:06.850 ","End":"10:08.110","Text":"can be whatever you like."},{"Start":"10:08.110 ","End":"10:11.200","Text":"You can just let c be anything then a and b are"},{"Start":"10:11.200 ","End":"10:15.445","Text":"determined from c. When I say c equals anything,"},{"Start":"10:15.445 ","End":"10:20.125","Text":"not quite because the contexts equals 0."},{"Start":"10:20.125 ","End":"10:24.030","Text":"I mean, it\u0027s possible to take c as 0,"},{"Start":"10:24.030 ","End":"10:26.300","Text":"will give me a solution but it\u0027s not the one I"},{"Start":"10:26.300 ","End":"10:28.715","Text":"want because that will be the trivial solution."},{"Start":"10:28.715 ","End":"10:33.395","Text":"Because if c is 0 then for me a b is 0 and if c and b is 0 then a is 0,"},{"Start":"10:33.395 ","End":"10:35.960","Text":"so anything except c equals 0."},{"Start":"10:35.960 ","End":"10:42.485","Text":"In fact, I chose c equals 1 in the solution I showed you earlier."},{"Start":"10:42.485 ","End":"10:45.740","Text":"That if you check by back substitution,"},{"Start":"10:45.740 ","End":"10:51.955","Text":"gives us b is minus 2 from this equation then from this one we get a is 1."},{"Start":"10:51.955 ","End":"10:58.365","Text":"Recall that a, b and c are what we wanted to plug in to this equation."},{"Start":"10:58.365 ","End":"11:04.570","Text":"This is what we get which really is a non-trivial solution for this."},{"Start":"11:04.570 ","End":"11:06.940","Text":"Like I said, if you choose any 1,"},{"Start":"11:06.940 ","End":"11:08.060","Text":"well they\u0027re all non-zero."},{"Start":"11:08.060 ","End":"11:10.340","Text":"Say this 1 then I can isolate."},{"Start":"11:10.340 ","End":"11:12.260","Text":"It\u0027s the easiest 1, 2,"},{"Start":"11:12.260 ","End":"11:16.370","Text":"3 as a linear combination of the others which shows that"},{"Start":"11:16.370 ","End":"11:21.035","Text":"even in the previous definition that 1 is a linear combination of the others."},{"Start":"11:21.035 ","End":"11:26.340","Text":"Anyway, I\u0027ve cleared my debt and we are really done."}],"ID":26715},{"Watched":false,"Name":"Solved Example - From Definition","Duration":"5m 17s","ChapterTopicVideoID":25907,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.220","Text":"This clip is a continuation of the previous clip where we"},{"Start":"00:05.220 ","End":"00:10.965","Text":"discussed linear dependence of a set of vectors."},{"Start":"00:10.965 ","End":"00:14.040","Text":"Just like in the example there,"},{"Start":"00:14.040 ","End":"00:15.450","Text":"we did it in 2 different ways."},{"Start":"00:15.450 ","End":"00:20.535","Text":"1 works straight from the definition where we look for"},{"Start":"00:20.535 ","End":"00:24.690","Text":"a solution to the equation where some"},{"Start":"00:24.690 ","End":"00:29.475","Text":"scalar times a plus another times b plus another time c is 0."},{"Start":"00:29.475 ","End":"00:32.189","Text":"We look for non-trivial solutions."},{"Start":"00:32.189 ","End":"00:33.570","Text":"We know that x equals 0,"},{"Start":"00:33.570 ","End":"00:36.750","Text":"y equals 0, z equals 0 will do it."},{"Start":"00:36.750 ","End":"00:39.960","Text":"But if that\u0027s the only solution, then they\u0027re independent."},{"Start":"00:39.960 ","End":"00:43.240","Text":"If we can find a non-trivial xyz,"},{"Start":"00:43.240 ","End":"00:47.580","Text":"then they will be dependent linearly."},{"Start":"00:47.630 ","End":"00:49.830","Text":"Just putting a, b,"},{"Start":"00:49.830 ","End":"00:54.125","Text":"and c replacing them by what they are here."},{"Start":"00:54.125 ","End":"00:59.120","Text":"0, of course, is the 0 matrix, so it\u0027s this."},{"Start":"00:59.120 ","End":"01:02.005","Text":"Now let\u0027s multiply the scalars."},{"Start":"01:02.005 ","End":"01:04.520","Text":"Everything here is multiplied by x."},{"Start":"01:04.520 ","End":"01:06.290","Text":"All these are multiplied by y."},{"Start":"01:06.290 ","End":"01:09.200","Text":"The next 1, they\u0027re multiplied by z."},{"Start":"01:09.200 ","End":"01:13.130","Text":"Now we add matrices."},{"Start":"01:13.130 ","End":"01:16.775","Text":"We add them, each particular position,"},{"Start":"01:16.775 ","End":"01:18.980","Text":"like the first row,"},{"Start":"01:18.980 ","End":"01:21.260","Text":"the first column, we add all those together."},{"Start":"01:21.260 ","End":"01:24.415","Text":"So this plus this plus this will be 0."},{"Start":"01:24.415 ","End":"01:27.260","Text":"First of all, I just add them and now we get a set of"},{"Start":"01:27.260 ","End":"01:31.580","Text":"4 equations so that this is equal to 0 and so on for the other 3."},{"Start":"01:31.580 ","End":"01:34.430","Text":"Here are the equations."},{"Start":"01:34.430 ","End":"01:38.855","Text":"We have 4 equations in 3 unknowns."},{"Start":"01:38.855 ","End":"01:44.210","Text":"Let\u0027s see if we get any solutions, particularly non-trivial."},{"Start":"01:44.210 ","End":"01:46.235","Text":"I know right away that x,"},{"Start":"01:46.235 ","End":"01:48.410","Text":"y, and z all 0 will solve it."},{"Start":"01:48.410 ","End":"01:51.970","Text":"You can see that, but we\u0027re looking for non-trivial solutions."},{"Start":"01:51.970 ","End":"01:56.075","Text":"We\u0027ll solve it using matrix form."},{"Start":"01:56.075 ","End":"02:01.235","Text":"Remember, if we have a homogeneous set of equations that are all 0\u0027s here,"},{"Start":"02:01.235 ","End":"02:03.475","Text":"we don\u0027t need the augmented matrix."},{"Start":"02:03.475 ","End":"02:04.850","Text":"Normally, if they weren\u0027t 0\u0027s,"},{"Start":"02:04.850 ","End":"02:06.020","Text":"we put a vertical bar,"},{"Start":"02:06.020 ","End":"02:07.910","Text":"and then we put the constants on the right."},{"Start":"02:07.910 ","End":"02:09.470","Text":"But in the homogeneous case,"},{"Start":"02:09.470 ","End":"02:11.735","Text":"we can dispense with that."},{"Start":"02:11.735 ","End":"02:14.330","Text":"I\u0027d like to get rid of the fractions,"},{"Start":"02:14.330 ","End":"02:16.640","Text":"the 2 1/2 and the 1/2 here."},{"Start":"02:16.640 ","End":"02:20.570","Text":"So I\u0027ll double the second and third rows."},{"Start":"02:20.570 ","End":"02:24.780","Text":"That gives us this matrix."},{"Start":"02:25.570 ","End":"02:33.905","Text":"Now we can do some row operations here to make all these entries 0 below the minus 1."},{"Start":"02:33.905 ","End":"02:40.320","Text":"We had 5 times this row to this row, et cetera."},{"Start":"02:40.320 ","End":"02:42.965","Text":"This is the matrix we now get."},{"Start":"02:42.965 ","End":"02:44.630","Text":"I\u0027ll do a bit of canceling."},{"Start":"02:44.630 ","End":"02:46.880","Text":"You can see that this row divides by 3,"},{"Start":"02:46.880 ","End":"02:50.074","Text":"this row divides by 7, by 5."},{"Start":"02:50.074 ","End":"02:53.395","Text":"This is how we write the row operations."},{"Start":"02:53.395 ","End":"02:57.030","Text":"This is what we get. Notice that the second,"},{"Start":"02:57.030 ","End":"03:01.455","Text":"third, and the fourth rows are all the same."},{"Start":"03:01.455 ","End":"03:05.990","Text":"What I\u0027m going to do now is subtract the second row from these."},{"Start":"03:05.990 ","End":"03:08.495","Text":"That will 0 those 2 out."},{"Start":"03:08.495 ","End":"03:17.370","Text":"Didn\u0027t bother to formally write R3 minus R2 into R3. You can see."},{"Start":"03:17.370 ","End":"03:22.035","Text":"Now, we can just cross these 2 out because they are just 0\u0027s."},{"Start":"03:22.035 ","End":"03:25.480","Text":"Now, we just have 2 equations in 3 unknowns."},{"Start":"03:25.480 ","End":"03:27.700","Text":"Let\u0027s go back to the equation form."},{"Start":"03:27.700 ","End":"03:30.780","Text":"Here we are, the 2 equations in 3 unknowns."},{"Start":"03:30.780 ","End":"03:37.475","Text":"Now, I like to put these boxes around the leading entries,"},{"Start":"03:37.475 ","End":"03:41.135","Text":"leading terms in each of the equations because these will be dependent."},{"Start":"03:41.135 ","End":"03:42.860","Text":"The other variables,"},{"Start":"03:42.860 ","End":"03:45.410","Text":"in this case just z, will be free."},{"Start":"03:45.410 ","End":"03:47.905","Text":"Z can be whatever we like."},{"Start":"03:47.905 ","End":"03:50.805","Text":"Well, not quite anything."},{"Start":"03:50.805 ","End":"03:53.895","Text":"Because if I let z equal 0,"},{"Start":"03:53.895 ","End":"03:55.960","Text":"then y is going to be 0."},{"Start":"03:55.960 ","End":"03:58.040","Text":"Then from here and here, I\u0027ll get x equals 0."},{"Start":"03:58.040 ","End":"04:00.745","Text":"So I\u0027ll get the trivial solution."},{"Start":"04:00.745 ","End":"04:03.600","Text":"I\u0027m looking for a non-trivial solution."},{"Start":"04:03.600 ","End":"04:05.580","Text":"So I\u0027ll take z not 0."},{"Start":"04:05.580 ","End":"04:07.935","Text":"For example, I\u0027ll take z equals 1."},{"Start":"04:07.935 ","End":"04:09.405","Text":"I\u0027m just saying,"},{"Start":"04:09.405 ","End":"04:13.170","Text":"anything can take z equals 17, or whatever."},{"Start":"04:13.170 ","End":"04:16.185","Text":"I\u0027ll take z equals 1 will be the easiest."},{"Start":"04:16.185 ","End":"04:18.560","Text":"Then using back substitution,"},{"Start":"04:18.560 ","End":"04:19.730","Text":"we plug in z here,"},{"Start":"04:19.730 ","End":"04:24.575","Text":"we\u0027ll get y, then we plug y and z in here, and we\u0027ll get x."},{"Start":"04:24.575 ","End":"04:27.050","Text":"This is what it comes out to be."},{"Start":"04:27.050 ","End":"04:29.470","Text":"It\u0027s easy enough to check."},{"Start":"04:29.470 ","End":"04:36.840","Text":"Of course, it\u0027s non-trivial as at least 1 that\u0027s not 0 vector all non-zero."},{"Start":"04:36.840 ","End":"04:38.990","Text":"I To remind you, the x, y,"},{"Start":"04:38.990 ","End":"04:42.890","Text":"and z are what we wanted to put here and here to see"},{"Start":"04:42.890 ","End":"04:47.410","Text":"if we could get a non-trivial solution to this, and we did."},{"Start":"04:47.410 ","End":"04:49.790","Text":"Explicitly, this is what we got."},{"Start":"04:49.790 ","End":"04:54.350","Text":"We got constants to put in front of each of these 3 vectors,"},{"Start":"04:54.350 ","End":"04:58.170","Text":"which are matrices and we got 0."},{"Start":"04:58.880 ","End":"05:06.230","Text":"That\u0027s that. Let\u0027s go on to an example in a different vector space with polynomials."},{"Start":"05:06.230 ","End":"05:11.725","Text":"I need to do a take 2 and remove that last sentence."},{"Start":"05:11.725 ","End":"05:15.240","Text":"Remember, I said there\u0027s 2 ways I want to solve this,"},{"Start":"05:15.240 ","End":"05:18.249","Text":"so let\u0027s go on to the second method."}],"ID":26717},{"Watched":false,"Name":"Solved Example - Additional Technique","Duration":"5m 3s","ChapterTopicVideoID":25906,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:07.710","Text":"Here we are again with the same 3 matrices considered as vectors in the vector space."},{"Start":"00:07.710 ","End":"00:16.570","Text":"The vector space to remind you was matrices 2 by 2 over the real numbers."},{"Start":"00:16.690 ","End":"00:21.060","Text":"The question is, are they linearly dependent or independent?"},{"Start":"00:21.060 ","End":"00:24.150","Text":"Well, of course, solution 1 showed us that they are"},{"Start":"00:24.150 ","End":"00:28.140","Text":"linearly dependent but we like to start again."},{"Start":"00:28.140 ","End":"00:34.995","Text":"Now, I believe that what I call these snake method we\u0027ve actually seen before."},{"Start":"00:34.995 ","End":"00:40.785","Text":"The 2 by 2 matrices are similar to the vector space,"},{"Start":"00:40.785 ","End":"00:43.485","Text":"4 or at least they can be represented."},{"Start":"00:43.485 ","End":"00:49.730","Text":"What we have to do is decide on a specific order that we\u0027re going to traverse"},{"Start":"00:49.730 ","End":"00:54.470","Text":"these 4 numbers and"},{"Start":"00:54.470 ","End":"00:59.840","Text":"the most customer is to go this way from top-left to bottom-right,"},{"Start":"00:59.840 ","End":"01:02.255","Text":"first, second, third and fourth,"},{"Start":"01:02.255 ","End":"01:07.745","Text":"so we would write this matrix like it was the following vector."},{"Start":"01:07.745 ","End":"01:10.820","Text":"We can always go back from here to here by just filling"},{"Start":"01:10.820 ","End":"01:14.315","Text":"them in from top left to bottom right."},{"Start":"01:14.315 ","End":"01:20.640","Text":"If I do that for each of these 3 matrices,"},{"Start":"01:20.740 ","End":"01:23.810","Text":"this is how they come out,"},{"Start":"01:23.810 ","End":"01:24.980","Text":"I want to show you the first 1,"},{"Start":"01:24.980 ","End":"01:27.800","Text":"you go like this means minus 1,"},{"Start":"01:27.800 ","End":"01:34.080","Text":"2.5 minus 4,4 and that\u0027s what we have here, flattening it out."},{"Start":"01:34.280 ","End":"01:39.335","Text":"Now we use a technique that we actually used for R^4"},{"Start":"01:39.335 ","End":"01:44.825","Text":"where we put these vectors 1on top of each other."},{"Start":"01:44.825 ","End":"01:50.030","Text":"I mean as rows of a matrix but we make it an"},{"Start":"01:50.030 ","End":"01:55.370","Text":"augmented matrix and that the side of each set of 4 numbers,"},{"Start":"01:55.370 ","End":"01:59.390","Text":"we put the name of the vector or matrix"},{"Start":"01:59.390 ","End":"02:04.040","Text":"or however you want to look at it and then we take this and"},{"Start":"02:04.040 ","End":"02:07.955","Text":"we\u0027re going to start doing row operations on this"},{"Start":"02:07.955 ","End":"02:13.655","Text":"and see if we get all zeros at some point."},{"Start":"02:13.655 ","End":"02:20.780","Text":"These are the row operations that will get 0s beneath this minus 1."},{"Start":"02:20.780 ","End":"02:24.320","Text":"I\u0027ll leave you to look at it and here we are after"},{"Start":"02:24.320 ","End":"02:28.010","Text":"doing that and remember that the right part,"},{"Start":"02:28.010 ","End":"02:30.095","Text":"the augmented part,"},{"Start":"02:30.095 ","End":"02:34.270","Text":"also you do the same row operations on those,"},{"Start":"02:34.270 ","End":"02:37.890","Text":"I add twice row 1, to row 3."},{"Start":"02:37.890 ","End":"02:42.370","Text":"Here I add twice A to the C and so on."},{"Start":"02:42.380 ","End":"02:45.375","Text":"Then I want to get a 0 here."},{"Start":"02:45.375 ","End":"02:48.630","Text":"I noticed that 4 times this is this,"},{"Start":"02:48.630 ","End":"02:51.680","Text":"if I just subtract 4 times this row from this row,"},{"Start":"02:51.680 ","End":"02:56.780","Text":"like I said here then we get this and note that"},{"Start":"02:56.780 ","End":"03:02.090","Text":"even though we were only really planning to get a 0 here,"},{"Start":"03:02.090 ","End":"03:06.050","Text":"we got lucky and got 0s here and here also"},{"Start":"03:06.050 ","End":"03:10.470","Text":"because 4 times this is this and 4 times this is this."},{"Start":"03:10.470 ","End":"03:13.130","Text":"We\u0027ve got a row of all 0s here,"},{"Start":"03:13.130 ","End":"03:20.790","Text":"which means that what\u0027s written on the other side of the dividing line is 0."},{"Start":"03:20.790 ","End":"03:25.100","Text":"We get an equation that I just copied this here,"},{"Start":"03:25.100 ","End":"03:29.540","Text":"that that expression is equal to 0 and I\u0027m just going"},{"Start":"03:29.540 ","End":"03:35.675","Text":"to open the brackets and rearrange."},{"Start":"03:35.675 ","End":"03:37.220","Text":"For example, let\u0027s look at the As,"},{"Start":"03:37.220 ","End":"03:43.320","Text":"we have here 2A and we have here minus 4A altogether minus 2A,"},{"Start":"03:43.320 ","End":"03:45.585","Text":"similarly for the Bs and Cs."},{"Start":"03:45.585 ","End":"03:49.365","Text":"We see that we have the equation we want something A,"},{"Start":"03:49.365 ","End":"03:53.459","Text":"something B, something C and not all 3 of them are 0."},{"Start":"03:53.459 ","End":"03:56.090","Text":"In fact, they\u0027re all 3 non-zero but I could have had"},{"Start":"03:56.090 ","End":"03:59.575","Text":"some zeros as long as not all of them as 0."},{"Start":"03:59.575 ","End":"04:07.480","Text":"That means that, these ABC as a set are linearly dependent."},{"Start":"04:07.480 ","End":"04:09.500","Text":"But just as an extra,"},{"Start":"04:09.500 ","End":"04:14.570","Text":"remember we have the other alternative way of looking at linear dependence where we say"},{"Start":"04:14.570 ","End":"04:20.535","Text":"that at least 1 of them is a linear combination of the others."},{"Start":"04:20.535 ","End":"04:25.550","Text":"For example, if I want to isolate A,"},{"Start":"04:25.550 ","End":"04:28.830","Text":"we could take the second,"},{"Start":"04:28.830 ","End":"04:32.860","Text":"and third terms to the other side and divide by minus 2 and that would give"},{"Start":"04:32.860 ","End":"04:37.370","Text":"us A as a combination of C and B and of course,"},{"Start":"04:37.370 ","End":"04:40.160","Text":"any non-zero constant here,"},{"Start":"04:40.160 ","End":"04:43.595","Text":"we could do it if you alternatively want."},{"Start":"04:43.595 ","End":"04:47.890","Text":"We can have B as the combination of A and C or C as"},{"Start":"04:47.890 ","End":"04:52.070","Text":"a combination of A and B. I\u0027m not saying that in every case"},{"Start":"04:52.070 ","End":"04:55.970","Text":"you can do it with every vector but the ones that"},{"Start":"04:55.970 ","End":"05:00.395","Text":"have a non-zero coefficient and there will be some of those,"},{"Start":"05:00.395 ","End":"05:04.230","Text":"those can be written in terms of the others."}],"ID":26716},{"Watched":false,"Name":"Sufficient Condition for Linear Independence","Duration":"4m 40s","ChapterTopicVideoID":25908,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"There is a small proposition,"},{"Start":"00:03.045 ","End":"00:07.050","Text":"which is very useful at times."},{"Start":"00:07.050 ","End":"00:11.820","Text":"It\u0027s sufficient condition for linear dependence that under"},{"Start":"00:11.820 ","End":"00:17.415","Text":"certain conditions we can conclude that set of vectors is dependent."},{"Start":"00:17.415 ","End":"00:22.230","Text":"There is a more general formulation for vector spaces in general,"},{"Start":"00:22.230 ","End":"00:23.760","Text":"but we won\u0027t bring that here."},{"Start":"00:23.760 ","End":"00:27.720","Text":"I\u0027ll just show you how it expresses itself in"},{"Start":"00:27.720 ","End":"00:32.670","Text":"the case of R^n and in 2 other examples that you\u0027ll see."},{"Start":"00:32.670 ","End":"00:39.020","Text":"In R^n, the proposition says that if our set contains more than n vectors,"},{"Start":"00:39.020 ","End":"00:46.970","Text":"where this n is the same as this n then the set is linearly independent."},{"Start":"00:46.970 ","End":"00:50.690","Text":"For example, in R^2,"},{"Start":"00:50.690 ","End":"00:52.160","Text":"which is 2D space,"},{"Start":"00:52.160 ","End":"00:54.140","Text":"so we have vectors with 2 components,"},{"Start":"00:54.140 ","End":"00:56.240","Text":"if we take more than 2, say,"},{"Start":"00:56.240 ","End":"00:58.370","Text":"3 of them as we have here,"},{"Start":"00:58.370 ","End":"01:02.390","Text":"then these I know are going to be linearly dependent."},{"Start":"01:02.390 ","End":"01:07.170","Text":"You don\u0027t even have to do all the computations because 3 is bigger than 2."},{"Start":"01:08.000 ","End":"01:13.680","Text":"In 3D space, if I have something bigger than 3,"},{"Start":"01:13.680 ","End":"01:22.510","Text":"say 4 vectors, then I know that these are going to be linearly dependent."},{"Start":"01:22.510 ","End":"01:24.490","Text":"These 4 are all different,"},{"Start":"01:24.490 ","End":"01:26.990","Text":"even though there\u0027s a parameter here B."},{"Start":"01:26.990 ","End":"01:28.700","Text":"There is another parameter here, C,"},{"Start":"01:28.700 ","End":"01:30.470","Text":"it doesn\u0027t matter what B and C are."},{"Start":"01:30.470 ","End":"01:33.780","Text":"This is still going to be different from all the rest."},{"Start":"01:34.520 ","End":"01:37.695","Text":"This 1 also has a 1."},{"Start":"01:37.695 ","End":"01:39.780","Text":"If I let B equals 1,"},{"Start":"01:39.780 ","End":"01:43.455","Text":"but still they\u0027re going to be different,"},{"Start":"01:43.455 ","End":"01:45.260","Text":"all 4 of them no matter what B and C are,"},{"Start":"01:45.260 ","End":"01:47.605","Text":"so they\u0027re going to be dependent."},{"Start":"01:47.605 ","End":"01:55.605","Text":"A third example of this is in our 4D space."},{"Start":"01:55.605 ","End":"01:58.020","Text":"For that I need at least 5 vectors."},{"Start":"01:58.020 ","End":"02:00.795","Text":"Here\u0027s an example with 5 vectors."},{"Start":"02:00.795 ","End":"02:06.245","Text":"I know without doing any computation that they are linearly dependent."},{"Start":"02:06.245 ","End":"02:12.295","Text":"Now, we can extend this proposition to other spaces."},{"Start":"02:12.295 ","End":"02:15.050","Text":"If a vector space is this,"},{"Start":"02:15.050 ","End":"02:20.930","Text":"the set of all m by n matrices over the reals,"},{"Start":"02:20.930 ","End":"02:28.865","Text":"then the rule is that any set of more than m times n vectors is dependent."},{"Start":"02:28.865 ","End":"02:33.775","Text":"Another specific example, let\u0027s take 2-by-2 matrices."},{"Start":"02:33.775 ","End":"02:38.055","Text":"Here M_n is 2 times 2 is 4."},{"Start":"02:38.055 ","End":"02:43.760","Text":"If I have more than 4 matrices and here I have 1, 2, 3, 4,"},{"Start":"02:43.760 ","End":"02:51.960","Text":"5, then these are going to be linearly dependent because this 5 is bigger than 2 times 2."},{"Start":"02:51.960 ","End":"02:58.160","Text":"It\u0027s similar to the previous 1 with n equals 4 because we already showed how we can"},{"Start":"02:58.160 ","End":"03:05.540","Text":"convert from 2-by-2 matrices to 4D vectors."},{"Start":"03:05.540 ","End":"03:07.715","Text":"Remember we had a snake method."},{"Start":"03:07.715 ","End":"03:11.160","Text":"Anyway, if you don\u0027t remember that it doesn\u0027t matter."},{"Start":"03:11.260 ","End":"03:18.065","Text":"I want to bring 1 more example from another kind of vector space of polynomials."},{"Start":"03:18.065 ","End":"03:21.375","Text":"Remember that when we write P_n of R,"},{"Start":"03:21.375 ","End":"03:27.725","Text":"this n is the degree or maximum degree of the polynomial."},{"Start":"03:27.725 ","End":"03:33.650","Text":"It\u0027s the set of all polynomials up to degree and then could be"},{"Start":"03:33.650 ","End":"03:39.815","Text":"less than n. The rule here is that if we have a set of more than n plus 1,"},{"Start":"03:39.815 ","End":"03:42.320","Text":"not the same as n, here is n plus 1,"},{"Start":"03:42.320 ","End":"03:45.475","Text":"then we\u0027re going to be linearly dependent."},{"Start":"03:45.475 ","End":"03:52.940","Text":"If I write P_2 of ours polynomials with real coefficients of degree at most 2,"},{"Start":"03:52.940 ","End":"03:59.540","Text":"basically it means the ax squared plus bx plus c,"},{"Start":"03:59.540 ","End":"04:01.475","Text":"the set of all those."},{"Start":"04:01.475 ","End":"04:03.620","Text":"A and B could be 0,"},{"Start":"04:03.620 ","End":"04:06.325","Text":"could be less degree less than 2."},{"Start":"04:06.325 ","End":"04:09.510","Text":"You can see where the n plus 1 comes in."},{"Start":"04:09.510 ","End":"04:10.740","Text":"Because of degree 2,"},{"Start":"04:10.740 ","End":"04:14.860","Text":"there are actually 3 real numbers here."},{"Start":"04:15.010 ","End":"04:17.450","Text":"Anyway, to be bigger than 3,"},{"Start":"04:17.450 ","End":"04:19.775","Text":"we have to be at least 4."},{"Start":"04:19.775 ","End":"04:23.225","Text":"Here indeed are 4, 1, 2,"},{"Start":"04:23.225 ","End":"04:27.070","Text":"3, 4 polynomials of degree up to 2."},{"Start":"04:27.070 ","End":"04:30.305","Text":"You know that these are going to be linearly dependent"},{"Start":"04:30.305 ","End":"04:34.215","Text":"without any computations because of this proposition."},{"Start":"04:34.215 ","End":"04:40.520","Text":"That\u0027s the last example and I\u0027ll finish the clip here."}],"ID":26718},{"Watched":false,"Name":"Exercise 1","Duration":"5m 42s","ChapterTopicVideoID":25909,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.720","Text":"In this exercise, we\u0027re given the following matrices in this space."},{"Start":"00:06.720 ","End":"00:14.535","Text":"This is sometimes written as M 2 by 2 over the reals."},{"Start":"00:14.535 ","End":"00:16.469","Text":"The 1 it\u0027s square matrices,"},{"Start":"00:16.469 ","End":"00:19.305","Text":"you just put the single number usually."},{"Start":"00:19.305 ","End":"00:24.135","Text":"Here they are 4 of them and there are 3 questions."},{"Start":"00:24.135 ","End":"00:29.910","Text":"First of all, to find out if they\u0027re linearly dependent or independent, and secondly,"},{"Start":"00:29.910 ","End":"00:36.105","Text":"if they\u0027re dependent, we can try and write each one as a linear combination of the rest."},{"Start":"00:36.105 ","End":"00:40.440","Text":"It won\u0027t be guaranteed because dependent means at least 1 could"},{"Start":"00:40.440 ","End":"00:44.980","Text":"be written as a combination of the others but doesn\u0027t mean that everyone of them."},{"Start":"00:44.980 ","End":"00:47.220","Text":"Anyway, we can try."},{"Start":"00:47.220 ","End":"00:52.740","Text":"Last question, if A belongs to the span of"},{"Start":"00:52.740 ","End":"00:58.130","Text":"B and C. Now these matrices,"},{"Start":"00:58.130 ","End":"01:01.700","Text":"but there are also vectors in the vector space,"},{"Start":"01:01.700 ","End":"01:04.790","Text":"and if you want to use our usual techniques,"},{"Start":"01:04.790 ","End":"01:08.470","Text":"there\u0027s a way of writing these as vectors,"},{"Start":"01:08.470 ","End":"01:10.860","Text":"not in the matrix form."},{"Start":"01:10.860 ","End":"01:15.980","Text":"What we do is we make a snake pattern, so I go 4,"},{"Start":"01:15.980 ","End":"01:17.870","Text":"1, 1, 5 here,"},{"Start":"01:17.870 ","End":"01:23.070","Text":"like so, like so, like so."},{"Start":"01:23.200 ","End":"01:25.805","Text":"Let me give you an example."},{"Start":"01:25.805 ","End":"01:29.720","Text":"If we had a larger matrix that suppose I don\u0027t know,"},{"Start":"01:29.720 ","End":"01:32.120","Text":"just any numbers 1, 2, 3, 4,"},{"Start":"01:32.120 ","End":"01:35.905","Text":"5, 6, 7, 8, 9,"},{"Start":"01:35.905 ","End":"01:38.835","Text":"then the order I would follow would be 1,"},{"Start":"01:38.835 ","End":"01:40.110","Text":"2, 3,"},{"Start":"01:40.110 ","End":"01:41.190","Text":"4, 5,"},{"Start":"01:41.190 ","End":"01:43.035","Text":"6, 7, 8,"},{"Start":"01:43.035 ","End":"01:46.470","Text":"9, left to right,"},{"Start":"01:46.470 ","End":"01:47.880","Text":"and then top to bottom."},{"Start":"01:47.880 ","End":"01:56.450","Text":"Now we can write these in an augmented matrix like this 4,"},{"Start":"01:56.450 ","End":"01:59.120","Text":"1, 1, 5 is this 4, 1, 1,"},{"Start":"01:59.120 ","End":"02:03.900","Text":"5 and then we also write the A, and so on."},{"Start":"02:03.900 ","End":"02:06.150","Text":"D is 1, 3 minus 1,"},{"Start":"02:06.150 ","End":"02:07.590","Text":"2, 1, 3 minus 1,"},{"Start":"02:07.590 ","End":"02:13.099","Text":"2 is D. I want to say that this is just by convention,"},{"Start":"02:13.099 ","End":"02:14.690","Text":"the order we do things in."},{"Start":"02:14.690 ","End":"02:16.760","Text":"You could actually do it in any order,"},{"Start":"02:16.760 ","End":"02:20.420","Text":"as long as you are completely consistent and always use that order,"},{"Start":"02:20.420 ","End":"02:24.140","Text":"you might decide to go. I don\u0027t know."},{"Start":"02:24.140 ","End":"02:28.220","Text":"Anyway, this is the usual pattern."},{"Start":"02:28.670 ","End":"02:34.630","Text":"Our goal now is to bring this matrix into"},{"Start":"02:34.630 ","End":"02:42.670","Text":"row echelon form and see if we get any rows of 0s in the restricted part."},{"Start":"02:42.670 ","End":"02:45.235","Text":"I mean, to the left of this partition."},{"Start":"02:45.235 ","End":"02:48.190","Text":"I did some row operations, I didn\u0027t write down,"},{"Start":"02:48.190 ","End":"02:50.680","Text":"but you can get a hint from here what I did,"},{"Start":"02:50.680 ","End":"02:52.860","Text":"I wanted these to be 0,"},{"Start":"02:52.860 ","End":"02:59.830","Text":"so this gives me a hint that I took twice the third row minus the first row here,"},{"Start":"02:59.830 ","End":"03:07.190","Text":"and 4 times this row minus this row here and we get this."},{"Start":"03:07.190 ","End":"03:10.425","Text":"Then I added the second row,"},{"Start":"03:10.425 ","End":"03:13.450","Text":"to the third row and subtracted it from the fourth row,"},{"Start":"03:13.450 ","End":"03:15.845","Text":"and got 0 here and here."},{"Start":"03:15.845 ","End":"03:22.520","Text":"But that\u0027s not all because I got 0s also here and here."},{"Start":"03:22.520 ","End":"03:26.570","Text":"There\u0027s only 2 non-zero rows."},{"Start":"03:26.570 ","End":"03:32.740","Text":"Anyway, it means that they are dependent and even got some equations."},{"Start":"03:32.740 ","End":"03:36.890","Text":"For example, if I have this written next to the 0s,"},{"Start":"03:36.890 ","End":"03:45.080","Text":"it means that 2 C minus A plus B. I forgot to write this,"},{"Start":"03:45.080 ","End":"03:47.660","Text":"of course, equals 0,"},{"Start":"03:47.660 ","End":"03:49.970","Text":"and similarly, from this,"},{"Start":"03:49.970 ","End":"03:53.350","Text":"I get that this is equal to 0."},{"Start":"03:53.350 ","End":"03:57.860","Text":"Now the fact that I got a row of 0s, in fact,"},{"Start":"03:57.860 ","End":"04:00.830","Text":"I got more than 1 just tells me yes,"},{"Start":"04:00.830 ","End":"04:02.555","Text":"that they\u0027re linearly independent."},{"Start":"04:02.555 ","End":"04:03.935","Text":"If that\u0027s all I cared about,"},{"Start":"04:03.935 ","End":"04:06.680","Text":"I needn\u0027t have bothered with putting in the A, B, C,"},{"Start":"04:06.680 ","End":"04:11.990","Text":"D. But I need to know more things because the questions also,"},{"Start":"04:11.990 ","End":"04:14.780","Text":"well, let\u0027s just continue."},{"Start":"04:14.780 ","End":"04:18.650","Text":"We were asked to see if we can write each 1 in"},{"Start":"04:18.650 ","End":"04:21.830","Text":"terms of the others is the linear combination,"},{"Start":"04:21.830 ","End":"04:25.895","Text":"and these 2 equations will help us to do that."},{"Start":"04:25.895 ","End":"04:29.780","Text":"There may be more than one way of going about it,"},{"Start":"04:29.780 ","End":"04:33.440","Text":"so for example, if I want to isolate A,"},{"Start":"04:33.440 ","End":"04:37.015","Text":"I could do it from here or from here."},{"Start":"04:37.015 ","End":"04:43.255","Text":"If I\u0027d use this, I just bring A to the right and I\u0027ve got that it\u0027s B plus 2C."},{"Start":"04:43.255 ","End":"04:49.800","Text":"But I could have done it from here and set it equal to 4D minus B, you have a choice."},{"Start":"04:49.850 ","End":"04:54.655","Text":"If I want to isolate B, extract B,"},{"Start":"04:54.655 ","End":"04:57.110","Text":"then again from here,"},{"Start":"04:57.110 ","End":"05:01.710","Text":"just have to bring the 2 C and the A to the other side."},{"Start":"05:02.180 ","End":"05:05.430","Text":"I can extract C from here."},{"Start":"05:05.430 ","End":"05:08.970","Text":"For D, I have to go to this 1 because there is no D"},{"Start":"05:08.970 ","End":"05:12.500","Text":"here so I bring A plus B to the other side,"},{"Start":"05:12.500 ","End":"05:13.940","Text":"divide by 4,"},{"Start":"05:13.940 ","End":"05:16.250","Text":"and this is what I have."},{"Start":"05:16.250 ","End":"05:19.400","Text":"If you go back and look what the last part was,"},{"Start":"05:19.400 ","End":"05:24.920","Text":"it asked if A is in the span of B and C,"},{"Start":"05:24.920 ","End":"05:31.460","Text":"and to be in the span of B and C means to be a linear combination of B and C,"},{"Start":"05:31.460 ","End":"05:34.370","Text":"which it is, we take this bit here."},{"Start":"05:34.370 ","End":"05:40.370","Text":"This one, it shows us that A is the linear combination of B and C and so is in the span."},{"Start":"05:40.370 ","End":"05:43.780","Text":"Done."}],"ID":26719},{"Watched":false,"Name":"Exercise 2","Duration":"12m 22s","ChapterTopicVideoID":25910,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.130","Text":"Before I start this exercise,"},{"Start":"00:02.130 ","End":"00:05.430","Text":"I would like to remind you what this means."},{"Start":"00:05.430 ","End":"00:07.275","Text":"What is P_3 of R?"},{"Start":"00:07.275 ","End":"00:13.380","Text":"It\u0027s the polynomials up to degree 3 with real number coefficients."},{"Start":"00:13.380 ","End":"00:18.970","Text":"A typical example would be a plus"},{"Start":"00:18.970 ","End":"00:28.525","Text":"bx plus cx squared plus dx cubed."},{"Start":"00:28.525 ","End":"00:31.380","Text":"Up to and including degree 3,"},{"Start":"00:31.380 ","End":"00:33.435","Text":"some of these of course could be 0,"},{"Start":"00:33.435 ","End":"00:37.005","Text":"that\u0027s this space is,"},{"Start":"00:37.005 ","End":"00:39.750","Text":"a, b, c, and d are real numbers."},{"Start":"00:39.750 ","End":"00:44.790","Text":"We have 4 polynomials from here, p_1, p_2,"},{"Start":"00:44.790 ","End":"00:48.255","Text":"p_3, p_4 as follows,"},{"Start":"00:48.255 ","End":"00:50.150","Text":"then there are some questions."},{"Start":"00:50.150 ","End":"00:57.160","Text":"First question is to see if they are linearly dependent or independent."},{"Start":"00:57.160 ","End":"01:00.175","Text":"If they are linearly dependent,"},{"Start":"01:00.175 ","End":"01:05.345","Text":"we could try and write each one of them as a linear combination of the rest."},{"Start":"01:05.345 ","End":"01:10.355","Text":"The reason I say try is that it\u0027s not guaranteed."},{"Start":"01:10.355 ","End":"01:14.040","Text":"If linearly dependent,"},{"Start":"01:14.040 ","End":"01:18.355","Text":"it just means that at least one can be written as a combination of the others,"},{"Start":"01:18.355 ","End":"01:21.695","Text":"but it doesn\u0027t mean that everyone can, in this case,"},{"Start":"01:21.695 ","End":"01:25.970","Text":"they will be able to because I cooked up this question."},{"Start":"01:25.970 ","End":"01:29.310","Text":"The last part is really just a test,"},{"Start":"01:29.310 ","End":"01:32.015","Text":"if you remember what the span is."},{"Start":"01:32.015 ","End":"01:37.805","Text":"This is the spanning subset of p_1 and p_4."},{"Start":"01:37.805 ","End":"01:46.800","Text":"It\u0027s the space of all linear combinations of p_1 and p_4."},{"Start":"01:46.800 ","End":"01:49.200","Text":"Something p_1 plus something p_4,"},{"Start":"01:49.200 ","End":"01:50.910","Text":"the set of all of those."},{"Start":"01:50.910 ","End":"01:55.040","Text":"For some reason this is a capital P. Forgive that,"},{"Start":"01:55.040 ","End":"02:02.405","Text":"should be little p. Let\u0027s get started with number 1."},{"Start":"02:02.405 ","End":"02:04.930","Text":"How do we proceed?"},{"Start":"02:04.930 ","End":"02:07.330","Text":"Now if it was normal vectors,"},{"Start":"02:07.330 ","End":"02:12.800","Text":"we would do this with matrices,"},{"Start":"02:12.800 ","End":"02:16.030","Text":"we\u0027d write an augmented matrix and then see if we"},{"Start":"02:16.030 ","End":"02:20.200","Text":"can do row operations and get a row of 0s."},{"Start":"02:20.200 ","End":"02:25.960","Text":"But how do I write these in regular vector form? The answer is simple."},{"Start":"02:25.960 ","End":"02:27.790","Text":"If I have something like this,"},{"Start":"02:27.790 ","End":"02:31.495","Text":"I can write it like a vector a, b,"},{"Start":"02:31.495 ","End":"02:38.810","Text":"c, d. Because you have to make sure your polynomial is written in the correct order."},{"Start":"02:38.830 ","End":"02:42.500","Text":"You could do it either from low to high or from high to low,"},{"Start":"02:42.500 ","End":"02:44.360","Text":"as long as you\u0027re consistent."},{"Start":"02:44.360 ","End":"02:47.675","Text":"Here, we\u0027re ordering them from low to high."},{"Start":"02:47.675 ","End":"02:50.255","Text":"I see that all of them are in order."},{"Start":"02:50.255 ","End":"02:56.145","Text":"You also have to watch out if the missing term that would indicate a 0."},{"Start":"02:56.145 ","End":"03:00.320","Text":"For example here the constant term is missing is 0."},{"Start":"03:00.320 ","End":"03:07.070","Text":"In short, after we\u0027ve written all these in the vector form,"},{"Start":"03:07.070 ","End":"03:08.480","Text":"this is what we get."},{"Start":"03:08.480 ","End":"03:10.415","Text":"I just noticed that here we have like 4,"},{"Start":"03:10.415 ","End":"03:17.390","Text":"1, 1, 5 and that is p_1, p_2 is 0."},{"Start":"03:17.390 ","End":"03:23.035","Text":"Note the 0, 11 minus 5, 3, and so on."},{"Start":"03:23.035 ","End":"03:27.100","Text":"That\u0027s just the standard way of doing it."},{"Start":"03:27.100 ","End":"03:29.195","Text":"As I said, watch out."},{"Start":"03:29.195 ","End":"03:30.935","Text":"They have to be in the right order."},{"Start":"03:30.935 ","End":"03:34.820","Text":"If something\u0027s missing, it\u0027s a 0."},{"Start":"03:34.820 ","End":"03:40.945","Text":"Let\u0027s see if we can get a row of 0s using row operations."},{"Start":"03:40.945 ","End":"03:44.355","Text":"First thing I can do,"},{"Start":"03:44.355 ","End":"03:49.095","Text":"and I\u0027m doing this quickly with shortcuts,"},{"Start":"03:49.095 ","End":"03:51.680","Text":"I want to get all of these to be 0."},{"Start":"03:51.680 ","End":"03:53.705","Text":"This is 0 already, that\u0027s nice."},{"Start":"03:53.705 ","End":"03:56.519","Text":"These two have to be 0."},{"Start":"03:57.650 ","End":"04:02.855","Text":"Here, take twice this row and subtract the top row."},{"Start":"04:02.855 ","End":"04:07.430","Text":"Then this is what will get and not to forget to do it to"},{"Start":"04:07.430 ","End":"04:14.855","Text":"the right column in the augmented twice p_3 minus p_1."},{"Start":"04:14.855 ","End":"04:18.665","Text":"Here we\u0027re going to multiply this by 4 and then subtract this."},{"Start":"04:18.665 ","End":"04:21.145","Text":"This is what we get."},{"Start":"04:21.145 ","End":"04:27.825","Text":"Now, what we want to do is get 0s below the 11 here."},{"Start":"04:27.825 ","End":"04:31.280","Text":"Clearly, we\u0027re going to add the second row to"},{"Start":"04:31.280 ","End":"04:35.675","Text":"the third row and subtract it from the fourth row."},{"Start":"04:35.675 ","End":"04:40.805","Text":"After we do that and also on the right-hand side,"},{"Start":"04:40.805 ","End":"04:44.305","Text":"what we get is we have to have 2 equations."},{"Start":"04:44.305 ","End":"04:51.530","Text":"We have that this will equal 0 and this is 0."},{"Start":"04:51.530 ","End":"04:55.670","Text":"But for sure we can say that the vectors"},{"Start":"04:55.670 ","End":"04:59.690","Text":"are linearly dependent because even if you have just one row of 0s,"},{"Start":"04:59.690 ","End":"05:02.050","Text":"without even looking at what\u0027s here."},{"Start":"05:02.050 ","End":"05:05.945","Text":"Let me just write equals 0, equals 0."},{"Start":"05:05.945 ","End":"05:08.675","Text":"Because for the second part,"},{"Start":"05:08.675 ","End":"05:14.135","Text":"and I\u0027ve just gone ahead and exposed the whole answer is to write"},{"Start":"05:14.135 ","End":"05:18.800","Text":"each one in terms of linear combination of the others."},{"Start":"05:18.800 ","End":"05:21.005","Text":"I could use either of these,"},{"Start":"05:21.005 ","End":"05:23.600","Text":"whichever I find easier."},{"Start":"05:23.600 ","End":"05:26.675","Text":"For example, if I want p_1,"},{"Start":"05:26.675 ","End":"05:29.540","Text":"I could have used this equation,"},{"Start":"05:29.540 ","End":"05:31.250","Text":"but I prefer to use this one,"},{"Start":"05:31.250 ","End":"05:32.960","Text":"bring it to the other side,"},{"Start":"05:32.960 ","End":"05:35.900","Text":"and we get p_2 plus 2p_3."},{"Start":"05:35.900 ","End":"05:38.540","Text":"If I want p_2 again,"},{"Start":"05:38.540 ","End":"05:41.630","Text":"I can use this equation and get this."},{"Start":"05:41.630 ","End":"05:46.125","Text":"For p_3, I have to use this one,"},{"Start":"05:46.125 ","End":"05:47.880","Text":"there\u0027s no p_3 here,"},{"Start":"05:47.880 ","End":"05:50.840","Text":"then I have to bring stuff over and divide by 2."},{"Start":"05:50.840 ","End":"05:53.720","Text":"To get p_4, I must use this one."},{"Start":"05:53.720 ","End":"05:57.380","Text":"I get p_1 plus p_2 on the other side then divide by 4."},{"Start":"05:57.380 ","End":"05:59.730","Text":"This is what I get."},{"Start":"06:00.040 ","End":"06:03.440","Text":"I might ask why I put this in a different color."},{"Start":"06:03.440 ","End":"06:12.120","Text":"That\u0027s for the last part which asked if p_2 is in the span of p_1 and p_3."},{"Start":"06:12.120 ","End":"06:16.190","Text":"That\u0027s the same thing as saying that it is a linear combination of them."},{"Start":"06:16.190 ","End":"06:18.890","Text":"Here it is which I got from here."},{"Start":"06:18.890 ","End":"06:22.245","Text":"But someone else might have preferred to use this one,"},{"Start":"06:22.245 ","End":"06:27.259","Text":"and you would have got then that p_2 is this."},{"Start":"06:27.259 ","End":"06:32.195","Text":"Either way, p_2 is in the span of p_1 and p_3."},{"Start":"06:32.195 ","End":"06:38.300","Text":"This exercise involves polynomials in this space."},{"Start":"06:38.300 ","End":"06:42.620","Text":"Let me just quickly tell you what this is in case you\u0027ve forgotten."},{"Start":"06:42.620 ","End":"06:48.350","Text":"It\u0027s the vector space of polynomials up to and including degree 3."},{"Start":"06:48.350 ","End":"06:54.785","Text":"A typical element here would be a constant part"},{"Start":"06:54.785 ","End":"07:03.649","Text":"and then something x and something x squared and something x cubed and no more."},{"Start":"07:03.649 ","End":"07:06.215","Text":"Some of them could be 0 of course,"},{"Start":"07:06.215 ","End":"07:08.895","Text":"but only up to degree 3,"},{"Start":"07:08.895 ","End":"07:12.340","Text":"and the coefficients are real numbers."},{"Start":"07:12.650 ","End":"07:15.390","Text":"We\u0027re given 4 of them, p_1,"},{"Start":"07:15.390 ","End":"07:18.720","Text":"p_2, p_3 and p_4."},{"Start":"07:18.720 ","End":"07:21.395","Text":"Then there are questions,"},{"Start":"07:21.395 ","End":"07:25.940","Text":"are they linearly dependent? We\u0027ll check."},{"Start":"07:25.940 ","End":"07:33.305","Text":"If so then we\u0027re going to try and write each one of the linear combination of the rest."},{"Start":"07:33.305 ","End":"07:35.210","Text":"The reason I say try,"},{"Start":"07:35.210 ","End":"07:38.030","Text":"and it\u0027s not guaranteed is that"},{"Start":"07:38.030 ","End":"07:40.970","Text":"linear dependency just means that at"},{"Start":"07:40.970 ","End":"07:44.780","Text":"least 1 can be written as a combination of the others,"},{"Start":"07:44.780 ","End":"07:47.620","Text":"but doesn\u0027t mean that everyone can."},{"Start":"07:47.620 ","End":"07:49.525","Text":"Well, I can tell you, in this case,"},{"Start":"07:49.525 ","End":"07:54.245","Text":"it will be the case that everyone can be written as a combination of the rest."},{"Start":"07:54.245 ","End":"07:57.395","Text":"Finally, this is really just to see if you remember what"},{"Start":"07:57.395 ","End":"08:00.690","Text":"span means there\u0027s p_2 belong to"},{"Start":"08:00.690 ","End":"08:07.350","Text":"the span of p_1 and p_3 just means is p_2 a linear combination of p_1 and p_3 really,"},{"Start":"08:07.350 ","End":"08:09.669","Text":"that\u0027s what it means."},{"Start":"08:10.250 ","End":"08:12.500","Text":"We want to get started,"},{"Start":"08:12.500 ","End":"08:16.730","Text":"and we want to use the technique with the augmented matrix."},{"Start":"08:16.730 ","End":"08:21.275","Text":"The only thing is that we\u0027re used to doing it with vectors, not with polynomials."},{"Start":"08:21.275 ","End":"08:24.590","Text":"The remedy for that is just like we had"},{"Start":"08:24.590 ","End":"08:27.695","Text":"a trick with matrices where we use the snake method,"},{"Start":"08:27.695 ","End":"08:30.320","Text":"with polynomial it\u0027s even simpler."},{"Start":"08:30.320 ","End":"08:34.730","Text":"We would express this as a vector just by taking"},{"Start":"08:34.730 ","End":"08:40.610","Text":"the coefficients and writing them as a vector."},{"Start":"08:40.610 ","End":"08:47.360","Text":"Now, it\u0027s important that you decide on a definite order."},{"Start":"08:47.360 ","End":"08:54.500","Text":"Let\u0027s say this one that we go from low to high degrees constants x is x squared,"},{"Start":"08:54.500 ","End":"08:57.230","Text":"x cubed, and it doesn\u0027t really matter,"},{"Start":"08:57.230 ","End":"08:58.820","Text":"but you have to be consistent."},{"Start":"08:58.820 ","End":"09:02.285","Text":"All of them have to be written from low to high."},{"Start":"09:02.285 ","End":"09:06.709","Text":"The other thing to notice is that if you have a missing one,"},{"Start":"09:06.709 ","End":"09:08.765","Text":"that will come out as a 0."},{"Start":"09:08.765 ","End":"09:12.170","Text":"I notice that p_2 has a missing constant term,"},{"Start":"09:12.170 ","End":"09:16.500","Text":"and we\u0027ll see that appear as a 0."},{"Start":"09:18.640 ","End":"09:22.070","Text":"Here is a matrix,"},{"Start":"09:22.070 ","End":"09:26.170","Text":"like p_1 is 4, 1, 1, 5."},{"Start":"09:26.170 ","End":"09:28.590","Text":"I guess we don\u0027t see a 1."},{"Start":"09:28.590 ","End":"09:30.125","Text":"I mean well, it\u0027s 1x,"},{"Start":"09:30.125 ","End":"09:32.370","Text":"so we just don\u0027t write the 1."},{"Start":"09:32.370 ","End":"09:34.950","Text":"Also the 0 like I said,"},{"Start":"09:34.950 ","End":"09:38.150","Text":"from p_2 is they\u0027re careful,"},{"Start":"09:38.150 ","End":"09:41.040","Text":"something\u0027s missing, it\u0027s a 0."},{"Start":"09:41.230 ","End":"09:45.275","Text":"On the right in the augmented part,"},{"Start":"09:45.275 ","End":"09:47.060","Text":"we write just p_1,"},{"Start":"09:47.060 ","End":"09:48.170","Text":"p_2, p_3, p_4."},{"Start":"09:48.170 ","End":"09:52.160","Text":"We wouldn\u0027t need to use this."},{"Start":"09:52.160 ","End":"09:56.750","Text":"We could use this restricted one if we just had to answer Question 1."},{"Start":"09:56.750 ","End":"10:02.540","Text":"When we start writing linear combinations then we\u0027ll need these also."},{"Start":"10:02.540 ","End":"10:10.145","Text":"We want to see if we\u0027re going to get any 0s if we bring it to row echelon form."},{"Start":"10:10.145 ","End":"10:12.485","Text":"Let\u0027s start row operations."},{"Start":"10:12.485 ","End":"10:14.465","Text":"Here, we have a 0 already."},{"Start":"10:14.465 ","End":"10:16.145","Text":"I want 0s here and here,"},{"Start":"10:16.145 ","End":"10:23.215","Text":"so twice this minus this will go here and 4 times this minus this here,"},{"Start":"10:23.215 ","End":"10:25.135","Text":"and this is what we get."},{"Start":"10:25.135 ","End":"10:29.305","Text":"Notice that I also do the operations on the right column."},{"Start":"10:29.305 ","End":"10:31.180","Text":"Feel free to pause and check"},{"Start":"10:31.180 ","End":"10:34.900","Text":"the computations because I\u0027m going through this fairly quickly."},{"Start":"10:34.900 ","End":"10:38.790","Text":"Next thing we want to do is continue to get 0s."},{"Start":"10:38.790 ","End":"10:41.475","Text":"We want 0s here and here."},{"Start":"10:41.475 ","End":"10:46.310","Text":"I just add this row to this row and subtract it from this row."},{"Start":"10:46.400 ","End":"10:53.380","Text":"It\u0027s easy to see that we get 0s here and here and this is what we get on the right."},{"Start":"10:53.380 ","End":"10:56.200","Text":"The fact that we have even 1 row of 0s,"},{"Start":"10:56.200 ","End":"11:02.850","Text":"never mind more than 1 tells us that we had linear dependency,"},{"Start":"11:02.850 ","End":"11:05.585","Text":"so that\u0027s the first part of 3."},{"Start":"11:05.585 ","End":"11:07.660","Text":"Parts 2 and 3,"},{"Start":"11:07.660 ","End":"11:14.795","Text":"I\u0027m going to use the fact that these must both be 0."},{"Start":"11:14.795 ","End":"11:21.190","Text":"I can just write that this equals 0 and this equals 0."},{"Start":"11:21.190 ","End":"11:25.340","Text":"Remember Part 2 was to try and write each of the polynomial p_1,"},{"Start":"11:25.340 ","End":"11:29.010","Text":"p_2, p_3, p_4 in terms of the rest of them."},{"Start":"11:29.230 ","End":"11:34.010","Text":"Here\u0027s the result. For p_1, p_2, p_3,"},{"Start":"11:34.010 ","End":"11:38.850","Text":"I used this equation here because it has p_1,"},{"Start":"11:38.850 ","End":"11:40.635","Text":"p_2, p_3 in it."},{"Start":"11:40.635 ","End":"11:43.970","Text":"Like if isolated or extracted p_1,"},{"Start":"11:43.970 ","End":"11:46.925","Text":"it comes out like this and so on, p_2, p_3."},{"Start":"11:46.925 ","End":"11:49.600","Text":"For p_4, I had to use this one,"},{"Start":"11:49.600 ","End":"11:53.000","Text":"I brought these 2 to the other side divided by 4."},{"Start":"11:53.000 ","End":"11:56.690","Text":"This is the answer to 2."},{"Start":"11:56.690 ","End":"12:01.400","Text":"The reason I\u0027ve put this one in color is that"},{"Start":"12:01.400 ","End":"12:06.560","Text":"this helps me with the third part which asks if p_2 is in"},{"Start":"12:06.560 ","End":"12:11.070","Text":"the span of p_1 and p_3 which I said just means that"},{"Start":"12:11.070 ","End":"12:16.550","Text":"p_2 is a linear combination of p_1 and p_3 which it is from here."},{"Start":"12:16.550 ","End":"12:22.650","Text":"It follows. The answer to that one is yes, and we\u0027re done."}],"ID":26720},{"Watched":false,"Name":"Exercise 3","Duration":"1m 11s","ChapterTopicVideoID":25911,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.875","Text":"In this exercise, we\u0027re given the following set of 3-dimensional vectors."},{"Start":"00:07.875 ","End":"00:11.610","Text":"These as 4 of them notice."},{"Start":"00:11.610 ","End":"00:14.580","Text":"The question is for which values of a, b,"},{"Start":"00:14.580 ","End":"00:23.790","Text":"and c is"},{"Start":"00:23.790 ","End":"00:26.145","Text":"as linearly dependent."},{"Start":"00:26.145 ","End":"00:33.809","Text":"Now this is easier than it looks because I\u0027m going to pull a theorem out of a hat."},{"Start":"00:33.809 ","End":"00:36.390","Text":"Well, it\u0027s a theorem we\u0027ve had before."},{"Start":"00:36.390 ","End":"00:44.625","Text":"It says that if we have n plus 1 vectors in R^n aren\u0027t necessarily linearly dependent."},{"Start":"00:44.625 ","End":"00:49.980","Text":"The theorem might have said more than n vectors."},{"Start":"00:50.240 ","End":"00:53.100","Text":"Here, n is 3."},{"Start":"00:53.100 ","End":"00:54.330","Text":"We had 4, 5,"},{"Start":"00:54.330 ","End":"00:57.835","Text":"6, anything more than 3 vectors."},{"Start":"00:57.835 ","End":"01:02.720","Text":"Then the theorem says they\u0027re linearly dependent."},{"Start":"01:02.720 ","End":"01:05.450","Text":"So to answer the question to which values of a, b,"},{"Start":"01:05.450 ","End":"01:07.100","Text":"and c for all values,"},{"Start":"01:07.100 ","End":"01:08.690","Text":"or any values of a, b,"},{"Start":"01:08.690 ","End":"01:11.760","Text":"and c, and we\u0027re done."}],"ID":26721},{"Watched":false,"Name":"Exercise 4","Duration":"3m 37s","ChapterTopicVideoID":25899,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.290","Text":"In this exercise, we have a vector space V over a field F. It\u0027s just general,"},{"Start":"00:07.290 ","End":"00:09.795","Text":"it doesn\u0027t have to be RN."},{"Start":"00:09.795 ","End":"00:12.570","Text":"We have 3 vectors, u, v,"},{"Start":"00:12.570 ","End":"00:17.910","Text":"and w. We\u0027re told that these 3 are linearly independent."},{"Start":"00:17.910 ","End":"00:20.315","Text":"Now the question is,"},{"Start":"00:20.315 ","End":"00:23.690","Text":"is this set also with 3 vectors,"},{"Start":"00:23.690 ","End":"00:25.520","Text":"u minus v, u minus w,"},{"Start":"00:25.520 ","End":"00:26.975","Text":"u plus v minus 2w,"},{"Start":"00:26.975 ","End":"00:30.125","Text":"is this set linearly dependent?"},{"Start":"00:30.125 ","End":"00:32.885","Text":"If the answer is yes,"},{"Start":"00:32.885 ","End":"00:35.930","Text":"then we have to try and write each of"},{"Start":"00:35.930 ","End":"00:40.670","Text":"these 3 vectors as a linear combination of the other 2."},{"Start":"00:40.670 ","End":"00:42.979","Text":"Even if these are dependent,"},{"Start":"00:42.979 ","End":"00:44.990","Text":"doesn\u0027t mean we can\u0027t do this."},{"Start":"00:44.990 ","End":"00:47.390","Text":"That would just guarantee that at least one of"},{"Start":"00:47.390 ","End":"00:49.925","Text":"them can be written as a combination of the others."},{"Start":"00:49.925 ","End":"00:55.755","Text":"But we\u0027re going to try to write each of the 3 if the answer comes out yes to part 1."},{"Start":"00:55.755 ","End":"00:59.860","Text":"How do we tackle this question?"},{"Start":"00:59.930 ","End":"01:03.345","Text":"Certainly we wouldn\u0027t have to give them names,"},{"Start":"01:03.345 ","End":"01:04.740","Text":"let\u0027s call them x, y,"},{"Start":"01:04.740 ","End":"01:07.875","Text":"and z for these 3 vectors."},{"Start":"01:07.875 ","End":"01:11.000","Text":"It turns out that if you look,"},{"Start":"01:11.000 ","End":"01:13.280","Text":"each of these is a combination of u, v,"},{"Start":"01:13.280 ","End":"01:17.155","Text":"and w and the coefficients,"},{"Start":"01:17.155 ","End":"01:21.830","Text":"it turns out that they are in some sense like coordinate vectors,"},{"Start":"01:21.830 ","End":"01:29.135","Text":"like this would be 1,"},{"Start":"01:29.135 ","End":"01:32.825","Text":"minus 1, 0 and that\u0027s this, that\u0027s x."},{"Start":"01:32.825 ","End":"01:35.110","Text":"As for y it\u0027s 1,"},{"Start":"01:35.110 ","End":"01:36.770","Text":"and then v is missing,"},{"Start":"01:36.770 ","End":"01:39.965","Text":"so that\u0027s the 0 and w is minus 1."},{"Start":"01:39.965 ","End":"01:42.035","Text":"Here 1, 1,"},{"Start":"01:42.035 ","End":"01:45.100","Text":"minus 2, that\u0027s z."},{"Start":"01:45.100 ","End":"01:52.700","Text":"Then we do row operations to try and bring this to row echelon form and"},{"Start":"01:52.700 ","End":"02:00.630","Text":"see if we get a row of 0s on the left in the restricted matrix."},{"Start":"02:00.630 ","End":"02:05.690","Text":"Subtract the top row from each of the second and the third and we get this."},{"Start":"02:05.690 ","End":"02:10.615","Text":"Of course you can pause the clip and check the computations."},{"Start":"02:10.615 ","End":"02:14.850","Text":"Next we want to have a 0 here,"},{"Start":"02:14.850 ","End":"02:18.695","Text":"and we\u0027ll get that if we subtract twice this from this,"},{"Start":"02:18.695 ","End":"02:22.580","Text":"don\u0027t forget to work on the right column as well."},{"Start":"02:22.580 ","End":"02:30.270","Text":"This is the result we get and notice that we do have a row of 0s,"},{"Start":"02:30.270 ","End":"02:34.930","Text":"which means that we can write that this is 0."},{"Start":"02:34.930 ","End":"02:40.250","Text":"Write it as 0 and then we do some simplification and put it in a nice box"},{"Start":"02:40.250 ","End":"02:46.000","Text":"and here we have a formula which connect x, y, and z."},{"Start":"02:46.000 ","End":"02:52.430","Text":"Since each of them appears with that and with a non-zero coefficient,"},{"Start":"02:52.430 ","End":"02:54.260","Text":"x has a coefficient of 1,"},{"Start":"02:54.260 ","End":"02:55.880","Text":"y has a coefficient of minus 2,"},{"Start":"02:55.880 ","End":"03:01.420","Text":"and so on, then that means we can extract each one in terms of the other 2."},{"Start":"03:01.420 ","End":"03:04.590","Text":"I\u0027ll just show you all 3 of them at once."},{"Start":"03:04.590 ","End":"03:06.200","Text":"The answer to part 1 was yes,"},{"Start":"03:06.200 ","End":"03:07.925","Text":"they are linearly dependent."},{"Start":"03:07.925 ","End":"03:11.975","Text":"We got that just from the 0s here without even looking at that."},{"Start":"03:11.975 ","End":"03:16.560","Text":"Then from here, we just one at a time extract,"},{"Start":"03:16.560 ","End":"03:22.070","Text":"so if we want x, we just put the 2y and the z on the other side."},{"Start":"03:22.070 ","End":"03:24.925","Text":"If we want z, similarly."},{"Start":"03:24.925 ","End":"03:30.800","Text":"If we want the y, I would put the 2y on the other side and divide by 2."},{"Start":"03:30.800 ","End":"03:32.300","Text":"I wrote it in decimals,"},{"Start":"03:32.300 ","End":"03:33.650","Text":"1 over 2 is a 1/2,"},{"Start":"03:33.650 ","End":"03:38.250","Text":"and this is what we get and that\u0027s the answer. We\u0027re done."}],"ID":26709},{"Watched":false,"Name":"Exercise 5","Duration":"2m 26s","ChapterTopicVideoID":25900,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.290","Text":"In this exercise, we have a vector space v over some field f,"},{"Start":"00:07.290 ","End":"00:11.635","Text":"could be r and could be real numbers, but not necessarily."},{"Start":"00:11.635 ","End":"00:14.040","Text":"We have 3 vectors, u, v,"},{"Start":"00:14.040 ","End":"00:17.545","Text":"w, which are linearly independent."},{"Start":"00:17.545 ","End":"00:21.480","Text":"The question is, is this set of 3 vectors,"},{"Start":"00:21.480 ","End":"00:24.975","Text":"u plus v, v plus w and w,"},{"Start":"00:24.975 ","End":"00:27.225","Text":"is it linearly dependent?"},{"Start":"00:27.225 ","End":"00:30.570","Text":"If so, well, we know"},{"Start":"00:30.570 ","End":"00:34.780","Text":"that at least 1 of them can be written as the linear combination of the others."},{"Start":"00:34.780 ","End":"00:39.270","Text":"We\u0027ll just try to write each 1 as a linear combination of the others."},{"Start":"00:39.270 ","End":"00:41.155","Text":"Maybe you won\u0027t be able to do all of them."},{"Start":"00:41.155 ","End":"00:45.510","Text":"Maybe it won\u0027t even be linearly dependent. Let\u0027s see."},{"Start":"00:45.610 ","End":"00:47.750","Text":"Let\u0027s give them names."},{"Start":"00:47.750 ","End":"00:51.120","Text":"We\u0027ll call them x, y, and z."},{"Start":"00:51.650 ","End":"00:58.280","Text":"It turns out that these coefficients, like 1, 1,"},{"Start":"00:58.280 ","End":"01:05.810","Text":"0 are actually coordinate vectors for these xyz in terms of uvw."},{"Start":"01:05.810 ","End":"01:13.730","Text":"What we can do is to put them into a matrix like here,"},{"Start":"01:13.730 ","End":"01:16.250","Text":"here\u0027s 1, 1, 0."},{"Start":"01:16.250 ","End":"01:17.825","Text":"There\u0027s no w here."},{"Start":"01:17.825 ","End":"01:23.210","Text":"Here. u is 0 plus 1 v plus 1 w that\u0027s a 0,"},{"Start":"01:23.210 ","End":"01:25.145","Text":"1, 1 for y,"},{"Start":"01:25.145 ","End":"01:28.215","Text":"and for z, it\u0027s 0,"},{"Start":"01:28.215 ","End":"01:33.490","Text":"u, 0, v, and 1 w."},{"Start":"01:34.070 ","End":"01:43.140","Text":"Now I want to bring this to row echelon form only look at already is in row echelon form."},{"Start":"01:45.470 ","End":"01:48.470","Text":"On the restricted part,"},{"Start":"01:48.470 ","End":"01:53.970","Text":"this part here, there are no 0 rows."},{"Start":"01:54.020 ","End":"01:57.000","Text":"It will never be, I mean,"},{"Start":"01:57.000 ","End":"02:02.220","Text":"it isn\u0027t linearly dependent."},{"Start":"02:02.220 ","End":"02:06.330","Text":"These 3 vectors are linearly independent."},{"Start":"02:07.580 ","End":"02:12.555","Text":"Because the answer to that part 1, is no,"},{"Start":"02:12.555 ","End":"02:19.040","Text":"they are not linearly dependent they\u0027re linearly independent,"},{"Start":"02:19.040 ","End":"02:26.700","Text":"then question 2 doesn\u0027t apply here so just forget about that and we\u0027re done."}],"ID":26710},{"Watched":false,"Name":"Exercise 6","Duration":"4m 3s","ChapterTopicVideoID":25901,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.750","Text":"In this exercise, we have a general vector space V over a field F,"},{"Start":"00:06.750 ","End":"00:08.490","Text":"it could be RN,"},{"Start":"00:08.490 ","End":"00:11.590","Text":"but it\u0027s more general."},{"Start":"00:12.560 ","End":"00:17.080","Text":"There\u0027s an extra letter here."},{"Start":"00:17.420 ","End":"00:21.510","Text":"Now we\u0027re given that there are 3 vectors, u, v,"},{"Start":"00:21.510 ","End":"00:23.220","Text":"and w in this space,"},{"Start":"00:23.220 ","End":"00:26.590","Text":"that are linearly independent."},{"Start":"00:26.900 ","End":"00:32.250","Text":"First question is, is this set linearly dependent?"},{"Start":"00:32.250 ","End":"00:34.320","Text":"There are 3 vectors in it."},{"Start":"00:34.320 ","End":"00:38.370","Text":"Notice that each of them is a different linear combination of u,"},{"Start":"00:38.370 ","End":"00:42.865","Text":"v, and w and then part 2,"},{"Start":"00:42.865 ","End":"00:45.605","Text":"if they are linearly dependent,"},{"Start":"00:45.605 ","End":"00:50.120","Text":"then we are to write each vector,"},{"Start":"00:50.120 ","End":"00:53.990","Text":"each of the 3 as a linear combination of the other 2."},{"Start":"00:53.990 ","End":"00:58.220","Text":"It\u0027s not always possible because in linearly dependent,"},{"Start":"00:58.220 ","End":"01:01.580","Text":"it just means that at least 1 of them is a combination of the rest."},{"Start":"01:01.580 ","End":"01:03.650","Text":"But I guess in this case,"},{"Start":"01:03.650 ","End":"01:06.010","Text":"it will turn out that all 3 of them are,"},{"Start":"01:06.010 ","End":"01:07.940","Text":"and I\u0027m guessing that the answer will be that yes,"},{"Start":"01:07.940 ","End":"01:10.930","Text":"they are linearly dependent."},{"Start":"01:10.930 ","End":"01:14.820","Text":"Let\u0027s start off by giving each of these 3 a name,"},{"Start":"01:14.820 ","End":"01:16.410","Text":"the first one I\u0027ll call x,"},{"Start":"01:16.410 ","End":"01:22.465","Text":"next one y, next one z and the strategy is as follows."},{"Start":"01:22.465 ","End":"01:28.385","Text":"It\u0027s as if the coefficients are coordinates."},{"Start":"01:28.385 ","End":"01:30.515","Text":"So when we look at x,"},{"Start":"01:30.515 ","End":"01:33.930","Text":"we think 1, 2, 3."},{"Start":"01:33.930 ","End":"01:35.575","Text":"When you look at y,"},{"Start":"01:35.575 ","End":"01:38.360","Text":"you think 4, 5,"},{"Start":"01:38.360 ","End":"01:43.805","Text":"6, and z would be 7, 8, 9."},{"Start":"01:43.805 ","End":"01:46.310","Text":"There is actually a logic behind this."},{"Start":"01:46.310 ","End":"01:49.550","Text":"It turns out that these really are the coordinate vectors of"},{"Start":"01:49.550 ","End":"01:54.680","Text":"these in the subspace spanned by u, v,"},{"Start":"01:54.680 ","End":"01:59.550","Text":"and w. But I don\u0027t want to get theoretical,"},{"Start":"01:59.550 ","End":"02:06.920","Text":"let\u0027s just take it as how to approach and now we use our usual strategy."},{"Start":"02:06.920 ","End":"02:13.030","Text":"We write these coordinates in a matrix,"},{"Start":"02:13.030 ","End":"02:15.830","Text":"it could be restricted or it could be augmented."},{"Start":"02:15.830 ","End":"02:18.590","Text":"If we just want to know linearly dependent,"},{"Start":"02:18.590 ","End":"02:22.550","Text":"we don\u0027t need the augmented in the x, y, z."},{"Start":"02:22.550 ","End":"02:27.620","Text":"Then we\u0027ll do row operations and if we get a 0 row, then they\u0027re dependent."},{"Start":"02:27.620 ","End":"02:33.125","Text":"But if we want to start writing specific combinations to know how they\u0027re dependent,"},{"Start":"02:33.125 ","End":"02:36.200","Text":"then we need that extra column."},{"Start":"02:36.200 ","End":"02:38.030","Text":"That\u0027s what I\u0027m doing,"},{"Start":"02:38.030 ","End":"02:41.070","Text":"and now let\u0027s do some row operations."},{"Start":"02:41.070 ","End":"02:43.180","Text":"Let\u0027s go through this quickly."},{"Start":"02:43.180 ","End":"02:45.280","Text":"You can always pause and check."},{"Start":"02:45.280 ","End":"02:47.260","Text":"I want 0s here and here,"},{"Start":"02:47.260 ","End":"02:51.820","Text":"so I just subtract the correct multiple of the top row from each of these,"},{"Start":"02:51.820 ","End":"02:53.080","Text":"and we get this."},{"Start":"02:53.080 ","End":"02:56.515","Text":"Don\u0027t forget to do it to the right-hand column also."},{"Start":"02:56.515 ","End":"02:58.960","Text":"Next I need a 0 here,"},{"Start":"02:58.960 ","End":"03:04.025","Text":"so I subtract twice this row from this row."},{"Start":"03:04.025 ","End":"03:13.765","Text":"Now this is already in row echelon form and we do have a row of 0s here."},{"Start":"03:13.765 ","End":"03:18.820","Text":"So already we know that the vectors are linearly dependent"},{"Start":"03:18.820 ","End":"03:23.860","Text":"and we can actually do some computations if we set this equal to 0."},{"Start":"03:23.860 ","End":"03:27.905","Text":"Simplify it, put a nice box around it."},{"Start":"03:27.905 ","End":"03:31.110","Text":"As I said, the answer to the first one is yes,"},{"Start":"03:31.110 ","End":"03:33.530","Text":"they are linearly dependent."},{"Start":"03:33.530 ","End":"03:35.180","Text":"But more than that,"},{"Start":"03:35.180 ","End":"03:40.610","Text":"since each of these variables appear with a non-zero coefficient,"},{"Start":"03:40.610 ","End":"03:45.180","Text":"we can extract each one in terms of the others and here they are."},{"Start":"03:45.180 ","End":"03:51.710","Text":"For example, x would just be put the 2y to the other side and subtract z and so on."},{"Start":"03:51.710 ","End":"03:54.605","Text":"In the case of y, we have to divide by 2,"},{"Start":"03:54.605 ","End":"03:56.360","Text":"I wrote it in decimals."},{"Start":"03:56.360 ","End":"04:00.285","Text":"You could also write 1.5x plus 1.5z whatever,"},{"Start":"04:00.285 ","End":"04:03.940","Text":"and that\u0027s the answer and we\u0027re done."}],"ID":26711},{"Watched":false,"Name":"Exercise 7","Duration":"3m 10s","ChapterTopicVideoID":25902,"CourseChapterTopicPlaylistID":246316,"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, our vector space is C^3,"},{"Start":"00:05.070 ","End":"00:08.595","Text":"3 dimensional complex space."},{"Start":"00:08.595 ","End":"00:11.715","Text":"The field, even if I didn\u0027t say so,"},{"Start":"00:11.715 ","End":"00:13.530","Text":"would be the complex numbers."},{"Start":"00:13.530 ","End":"00:16.710","Text":"That\u0027s the default unless stated otherwise."},{"Start":"00:16.710 ","End":"00:21.240","Text":"It\u0027s C^3 over C. Here we have"},{"Start":"00:21.240 ","End":"00:28.420","Text":"just 2 vectors in this space and we want to know if they\u0027re linearly independent."},{"Start":"00:29.270 ","End":"00:32.340","Text":"I claim the answer is no."},{"Start":"00:32.340 ","End":"00:37.650","Text":"I\u0027ve done the exercise already and they are linearly dependent. Let\u0027s see why."},{"Start":"00:37.650 ","End":"00:41.420","Text":"We\u0027ll do it with matrices may be a bit of an overkill,"},{"Start":"00:41.420 ","End":"00:44.065","Text":"but I like to use matrices."},{"Start":"00:44.065 ","End":"00:46.700","Text":"Here they are."},{"Start":"00:46.700 ","End":"00:48.620","Text":"This one is the top row,"},{"Start":"00:48.620 ","End":"00:49.685","Text":"this one\u0027s the bottom row."},{"Start":"00:49.685 ","End":"00:51.560","Text":"I didn\u0027t need to give them names."},{"Start":"00:51.560 ","End":"00:52.610","Text":"I could have said,"},{"Start":"00:52.610 ","End":"00:54.725","Text":"okay, let this be x,"},{"Start":"00:54.725 ","End":"00:56.145","Text":"let this be y,"},{"Start":"00:56.145 ","End":"01:00.560","Text":"and then use an augmented matrix with x and y."},{"Start":"01:00.560 ","End":"01:05.780","Text":"But I don\u0027t need all that because I just want to know they are linearly dependent or not."},{"Start":"01:05.780 ","End":"01:09.525","Text":"I don\u0027t need to know what the relation is,"},{"Start":"01:09.525 ","End":"01:13.470","Text":"how to get a linear combination to be 0."},{"Start":"01:13.470 ","End":"01:15.620","Text":"I don\u0027t need that."},{"Start":"01:15.620 ","End":"01:23.075","Text":"Erased it. Let\u0027s see if we can bring it into row echelon form."},{"Start":"01:23.075 ","End":"01:27.020","Text":"What I would normally do would be to try and get a 0 here by"},{"Start":"01:27.020 ","End":"01:31.735","Text":"multiplying the top row by i plus 1 and then subtracting."},{"Start":"01:31.735 ","End":"01:33.980","Text":"With the proper notation,"},{"Start":"01:33.980 ","End":"01:37.460","Text":"it\u0027s row 2 minus i plus 1 row 1,"},{"Start":"01:37.460 ","End":"01:40.265","Text":"put the result in row 2."},{"Start":"01:40.265 ","End":"01:44.600","Text":"If we do that, for sure we get 0 here, that\u0027s why we did it."},{"Start":"01:44.600 ","End":"01:49.245","Text":"Then this minus i plus 1 times i,"},{"Start":"01:49.245 ","End":"01:53.505","Text":"and then minus 2 less i plus 1 times this."},{"Start":"01:53.505 ","End":"01:56.705","Text":"We just have to do a bit of simplification."},{"Start":"01:56.705 ","End":"02:00.750","Text":"Open brackets here and here."},{"Start":"02:00.750 ","End":"02:05.090","Text":"Remember that i squared is negative 1."},{"Start":"02:05.090 ","End":"02:08.755","Text":"That\u0027s the whole idea with complex numbers."},{"Start":"02:08.755 ","End":"02:12.240","Text":"Look, i minus i is 0,"},{"Start":"02:12.240 ","End":"02:14.745","Text":"i squared is minus 1."},{"Start":"02:14.745 ","End":"02:16.500","Text":"Minus i squared is plus 1."},{"Start":"02:16.500 ","End":"02:20.345","Text":"This one is a 0, this one also comes out 0."},{"Start":"02:20.345 ","End":"02:22.640","Text":"We have a row of 0s,"},{"Start":"02:22.640 ","End":"02:29.090","Text":"which means now we are done and we can just say yes, they\u0027re linearly dependent."},{"Start":"02:29.090 ","End":"02:30.845","Text":"But I could even say more."},{"Start":"02:30.845 ","End":"02:33.020","Text":"If I look at this equation here,"},{"Start":"02:33.020 ","End":"02:37.630","Text":"this is what gave me a 0 in the second row."},{"Start":"02:37.630 ","End":"02:41.985","Text":"You could actually say that this holds."},{"Start":"02:41.985 ","End":"02:45.580","Text":"If you gave the original vector\u0027s name,"},{"Start":"02:45.580 ","End":"02:49.745","Text":"let\u0027s say you call the first one x and the second one y,"},{"Start":"02:49.745 ","End":"02:52.430","Text":"you would get this relationship,"},{"Start":"02:52.430 ","End":"02:57.470","Text":"which you could also say that y is equal to i"},{"Start":"02:57.470 ","End":"03:02.900","Text":"plus 1 times x and get 1 as the combination of the other."},{"Start":"03:02.900 ","End":"03:05.715","Text":"Anyway, that\u0027s beyond what was asked."},{"Start":"03:05.715 ","End":"03:10.660","Text":"The answer is just yes. We\u0027re done."}],"ID":26712},{"Watched":false,"Name":"Exercise 8","Duration":"1m 34s","ChapterTopicVideoID":25903,"CourseChapterTopicPlaylistID":246316,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.560 ","End":"00:06.270","Text":"Now this exercise is very similar to the previous 1."},{"Start":"00:06.270 ","End":"00:11.880","Text":"The difference being that this time the vector space is C3,"},{"Start":"00:11.880 ","End":"00:14.820","Text":"but not the default field,"},{"Start":"00:14.820 ","End":"00:19.560","Text":"which would be C, but over the real numbers."},{"Start":"00:19.560 ","End":"00:25.140","Text":"Certainly you can multiply a real number by a complex number to get a complex number."},{"Start":"00:25.140 ","End":"00:30.435","Text":"Often, the real numbers are scalars for complex spaces."},{"Start":"00:30.435 ","End":"00:32.885","Text":"Let\u0027s see if it makes a difference."},{"Start":"00:32.885 ","End":"00:35.690","Text":"Last time we got the answer was yes,"},{"Start":"00:35.690 ","End":"00:37.820","Text":"they are dependent or no,"},{"Start":"00:37.820 ","End":"00:40.490","Text":"they\u0027re not linearly independent."},{"Start":"00:40.550 ","End":"00:42.945","Text":"I\u0027m going to give you a spoiler,"},{"Start":"00:42.945 ","End":"00:45.440","Text":"the answer is going to be different this time."},{"Start":"00:45.440 ","End":"00:53.265","Text":"It\u0027s not the same as having C3 over C. We\u0027ve got the same matrix as before."},{"Start":"00:53.265 ","End":"00:56.830","Text":"We want to try and bring this to echelon form,"},{"Start":"00:56.830 ","End":"01:04.505","Text":"but we can\u0027t because our scalars have to come from the real numbers."},{"Start":"01:04.505 ","End":"01:09.780","Text":"There\u0027s nothing I can multiply this by to give"},{"Start":"01:09.780 ","End":"01:15.815","Text":"me an i plus 1 here because I only have real numbers at my disposal."},{"Start":"01:15.815 ","End":"01:19.590","Text":"This is as far as we can go."},{"Start":"01:19.730 ","End":"01:25.310","Text":"That set over the real numbers can\u0027t bring it to row echelon form,"},{"Start":"01:25.310 ","End":"01:27.635","Text":"so can\u0027t get a row of 0s,"},{"Start":"01:27.635 ","End":"01:35.040","Text":"and so the vectors are linearly independent in this case. We\u0027re done."}],"ID":26713}],"Thumbnail":null,"ID":246316},{"Name":"Vector Basis","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Sufficient condition for Finding Basis","Duration":"19m 36s","ChapterTopicVideoID":25913,"CourseChapterTopicPlaylistID":246317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.260","Text":"We\u0027re back from the break where we just worked hard to"},{"Start":"00:04.260 ","End":"00:09.330","Text":"prove that a certain set was a basis of vector space."},{"Start":"00:09.330 ","End":"00:11.190","Text":"To make life easier,"},{"Start":"00:11.190 ","End":"00:13.140","Text":"I\u0027m going to bring you a proposition."},{"Start":"00:13.140 ","End":"00:16.740","Text":"A proposition is a mini theorem or a claim,"},{"Start":"00:16.740 ","End":"00:20.730","Text":"which will be a sufficient condition for a basis."},{"Start":"00:20.730 ","End":"00:23.970","Text":"We\u0027re still with famous vector spaces we call"},{"Start":"00:23.970 ","End":"00:28.420","Text":"famous and it\u0027s going to save us a lot of work."},{"Start":"00:28.820 ","End":"00:31.950","Text":"There\u0027ll be 3 parts."},{"Start":"00:31.950 ","End":"00:34.394","Text":"I\u0027ll give you all 3 of them at once."},{"Start":"00:34.394 ","End":"00:43.250","Text":"Part 1 says that if we have exactly n linearly dependent vectors in R^n,"},{"Start":"00:43.250 ","End":"00:46.040","Text":"n dimensional because that\u0027s what R^n is,"},{"Start":"00:46.040 ","End":"00:49.895","Text":"then this is a basis of R^n."},{"Start":"00:49.895 ","End":"00:51.470","Text":"Now what have we done here?"},{"Start":"00:51.470 ","End":"00:54.920","Text":"Already we had to prove linearly independent"},{"Start":"00:54.920 ","End":"01:02.085","Text":"but the second part which was the set had to spend R^n we\u0027ve omitted."},{"Start":"01:02.085 ","End":"01:06.515","Text":"Instead of that all we require is that there\u0027ll be the right number in the set."},{"Start":"01:06.515 ","End":"01:10.055","Text":"It\u0027s got to be exactly n of them and we have exactly n of them,"},{"Start":"01:10.055 ","End":"01:14.440","Text":"then all we need is linearly independence without the spanning part."},{"Start":"01:14.440 ","End":"01:19.865","Text":"Similarly for this, for the space of m by n matrices,"},{"Start":"01:19.865 ","End":"01:22.920","Text":"for R^n we need exactly n vectors here,"},{"Start":"01:22.920 ","End":"01:26.080","Text":"we need m n vectors also linearly independent."},{"Start":"01:26.080 ","End":"01:29.750","Text":"Then there are basis and we don\u0027t have to check the spanning path."},{"Start":"01:29.750 ","End":"01:33.100","Text":"For the case of polynomials of degree up to n,"},{"Start":"01:33.100 ","End":"01:36.799","Text":"don\u0027t get confused is not n, it\u0027s n plus 1."},{"Start":"01:36.799 ","End":"01:41.680","Text":"If we have n plus 1 linearly independent polynomials of degree less than or equal to n,"},{"Start":"01:41.680 ","End":"01:45.630","Text":"then that will be a basis and we don\u0027t have to prove the spanning part."},{"Start":"01:45.630 ","End":"01:48.735","Text":"I\u0027ll give you an example of each."},{"Start":"01:48.735 ","End":"01:52.660","Text":"For R^n I\u0027ll give you 2 examples 1 from R^2 and 1 from R^3."},{"Start":"01:52.660 ","End":"01:56.825","Text":"I mean this is a whole class of examples really because n is any natural number."},{"Start":"01:56.825 ","End":"02:02.160","Text":"Here we\u0027ll take the 2 vectors, 1,2 and 3,4."},{"Start":"02:02.160 ","End":"02:04.320","Text":"First of all it\u0027s the right number."},{"Start":"02:04.320 ","End":"02:06.255","Text":"If we count them there\u0027s 2 of them."},{"Start":"02:06.255 ","End":"02:08.010","Text":"We already said that for R^n,"},{"Start":"02:08.010 ","End":"02:11.650","Text":"we want n so for R_2 we want 2 of them."},{"Start":"02:11.650 ","End":"02:15.200","Text":"All they have to show is that they\u0027re linearly independent."},{"Start":"02:15.200 ","End":"02:17.480","Text":"Now, I\u0027m going to skip this part."},{"Start":"02:17.480 ","End":"02:21.530","Text":"I don\u0027t want to waste a lot of time with tedious routine stuff."},{"Start":"02:21.530 ","End":"02:26.015","Text":"I\u0027ll leave you to check these 2 are linearly independent."},{"Start":"02:26.015 ","End":"02:30.000","Text":"Then we\u0027ll know that there are basis of R^2."},{"Start":"02:30.380 ","End":"02:38.410","Text":"Here\u0027s an example, it\u0027s also R^n but this time R^3 so we\u0027ll need exactly 3 vectors."},{"Start":"02:38.660 ","End":"02:42.530","Text":"If I can show that these are linearly independent,"},{"Start":"02:42.530 ","End":"02:46.400","Text":"then I can forget about the spanning path because it\u0027s got the right number,"},{"Start":"02:46.400 ","End":"02:47.960","Text":"it\u0027s got exactly 3 of them."},{"Start":"02:47.960 ","End":"02:52.040","Text":"That\u0027s the same 3 is from here and that\u0027s what the first part said."},{"Start":"02:52.040 ","End":"02:54.755","Text":"Again, I\u0027m not going to check that these are"},{"Start":"02:54.755 ","End":"02:59.110","Text":"linearly independent I\u0027ll leave that for you to check."},{"Start":"02:59.110 ","End":"03:03.379","Text":"When we\u0027ve done that then we can tell straight away that they\u0027re going to be a basis."},{"Start":"03:03.379 ","End":"03:09.190","Text":"Now let\u0027s take an example from the space of m by n matrices."},{"Start":"03:09.190 ","End":"03:12.950","Text":"The example I\u0027ll take will be m is 2 and n is 2,"},{"Start":"03:12.950 ","End":"03:16.680","Text":"so we have 2 by 2 matrices."},{"Start":"03:16.850 ","End":"03:24.975","Text":"According to part 2 I need m times n. That would be 2 times 2 is 4."},{"Start":"03:24.975 ","End":"03:28.700","Text":"If I have 4 linearly independent matrices,"},{"Start":"03:28.700 ","End":"03:32.945","Text":"then there will be a basis and I don\u0027t have to check the spanning path."},{"Start":"03:32.945 ","End":"03:39.260","Text":"Again, I\u0027m not going to do the routine checking that these are linearly independent,"},{"Start":"03:39.260 ","End":"03:44.270","Text":"it\u0027s a far bit of work but it\u0027s routine and I\u0027ll leave you to check that."},{"Start":"03:44.270 ","End":"03:47.010","Text":"I\u0027ve checked it so I can tell you,"},{"Start":"03:47.010 ","End":"03:48.450","Text":"if you want to just take my word for it,"},{"Start":"03:48.450 ","End":"03:50.810","Text":"these 4 are linearly independent."},{"Start":"03:50.810 ","End":"03:53.345","Text":"Now we need 1 more example."},{"Start":"03:53.345 ","End":"03:56.635","Text":"I want an example with polynomials."},{"Start":"03:56.635 ","End":"04:00.120","Text":"Here we\u0027ll take n to be 2."},{"Start":"04:00.120 ","End":"04:07.065","Text":"It\u0027s the polynomials of degree up to 2 and we will then need 3 of them."},{"Start":"04:07.065 ","End":"04:10.715","Text":"In this case, we actually do have 3 polynomials."},{"Start":"04:10.715 ","End":"04:13.610","Text":"Polynomial 1, the polynomial x plus 1,"},{"Start":"04:13.610 ","End":"04:16.355","Text":"and the polynomial 1 plus x plus x squared."},{"Start":"04:16.355 ","End":"04:18.295","Text":"There\u0027s 3 of them."},{"Start":"04:18.295 ","End":"04:22.520","Text":"All I need for them to be a basis."},{"Start":"04:22.520 ","End":"04:25.340","Text":"Instead of showing that they span the space,"},{"Start":"04:25.340 ","End":"04:27.940","Text":"I just needed to be the right number which in"},{"Start":"04:27.940 ","End":"04:31.390","Text":"this case is 2 plus 1 is 3 and there are 3 of them."},{"Start":"04:31.390 ","End":"04:36.710","Text":"The linearly independence, I\u0027ll leave that to you to do as an exercise."},{"Start":"04:36.710 ","End":"04:40.805","Text":"Now, I want to just give you as a summary,"},{"Start":"04:40.805 ","End":"04:44.840","Text":"what I\u0027m just going to tell you is what I said before is that,"},{"Start":"04:44.840 ","End":"04:48.470","Text":"we\u0027ve replaced the part about spanning"},{"Start":"04:48.470 ","End":"04:54.010","Text":"by just having the right number of linearly independent vectors."},{"Start":"04:54.010 ","End":"04:58.830","Text":"Here they are to show you all 3 parts at 1 instead of exposing them 1 at a time."},{"Start":"04:58.830 ","End":"05:06.340","Text":"If we have a set of n linearly independent vectors in R^n,"},{"Start":"05:06.340 ","End":"05:10.760","Text":"then what we said before is that they are a basis."},{"Start":"05:10.760 ","End":"05:17.575","Text":"Now basis means 2 things linearly independent which we have already and spanning."},{"Start":"05:17.575 ","End":"05:24.710","Text":"Because of what it says above that they are a basis of R^n in particular, they span R^n."},{"Start":"05:24.710 ","End":"05:28.265","Text":"Something is a basis of a space it spans that space."},{"Start":"05:28.265 ","End":"05:35.015","Text":"Similarly, if we have the right number m times n of m by n matrices,"},{"Start":"05:35.015 ","End":"05:38.480","Text":"then what it said above was that there are basis"},{"Start":"05:38.480 ","End":"05:42.530","Text":"of the space and if they\u0027re basis then in particular they span."},{"Start":"05:42.530 ","End":"05:43.940","Text":"Finally, for the polynomials,"},{"Start":"05:43.940 ","End":"05:50.105","Text":"if I take m plus 1 of them of degree less than or equal to n which is this base,"},{"Start":"05:50.105 ","End":"05:56.030","Text":"then this set will be a basis for this and therefore it will span that."},{"Start":"05:56.030 ","End":"06:00.350","Text":"That\u0027s a corollary but we just replaced being a basis"},{"Start":"06:00.350 ","End":"06:07.170","Text":"by the part and it spans which is less and so logically it follows."},{"Start":"06:08.900 ","End":"06:11.750","Text":"That\u0027s that."},{"Start":"06:11.750 ","End":"06:14.600","Text":"Next we\u0027ll be moving on to a proposition a bit similar to"},{"Start":"06:14.600 ","End":"06:17.720","Text":"the previous proposition but from the opposite direction."},{"Start":"06:17.720 ","End":"06:19.495","Text":"You\u0027ll see what I mean."},{"Start":"06:19.495 ","End":"06:21.200","Text":"I just copied again,"},{"Start":"06:21.200 ","End":"06:22.760","Text":"I called it review of"},{"Start":"06:22.760 ","End":"06:29.045","Text":"the previous proposition which was a sufficient condition for something to be a basis."},{"Start":"06:29.045 ","End":"06:35.570","Text":"I just abbreviated LI is the standard abbreviation for linearly independent and well,"},{"Start":"06:35.570 ","End":"06:40.840","Text":"we\u0027re at it LD is the standard abbreviation for linearly dependent."},{"Start":"06:40.840 ","End":"06:46.115","Text":"Whereas previously we had a sufficient condition for a set to be a basis."},{"Start":"06:46.115 ","End":"06:50.880","Text":"Here we\u0027re going to have a sufficient condition for a set not to be a basis."},{"Start":"06:50.880 ","End":"06:52.550","Text":"this is not great English,"},{"Start":"06:52.550 ","End":"06:56.105","Text":"but condition for not a basis, but you know what I mean."},{"Start":"06:56.105 ","End":"06:58.490","Text":"Not just contrast them."},{"Start":"06:58.490 ","End":"07:00.845","Text":"Here we had a set of"},{"Start":"07:00.845 ","End":"07:07.935","Text":"n linearly independent n-dimensional vectors and we said that it is a base of R^n."},{"Start":"07:07.935 ","End":"07:12.120","Text":"Here I\u0027m telling you that if we have a set of less than"},{"Start":"07:12.120 ","End":"07:18.500","Text":"n doesn\u0027t matter linearly independent or otherwise any set of less than n,"},{"Start":"07:18.500 ","End":"07:22.610","Text":"n-dimensional vectors is not going to be a basis"},{"Start":"07:22.610 ","End":"07:28.175","Text":"for R^n regardless of say, the linearly independence."},{"Start":"07:28.175 ","End":"07:33.900","Text":"The second one very similar instead of taking a set of m n,"},{"Start":"07:33.900 ","End":"07:41.795","Text":"we\u0027re going to take a set of less than m n of these m by n matrices,"},{"Start":"07:41.795 ","End":"07:44.525","Text":"then it is not going to be a basis."},{"Start":"07:44.525 ","End":"07:47.735","Text":"Doesn\u0027t matter again, about linearly independence."},{"Start":"07:47.735 ","End":"07:54.420","Text":"You can\u0027t have less than m n elements in a basis."},{"Start":"07:54.420 ","End":"07:56.940","Text":"Part 3 also instead of m plus 1,"},{"Start":"07:56.940 ","End":"07:59.055","Text":"I\u0027ll take less than."},{"Start":"07:59.055 ","End":"08:06.050","Text":"If we have less than n plus 1 polynomials of degree less than or equal to n,"},{"Start":"08:06.050 ","End":"08:13.070","Text":"then it\u0027s not going to be a basis for P_n of R. Just like we gave a few examples here,"},{"Start":"08:13.070 ","End":"08:15.785","Text":"let\u0027s give a few examples for this,"},{"Start":"08:15.785 ","End":"08:25.175","Text":"not exactly an opposite proposition but the other side condition for not being a basis."},{"Start":"08:25.175 ","End":"08:27.830","Text":"Let\u0027s take an example from here."},{"Start":"08:27.830 ","End":"08:32.595","Text":"An example of R^2 which is this where n is 2."},{"Start":"08:32.595 ","End":"08:38.460","Text":"If we take a set of less than 2 and this set A which contains"},{"Start":"08:38.460 ","End":"08:44.090","Text":"just a single vector is not going to be"},{"Start":"08:44.090 ","End":"08:50.800","Text":"a basis of R_2 because 1 is less than 2 and we need at least 2 members."},{"Start":"08:50.800 ","End":"08:52.635","Text":"I\u0027m still with R^n,"},{"Start":"08:52.635 ","End":"08:55.920","Text":"but we\u0027ll take n equals 3 this time."},{"Start":"08:55.920 ","End":"08:58.570","Text":"If we have less than 3,"},{"Start":"08:58.570 ","End":"09:05.445","Text":"and here we have a set with just 2 vectors of dimension 3,"},{"Start":"09:05.445 ","End":"09:13.500","Text":"there are not going to be a basis of R^3 because number members 2 is less than 3."},{"Start":"09:13.500 ","End":"09:16.750","Text":"Next an example with matrices,"},{"Start":"09:16.750 ","End":"09:19.520","Text":"where we\u0027re going to take m and n as each being 2."},{"Start":"09:19.520 ","End":"09:22.430","Text":"We\u0027re talking about 2 by 2 matrices."},{"Start":"09:22.430 ","End":"09:24.320","Text":"For 2 by 2 matrices,"},{"Start":"09:24.320 ","End":"09:31.170","Text":"for a basis we need 2 times 2 is 4 and less than 4 will not do."},{"Start":"09:31.170 ","End":"09:34.050","Text":"Here we have less than 4 because we have 1,"},{"Start":"09:34.050 ","End":"09:36.510","Text":"2, 3, less than 4."},{"Start":"09:36.510 ","End":"09:39.815","Text":"Without even looking at what these matrices are,"},{"Start":"09:39.815 ","End":"09:42.275","Text":"we don\u0027t care about linear independence."},{"Start":"09:42.275 ","End":"09:48.840","Text":"There\u0027s not enough of them so it is not going to be a basis."},{"Start":"09:48.840 ","End":"09:51.410","Text":"Finally, an example of polynomials,"},{"Start":"09:51.410 ","End":"09:53.330","Text":"n equals 3,"},{"Start":"09:53.330 ","End":"09:55.850","Text":"3 plus 1 is 4,"},{"Start":"09:55.850 ","End":"09:58.470","Text":"and I only have 3 of them."},{"Start":"09:58.470 ","End":"10:04.565","Text":"This is not going to be a basis for polynomials of degree 3 or less."},{"Start":"10:04.565 ","End":"10:07.255","Text":"That completes the examples."},{"Start":"10:07.255 ","End":"10:12.065","Text":"Moving on our next topic is something called the standard basis"},{"Start":"10:12.065 ","End":"10:16.970","Text":"but each vector space from stock of"},{"Start":"10:16.970 ","End":"10:22.920","Text":"famous vector spaces has its own special standard basis."},{"Start":"10:22.920 ","End":"10:25.175","Text":"Vector space will typically have"},{"Start":"10:25.175 ","End":"10:31.474","Text":"many possibilities for a basis but only 1 will be the standard basis."},{"Start":"10:31.474 ","End":"10:36.165","Text":"As for famous, we have 3 main kinds."},{"Start":"10:36.165 ","End":"10:41.360","Text":"They are R^n which is a family and goes from 1,"},{"Start":"10:41.360 ","End":"10:45.230","Text":"2, 3, 4, 5, etc, n-dimensional space."},{"Start":"10:45.230 ","End":"10:51.830","Text":"Then we also have matrices over the real numbers of size m by n. Again,"},{"Start":"10:51.830 ","End":"10:54.170","Text":"m and n can be natural numbers."},{"Start":"10:54.170 ","End":"10:59.660","Text":"Also we have the space of polynomials of degree up to and including"},{"Start":"10:59.660 ","End":"11:08.345","Text":"n. I\u0027m not going to show you in full generality the standard basis of R^n for example."},{"Start":"11:08.345 ","End":"11:14.720","Text":"We\u0027ll take the example of R^4 and you\u0027ll be able to figure out how to generalize that."},{"Start":"11:14.720 ","End":"11:19.540","Text":"Otherwise it\u0027s going to be too much dot dot dot and so on."},{"Start":"11:19.540 ","End":"11:26.980","Text":"For this we\u0027ll take 2 by 3 matrices and for n here we\u0027ll take 4."},{"Start":"11:27.600 ","End":"11:30.475","Text":"We\u0027ll begin with R^4,"},{"Start":"11:30.475 ","End":"11:35.680","Text":"and I\u0027ll tell you that the standard basis is the following set."},{"Start":"11:35.680 ","End":"11:38.635","Text":"We\u0027ll give it a name, we\u0027ll call it E,"},{"Start":"11:38.635 ","End":"11:42.385","Text":"and it contains 4 vectors,"},{"Start":"11:42.385 ","End":"11:44.440","Text":"each of them 4-dimensional."},{"Start":"11:44.440 ","End":"11:47.680","Text":"Notice that there\u0027s a 1 and 3 zeros,"},{"Start":"11:47.680 ","End":"11:51.535","Text":"and each of them with the 1 is in a different place each time"},{"Start":"11:51.535 ","End":"11:56.020","Text":"and this is the standard basis of R^4."},{"Start":"11:56.020 ","End":"12:00.325","Text":"I think it might have previously mentioned the standard basis of R^3,"},{"Start":"12:00.325 ","End":"12:01.930","Text":"where we have 3 vectors,"},{"Start":"12:01.930 ","End":"12:05.140","Text":"1,0,0, 0,1,0, 0,0,1."},{"Start":"12:05.140 ","End":"12:13.150","Text":"I think you can get the idea in general of what it would be for any n if I said R^6,"},{"Start":"12:13.150 ","End":"12:16.990","Text":"you\u0027d know how to give me 6 vectors with a 1 in"},{"Start":"12:16.990 ","End":"12:23.995","Text":"each place and 5 zeros. This is pretty general."},{"Start":"12:23.995 ","End":"12:28.960","Text":"Notice that it has 4 members."},{"Start":"12:28.960 ","End":"12:32.575","Text":"In other words, the basis has size 4,"},{"Start":"12:32.575 ","End":"12:40.885","Text":"and this 4 is this 4 from here. So that In general,"},{"Start":"12:40.885 ","End":"12:48.290","Text":"the standard basis of R^n will have n members."},{"Start":"12:48.290 ","End":"12:53.410","Text":"Number 2 is the m by n matrices,"},{"Start":"12:53.410 ","End":"12:57.520","Text":"and we\u0027re going to exemplify this with a 2 by 3."},{"Start":"12:57.520 ","End":"13:04.840","Text":"So the standard basis of this 2 by 3 matrices over the real numbers."},{"Start":"13:04.840 ","End":"13:08.485","Text":"The following set, I also call it E,"},{"Start":"13:08.485 ","End":"13:10.600","Text":"and it contains 6 members."},{"Start":"13:10.600 ","End":"13:13.600","Text":"I just wrote it in 2 rows of 3."},{"Start":"13:13.600 ","End":"13:16.390","Text":"Anyway, the 6 members,"},{"Start":"13:16.390 ","End":"13:18.985","Text":"each of them is a 2 by 3 matrix,"},{"Start":"13:18.985 ","End":"13:23.260","Text":"and they all have 5 zeros and a 1,"},{"Start":"13:23.260 ","End":"13:24.490","Text":"and a 1, as you see,"},{"Start":"13:24.490 ","End":"13:28.240","Text":"just keeps moving to different places."},{"Start":"13:28.240 ","End":"13:38.965","Text":"The 6 possibilities here and these are the 6 members of the standard basis."},{"Start":"13:38.965 ","End":"13:42.595","Text":"The 6 comes from 2 times 3"},{"Start":"13:42.595 ","End":"13:51.670","Text":"and in general, the set of m by n matrices,"},{"Start":"13:51.670 ","End":"13:59.860","Text":"we want to generalize from the 2 by 3 to generally m by n. I think it\u0027s"},{"Start":"13:59.860 ","End":"14:08.500","Text":"fairly clear that if we took the space of m by n matrices,"},{"Start":"14:08.500 ","End":"14:14.785","Text":"the standard basis would have m times n members,"},{"Start":"14:14.785 ","End":"14:18.895","Text":"and we need a third example for the polynomials."},{"Start":"14:18.895 ","End":"14:22.615","Text":"For the polynomials, I\u0027ll take the example"},{"Start":"14:22.615 ","End":"14:27.490","Text":"P_3 polynomials of degree up to and including 3,"},{"Start":"14:27.490 ","End":"14:32.230","Text":"and the standard basis is just all the monomials,"},{"Start":"14:32.230 ","End":"14:36.610","Text":"which are just powers of x up to and including 3."},{"Start":"14:36.610 ","End":"14:40.060","Text":"It\u0027s x^0, x^1,"},{"Start":"14:40.060 ","End":"14:41.455","Text":"x^2, x^3,"},{"Start":"14:41.455 ","End":"14:44.110","Text":"or 1, x, x squared, x cubed."},{"Start":"14:44.110 ","End":"14:48.145","Text":"Notice that although it says 3 here,"},{"Start":"14:48.145 ","End":"14:49.570","Text":"they\u0027re actually 1, 2,"},{"Start":"14:49.570 ","End":"14:51.355","Text":"3, 4 of them."},{"Start":"14:51.355 ","End":"14:56.290","Text":"The 4 being the 3 plus 1 because the exponents are from 1-3,"},{"Start":"14:56.290 ","End":"14:58.240","Text":"but there\u0027s also the exponent 0,"},{"Start":"14:58.240 ","End":"15:01.825","Text":"the constant, so you get 4 members."},{"Start":"15:01.825 ","End":"15:05.140","Text":"In general, if instead of 3 I had n,"},{"Start":"15:05.140 ","End":"15:12.775","Text":"then the basis for the polynomials P_n of r would have n plus 1 members."},{"Start":"15:12.775 ","End":"15:15.670","Text":"The standard basis would be 1, x,"},{"Start":"15:15.670 ","End":"15:21.805","Text":"x squared and so on, up to x^n."},{"Start":"15:21.805 ","End":"15:28.510","Text":"We really are leading up to the concept of the dimension of a vector space,"},{"Start":"15:28.510 ","End":"15:31.310","Text":"and that will be in just a moment."},{"Start":"15:31.470 ","End":"15:35.620","Text":"Here we are finally talking about dimension and just going to"},{"Start":"15:35.620 ","End":"15:41.725","Text":"give it a brief introduction of vector spaces."},{"Start":"15:41.725 ","End":"15:45.130","Text":"We\u0027re going to concentrate on the famous ones."},{"Start":"15:45.130 ","End":"15:48.055","Text":"Before I can really talk about dimension,"},{"Start":"15:48.055 ","End":"15:50.470","Text":"there\u0027s a theorem I have to mention,"},{"Start":"15:50.470 ","End":"15:51.850","Text":"I am not going to prove it or anything,"},{"Start":"15:51.850 ","End":"15:53.575","Text":"I\u0027m just going to state it"},{"Start":"15:53.575 ","End":"15:56.905","Text":"but if you have a vector space,"},{"Start":"15:56.905 ","End":"16:02.590","Text":"it could have many bases."},{"Start":"16:02.590 ","End":"16:06.280","Text":"That\u0027s the plural of basis is bases."},{"Start":"16:06.280 ","End":"16:08.935","Text":"If you take any 2 of them,"},{"Start":"16:08.935 ","End":"16:12.895","Text":"the theorem says that they\u0027re going to have the same number of elements."},{"Start":"16:12.895 ","End":"16:15.040","Text":"If I have, I don\u0027t know,"},{"Start":"16:15.040 ","End":"16:21.580","Text":"5 members of a basis of a vector space and then I\u0027d find another basis,"},{"Start":"16:21.580 ","End":"16:26.710","Text":"it will also have 5 members and In our case with the famous ones,"},{"Start":"16:26.710 ","End":"16:31.930","Text":"we can count the ones in the standard basis,"},{"Start":"16:31.930 ","End":"16:35.260","Text":"and that\u0027ll be the same number as in any basis."},{"Start":"16:35.260 ","End":"16:39.880","Text":"Anyway, this theorem paves way for the definition that if"},{"Start":"16:39.880 ","End":"16:43.915","Text":"any 2 bases have the same number of elements,"},{"Start":"16:43.915 ","End":"16:48.370","Text":"then the number of elements in a basis is well-defined."},{"Start":"16:48.370 ","End":"16:49.780","Text":"It\u0027s a definite number,"},{"Start":"16:49.780 ","End":"16:53.140","Text":"and that\u0027s what we call the dimension."},{"Start":"16:53.140 ","End":"16:57.670","Text":"In other words, you take a basis for the vector space and count the number of"},{"Start":"16:57.670 ","End":"17:03.400","Text":"elements in it and because the theorem doesn\u0027t matter which basis you take,"},{"Start":"17:03.400 ","End":"17:06.250","Text":"like I said, with famous vector spaces,"},{"Start":"17:06.250 ","End":"17:08.740","Text":"typically we will take the standard basis,"},{"Start":"17:08.740 ","End":"17:10.735","Text":"but you could take any basis."},{"Start":"17:10.735 ","End":"17:13.930","Text":"There\u0027s also a notation for dimension."},{"Start":"17:13.930 ","End":"17:17.215","Text":"We have a vector space V,"},{"Start":"17:17.215 ","End":"17:22.195","Text":"then the dimension we just use the first 3 letters of the word dimension,"},{"Start":"17:22.195 ","End":"17:30.470","Text":"dim of V, dimension of V. V could be R^n or anything."},{"Start":"17:30.690 ","End":"17:34.120","Text":"Now as I said, it\u0027s easy to find the dimension for"},{"Start":"17:34.120 ","End":"17:40.750","Text":"the famous vector spaces because we know that they have a standard basis,"},{"Start":"17:40.750 ","End":"17:44.815","Text":"and then we can use that to figure out the dimensions."},{"Start":"17:44.815 ","End":"17:49.585","Text":"I\u0027ll give the 3 main examples that we have."},{"Start":"17:49.585 ","End":"17:52.165","Text":"1 of them is R^n,"},{"Start":"17:52.165 ","End":"17:55.030","Text":"not an example, it\u0027s an infinite number of examples."},{"Start":"17:55.030 ","End":"17:56.575","Text":"N is a natural number,"},{"Start":"17:56.575 ","End":"18:02.170","Text":"but the dimension of R^n is n. Because earlier on,"},{"Start":"18:02.170 ","End":"18:07.510","Text":"if you just look back to the section on what we did on standard bases,"},{"Start":"18:07.510 ","End":"18:13.150","Text":"we found that the standard basis of R^n has n elements."},{"Start":"18:13.150 ","End":"18:15.550","Text":"Remember we did our 4, we had 0,"},{"Start":"18:15.550 ","End":"18:17.680","Text":"0, 1 and 0,"},{"Start":"18:17.680 ","End":"18:19.345","Text":"1, 0, 0, and all that."},{"Start":"18:19.345 ","End":"18:21.235","Text":"There were 4 of them anyway."},{"Start":"18:21.235 ","End":"18:28.000","Text":"We mentioned that this standard basis has n elements. Any basis are n elements,"},{"Start":"18:28.000 ","End":"18:31.750","Text":"so the dimension of R^n is n. This sort of make"},{"Start":"18:31.750 ","End":"18:35.605","Text":"some kind of sense that you would think N-dimensional space,"},{"Start":"18:35.605 ","End":"18:41.815","Text":"what we call R^n really does have dimension n according to this definition"},{"Start":"18:41.815 ","End":"18:46.855","Text":"and in the case of the space of m by n matrices,"},{"Start":"18:46.855 ","End":"18:50.020","Text":"we discovered that the standard basis,"},{"Start":"18:50.020 ","End":"18:53.500","Text":"if you look back, had m, n elements."},{"Start":"18:53.500 ","End":"18:58.165","Text":"If the standard basis has this and any basis has this number of elements,"},{"Start":"18:58.165 ","End":"19:03.055","Text":"so the dimension of this space of matrices is m times"},{"Start":"19:03.055 ","End":"19:09.325","Text":"n. In the case of polynomials up to including degree n,"},{"Start":"19:09.325 ","End":"19:15.190","Text":"we saw that the standard basis has n plus 1 members."},{"Start":"19:15.190 ","End":"19:23.305","Text":"Any basis will have n plus 1 members and that it will be the dimension of this space of"},{"Start":"19:23.305 ","End":"19:25.660","Text":"polynomials up to and including degree"},{"Start":"19:25.660 ","End":"19:31.690","Text":"n. That\u0027s really all I want to say at this point about dimension,"},{"Start":"19:31.690 ","End":"19:36.740","Text":"and we are done here."}],"ID":26723},{"Watched":false,"Name":"Exercise 1","Duration":"3m 54s","ChapterTopicVideoID":25914,"CourseChapterTopicPlaylistID":246317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:07.575","Text":"This exercise deals with the space M_2 by 2 over R,"},{"Start":"00:07.575 ","End":"00:11.340","Text":"which is the set of 2 by 2 matrices over R. When it\u0027s"},{"Start":"00:11.340 ","End":"00:15.600","Text":"a square matrix we sometimes just write it as M_2."},{"Start":"00:15.600 ","End":"00:18.720","Text":"There are 3 parts,"},{"Start":"00:18.720 ","End":"00:22.110","Text":"and each part we are given a set of matrices and we have to"},{"Start":"00:22.110 ","End":"00:27.060","Text":"decide if it\u0027s a basis or not for this space."},{"Start":"00:27.060 ","End":"00:36.480","Text":"We\u0027ll start with Part 1 and notice there are 3 members of this set,"},{"Start":"00:36.480 ","End":"00:38.685","Text":"and there\u0027s a theorem"},{"Start":"00:38.685 ","End":"00:48.740","Text":"that in general if we have n by n matrices over R,"},{"Start":"00:48.740 ","End":"00:52.615","Text":"it can\u0027t contain less than n squared matrices."},{"Start":"00:52.615 ","End":"00:55.880","Text":"More specifically, if you have less than n squared,"},{"Start":"00:55.880 ","End":"00:57.200","Text":"then they can\u0027t span,"},{"Start":"00:57.200 ","End":"00:59.810","Text":"if they can\u0027t be a basis."},{"Start":"00:59.810 ","End":"01:04.715","Text":"In our case, n squared is 2 squared which is 4,"},{"Start":"01:04.715 ","End":"01:13.245","Text":"and here we have 3 elements which is less than 4 and so they can\u0027t be a basis."},{"Start":"01:13.245 ","End":"01:16.700","Text":"Let\u0027s move on to the next one and this time,"},{"Start":"01:16.700 ","End":"01:19.010","Text":"if you count to see there\u0027s 1, 2, 3, 4,"},{"Start":"01:19.010 ","End":"01:26.360","Text":"5 a basis for this space in general,"},{"Start":"01:26.360 ","End":"01:34.355","Text":"n by n matrices also can\u0027t contain more than n squared matrices because in that case,"},{"Start":"01:34.355 ","End":"01:40.210","Text":"they\u0027re not going to be linearly independent and so there won\u0027t be a basis."},{"Start":"01:40.400 ","End":"01:48.240","Text":"In our case n is 2 and we have more than 2 squared,"},{"Start":"01:48.240 ","End":"01:52.950","Text":"it\u0027s 5 so we don\u0027t have a basis."},{"Start":"01:52.950 ","End":"01:56.450","Text":"In the third part we actually have the right number,"},{"Start":"01:56.450 ","End":"02:03.050","Text":"we have 2 squared which is 4 and there\u0027s a theorem that if you have the right number,"},{"Start":"02:03.050 ","End":"02:05.850","Text":"in this case n"},{"Start":"02:05.850 ","End":"02:09.590","Text":"squared the theorem helps"},{"Start":"02:09.590 ","End":"02:13.925","Text":"us by saying that we just have to prove the linearly independent path."},{"Start":"02:13.925 ","End":"02:19.830","Text":"Normally a basis has to span and be linearly independent,"},{"Start":"02:19.830 ","End":"02:24.790","Text":"and this theorem says, you can just show the any linear independent part."},{"Start":"02:25.270 ","End":"02:28.250","Text":"How do we show that?"},{"Start":"02:28.250 ","End":"02:31.505","Text":"You might recall we\u0027ve previously done this,"},{"Start":"02:31.505 ","End":"02:39.955","Text":"so we can flatten out the matrices by taking them in the snake order this,"},{"Start":"02:39.955 ","End":"02:42.390","Text":"so this 1 would be like 0, 1, 1,"},{"Start":"02:42.390 ","End":"02:45.700","Text":"1 then 0, 0, 1, 1."},{"Start":"02:53.360 ","End":"02:59.160","Text":"But if we take them as in the snake form,"},{"Start":"02:59.160 ","End":"03:01.560","Text":"let\u0027s take 1 of them and go back."},{"Start":"03:01.560 ","End":"03:05.130","Text":"The first 1 would be 1, 1, 1, 1,"},{"Start":"03:05.130 ","End":"03:10.590","Text":"and that\u0027s here 1, 1, 1,"},{"Start":"03:10.590 ","End":"03:13.040","Text":"1 and similarly for all the rest,"},{"Start":"03:13.040 ","End":"03:17.015","Text":"and we have to see if we do row operations,"},{"Start":"03:17.015 ","End":"03:20.995","Text":"whether we get a 0 or not."},{"Start":"03:20.995 ","End":"03:24.710","Text":"Now we happen to be in luck we don\u0027t have to bring it to"},{"Start":"03:24.710 ","End":"03:28.865","Text":"row echelon form because it already is in row echelon form."},{"Start":"03:28.865 ","End":"03:36.690","Text":"As you can see all zeros below the diagonal will certainly make it echelon,"},{"Start":"03:36.690 ","End":"03:40.540","Text":"it\u0027s even upper triangular."},{"Start":"03:40.880 ","End":"03:44.730","Text":"There are no rows of zeros,"},{"Start":"03:44.730 ","End":"03:49.095","Text":"and that means it\u0027s linearly independent and so,"},{"Start":"03:49.095 ","End":"03:55.000","Text":"yes, the 4 vectors form a basis and we\u0027re done."}],"ID":26724},{"Watched":false,"Name":"Exercise 2","Duration":"5m 13s","ChapterTopicVideoID":25912,"CourseChapterTopicPlaylistID":246317,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.565","Text":"This exercise deals with the space P2 over R,"},{"Start":"00:05.565 ","End":"00:13.825","Text":"meaning the space of polynomials with real coefficients of degree up to 2."},{"Start":"00:13.825 ","End":"00:17.880","Text":"Here we have 3 sets of polynomials."},{"Start":"00:17.880 ","End":"00:23.140","Text":"We have to decide for each of them whether it\u0027s a basis or not."},{"Start":"00:23.630 ","End":"00:28.530","Text":"What\u0027s going to help us are some theorems that we know about"},{"Start":"00:28.530 ","End":"00:33.885","Text":"how many elements there should be in a basis and so on."},{"Start":"00:33.885 ","End":"00:38.985","Text":"Now, there\u0027s a theorem that a basis for P_n,"},{"Start":"00:38.985 ","End":"00:40.380","Text":"and in general,"},{"Start":"00:40.380 ","End":"00:42.190","Text":"or in our case n is 2,"},{"Start":"00:42.190 ","End":"00:46.385","Text":"can\u0027t contain less than n plus 1 polynomials."},{"Start":"00:46.385 ","End":"00:48.320","Text":"If there\u0027s too few,"},{"Start":"00:48.320 ","End":"00:52.120","Text":"they can\u0027t span and so they can\u0027t be a basis."},{"Start":"00:52.120 ","End":"00:58.300","Text":"In our case, 2 plus 1 is 3."},{"Start":"00:58.760 ","End":"01:05.700","Text":"We have only 2 here and 2 is less than 3,"},{"Start":"01:05.700 ","End":"01:07.980","Text":"and so these 2 don\u0027t span."},{"Start":"01:07.980 ","End":"01:12.580","Text":"They\u0027re not a basis and the answer to 1 is no, not a basis."},{"Start":"01:12.580 ","End":"01:14.570","Text":"In the second set,"},{"Start":"01:14.570 ","End":"01:15.800","Text":"we actually have too many,"},{"Start":"01:15.800 ","End":"01:17.000","Text":"there\u0027s 4 of them,"},{"Start":"01:17.000 ","End":"01:19.160","Text":"which is bigger than 3,"},{"Start":"01:19.160 ","End":"01:23.630","Text":"and there\u0027s another theorem that says that if we have more than n plus 1,"},{"Start":"01:23.630 ","End":"01:25.460","Text":"in our case, n is 2,"},{"Start":"01:25.460 ","End":"01:27.290","Text":"so n plus 1 is 3,"},{"Start":"01:27.290 ","End":"01:30.660","Text":"and we have 4 of them which is bigger than 3,"},{"Start":"01:30.660 ","End":"01:32.310","Text":"again, it can\u0027t be a basis."},{"Start":"01:32.310 ","End":"01:34.235","Text":"But this time for a different reason,"},{"Start":"01:34.235 ","End":"01:36.725","Text":"they won\u0027t be linearly independent."},{"Start":"01:36.725 ","End":"01:38.510","Text":"If you have too few, they don\u0027t span,"},{"Start":"01:38.510 ","End":"01:41.345","Text":"if you have too many, they are not linearly independent."},{"Start":"01:41.345 ","End":"01:46.100","Text":"Although in both cases we could have used a shortcut and say that a basis has to"},{"Start":"01:46.100 ","End":"01:52.075","Text":"contain exactly n plus 1 and ruled them out in 1 go."},{"Start":"01:52.075 ","End":"01:54.120","Text":"Let\u0027s see what happens in the third case."},{"Start":"01:54.120 ","End":"01:56.510","Text":"It appears to have the correct number,"},{"Start":"01:56.510 ","End":"01:58.490","Text":"2 plus 1 is 3,"},{"Start":"01:58.490 ","End":"02:00.740","Text":"and we do have 1, 2,"},{"Start":"02:00.740 ","End":"02:05.240","Text":"3 members in this."},{"Start":"02:05.240 ","End":"02:09.995","Text":"I just forgot the last sentence in the previous part."},{"Start":"02:09.995 ","End":"02:12.844","Text":"Now, in Part 3,"},{"Start":"02:12.844 ","End":"02:14.990","Text":"we\u0027re going to use a theorem to help us,"},{"Start":"02:14.990 ","End":"02:17.200","Text":"but let me just give you some background."},{"Start":"02:17.200 ","End":"02:24.350","Text":"Normally, if I have a set of polynomials here to show that it\u0027s a basis,"},{"Start":"02:24.350 ","End":"02:28.565","Text":"I\u0027ll have to show that these, in this case,"},{"Start":"02:28.565 ","End":"02:36.220","Text":"both are linearly independent and that they span the vector space P_n."},{"Start":"02:36.220 ","End":"02:39.545","Text":"Sorry, this is not a definition."},{"Start":"02:39.545 ","End":"02:44.150","Text":"Sorry, this is actually a theorem."},{"Start":"02:44.150 ","End":"02:49.600","Text":"What the theorem does is it makes life easier."},{"Start":"02:49.600 ","End":"02:54.790","Text":"Normally, we would have to show that these vectors or"},{"Start":"02:54.790 ","End":"03:00.450","Text":"polynomials span and are linearly independent, like I said."},{"Start":"03:00.450 ","End":"03:05.110","Text":"The theorem says that we just have to check the linearly independent."},{"Start":"03:05.110 ","End":"03:10.790","Text":"We can save the time of showing the part about spanning P_n."},{"Start":"03:10.790 ","End":"03:14.340","Text":"In our case, we do have, n is 2,"},{"Start":"03:14.340 ","End":"03:16.860","Text":"we have 3 polynomials,"},{"Start":"03:16.860 ","End":"03:19.960","Text":"so let\u0027s just see."},{"Start":"03:20.330 ","End":"03:24.195","Text":"How do we show that our polynomial are linearly independent?"},{"Start":"03:24.195 ","End":"03:28.590","Text":"I\u0027ve lost them. Let\u0027s scroll back. Here they are."},{"Start":"03:28.590 ","End":"03:34.520","Text":"What we do is we give the vector representation of these."},{"Start":"03:34.820 ","End":"03:37.890","Text":"I\u0027ll just do 1 of them, let\u0027s say the middle 1."},{"Start":"03:37.890 ","End":"03:43.030","Text":"We can write it as 4, 5, 6."},{"Start":"03:43.030 ","End":"03:47.365","Text":"To do this, you have to make sure that they\u0027re all in the right order."},{"Start":"03:47.365 ","End":"03:49.300","Text":"We have to agree on an order,"},{"Start":"03:49.300 ","End":"03:51.655","Text":"let\u0027s say increasing order of powers,"},{"Start":"03:51.655 ","End":"03:55.380","Text":"and put them in order and make sure that there\u0027s none missing."},{"Start":"03:55.380 ","End":"03:56.500","Text":"If there\u0027s a missing term,"},{"Start":"03:56.500 ","End":"03:57.790","Text":"you put a 0."},{"Start":"03:57.790 ","End":"04:02.340","Text":"That way we get 3 vectors, 1 for each."},{"Start":"04:02.340 ","End":"04:04.020","Text":"Here we have 1, 2, 3, then 4, 5,"},{"Start":"04:04.020 ","End":"04:06.045","Text":"6, then 7, 8, 10."},{"Start":"04:06.045 ","End":"04:13.815","Text":"I\u0027m going to take those numbers and put them in a matrix. Here we are."},{"Start":"04:13.815 ","End":"04:15.660","Text":"I remembered the numbers, it\u0027s 1,2,3,"},{"Start":"04:15.660 ","End":"04:16.830","Text":"4, 5, 6, 7, 8,"},{"Start":"04:16.830 ","End":"04:17.970","Text":"and then not 9,"},{"Start":"04:17.970 ","End":"04:21.045","Text":"but 10. Let\u0027s see."},{"Start":"04:21.045 ","End":"04:27.910","Text":"If we bring this to row echelon form will we get a row of 0\u0027s or not?"},{"Start":"04:28.250 ","End":"04:31.965","Text":"Let\u0027s just do the row operations."},{"Start":"04:31.965 ","End":"04:36.795","Text":"Subtract 4 times this from this to get a 0 here."},{"Start":"04:36.795 ","End":"04:39.740","Text":"Subtract 7 times this row from this row,"},{"Start":"04:39.740 ","End":"04:42.315","Text":"and that gives us this 0."},{"Start":"04:42.315 ","End":"04:44.520","Text":"Now, we need a 0 here,"},{"Start":"04:44.520 ","End":"04:49.615","Text":"so let\u0027s subtract twice the second row from the last row."},{"Start":"04:49.615 ","End":"04:51.690","Text":"That gives us this,"},{"Start":"04:51.690 ","End":"04:56.225","Text":"and this is in row echelon form."},{"Start":"04:56.225 ","End":"04:59.030","Text":"But there are no 0 rows,"},{"Start":"04:59.030 ","End":"05:06.410","Text":"which means that these 3 vectors or the 3 polynomials are linearly independent."},{"Start":"05:06.410 ","End":"05:08.480","Text":"By that theorem, they\u0027re a basis."},{"Start":"05:08.480 ","End":"05:11.149","Text":"We don\u0027t have to show that they span."},{"Start":"05:11.149 ","End":"05:14.160","Text":"We are done."}],"ID":26722}],"Thumbnail":null,"ID":246317},{"Name":"Solution Space of Homogenous SLE","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"14m 20s","ChapterTopicVideoID":25917,"CourseChapterTopicPlaylistID":246318,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/25917.jpeg","UploadDate":"2021-06-28T11:54:36.6770000","DurationForVideoObject":"PT14M20S","Description":null,"MetaTitle":"Exercise 1 - Solution Space of Homogenous SLE: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Solution Space of Homogenous SLE practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/linear-algebra/general-vector-spaces/solution-space-of-homogenous-sle/vid26727","VideoComments":[],"Subtitles":[{"Start":"00:00.090 ","End":"00:03.640","Text":"In this exercise, which is fairly lengthy,"},{"Start":"00:03.640 ","End":"00:06.355","Text":"it doesn\u0027t even fit in on the screen, that\u0027s okay,"},{"Start":"00:06.355 ","End":"00:10.690","Text":"we have 3 systems of linear equations, 1, 2,"},{"Start":"00:10.690 ","End":"00:14.770","Text":"and 3, but they\u0027re not just any old systems, they are homogeneous."},{"Start":"00:14.770 ","End":"00:19.655","Text":"I perhaps should mention that it\u0027s important enough."},{"Start":"00:19.655 ","End":"00:24.040","Text":"Now, 1 of the things about homogeneous linear equations is"},{"Start":"00:24.040 ","End":"00:28.060","Text":"that the set of solutions is a vector subspace."},{"Start":"00:28.060 ","End":"00:33.100","Text":"In this case, we are in R4 because we have x, y, z,"},{"Start":"00:33.100 ","End":"00:38.520","Text":"and w. We always have 0 as a solution."},{"Start":"00:38.520 ","End":"00:39.750","Text":"If I put x, y, z,"},{"Start":"00:39.750 ","End":"00:40.980","Text":"and w all 0,"},{"Start":"00:40.980 ","End":"00:42.330","Text":"certainly that will work."},{"Start":"00:42.330 ","End":"00:45.320","Text":"But in general, we get a subspace."},{"Start":"00:45.320 ","End":"00:53.765","Text":"Now, the space of solutions here is going to be called W. Here we\u0027ll call it U,"},{"Start":"00:53.765 ","End":"01:02.780","Text":"and here we\u0027ll call it V. We want to find a basis for each of these."},{"Start":"01:02.780 ","End":"01:03.860","Text":"That\u0027s the first part."},{"Start":"01:03.860 ","End":"01:05.750","Text":"Find the basis for W,"},{"Start":"01:05.750 ","End":"01:08.810","Text":"for U, and for V and the dimension."},{"Start":"01:08.810 ","End":"01:11.770","Text":"Once you have a basis the dimension is easy."},{"Start":"01:11.770 ","End":"01:21.320","Text":"Then after that, we want to find a basis for the sum of these 2 sub-spaces."},{"Start":"01:21.320 ","End":"01:25.400","Text":"Hope you remember what a sum of 2 spaces is."},{"Start":"01:25.400 ","End":"01:30.870","Text":"If not, maybe I\u0027ll remind you what U plus V is,"},{"Start":"01:30.870 ","End":"01:34.850","Text":"is the set of all little u plus little v,"},{"Start":"01:34.850 ","End":"01:40.185","Text":"where u comes from u and v comes from v,"},{"Start":"01:40.185 ","End":"01:42.285","Text":"in case you\u0027ve forgotten."},{"Start":"01:42.285 ","End":"01:48.050","Text":"We also want the dimension of the intersection."},{"Start":"01:48.050 ","End":"01:53.825","Text":"The sum of 2 vector spaces is a vector space and so is the intersection."},{"Start":"01:53.825 ","End":"01:58.190","Text":"Then finally, we\u0027re going to find a basis"},{"Start":"01:58.190 ","End":"02:02.660","Text":"for this intersection of U and V. That\u0027s a lot of work."},{"Start":"02:02.660 ","End":"02:06.295","Text":"Let\u0027s get started on a fresh page."},{"Start":"02:06.295 ","End":"02:08.630","Text":"In a Part 1,"},{"Start":"02:08.630 ","End":"02:12.230","Text":"we want to find the solution space of this."},{"Start":"02:12.230 ","End":"02:15.245","Text":"We\u0027ll do it with matrices."},{"Start":"02:15.245 ","End":"02:17.660","Text":"This is the corresponding matrix."},{"Start":"02:17.660 ","End":"02:20.480","Text":"Remember with homogeneous, we don\u0027t need the augmented matrix."},{"Start":"02:20.480 ","End":"02:23.350","Text":"We don\u0027t need a column of 0s."},{"Start":"02:23.350 ","End":"02:25.960","Text":"As to row operations,"},{"Start":"02:25.960 ","End":"02:28.925","Text":"lets make 0 here and here."},{"Start":"02:28.925 ","End":"02:31.880","Text":"We subtract 3 times this row from"},{"Start":"02:31.880 ","End":"02:35.900","Text":"the second row and we add 5 times this row to the last row,"},{"Start":"02:35.900 ","End":"02:38.060","Text":"and then we\u0027ll get this,"},{"Start":"02:38.060 ","End":"02:39.560","Text":"still not echelon form,"},{"Start":"02:39.560 ","End":"02:43.415","Text":"I need a 0 where the 8 is."},{"Start":"02:43.415 ","End":"02:48.170","Text":"We just have to add twice this row to this row, we get a 0 here."},{"Start":"02:48.170 ","End":"02:52.270","Text":"As a matter of fact, the whole row comes out to be 0."},{"Start":"02:52.270 ","End":"02:58.955","Text":"Now what I want to do is go back from the matrix form to the equation form."},{"Start":"02:58.955 ","End":"03:04.545","Text":"Here they are; 2 equations and 4 unknowns."},{"Start":"03:04.545 ","End":"03:08.280","Text":"Z and w are the free variables,"},{"Start":"03:08.280 ","End":"03:14.810","Text":"and then x and y are called pivot leading variables."},{"Start":"03:14.810 ","End":"03:16.420","Text":"Anyway, they\u0027re constrained."},{"Start":"03:16.420 ","End":"03:22.580","Text":"They depend on z and w. We let z and w be a parameter. No, change of mind."},{"Start":"03:22.580 ","End":"03:24.050","Text":"We don\u0027t want the general solution,"},{"Start":"03:24.050 ","End":"03:25.205","Text":"we just want the basis."},{"Start":"03:25.205 ","End":"03:27.035","Text":"We\u0027ll use the method of the,"},{"Start":"03:27.035 ","End":"03:29.990","Text":"I called it the wandering 1s."},{"Start":"03:29.990 ","End":"03:35.695","Text":"Each time we let 1 of the free variables be 1 and the others 0."},{"Start":"03:35.695 ","End":"03:37.170","Text":"First-time around we let,"},{"Start":"03:37.170 ","End":"03:40.080","Text":"w be 1 and z be 0."},{"Start":"03:40.080 ","End":"03:42.410","Text":"Once I have w and z,"},{"Start":"03:42.410 ","End":"03:44.930","Text":"everything else follows from back substitution."},{"Start":"03:44.930 ","End":"03:46.460","Text":"From z and w here,"},{"Start":"03:46.460 ","End":"03:48.535","Text":"we get y here."},{"Start":"03:48.535 ","End":"03:50.160","Text":"Then once we have y, z,"},{"Start":"03:50.160 ","End":"03:52.655","Text":"and w, we plugin here and we get x."},{"Start":"03:52.655 ","End":"03:57.180","Text":"Anyway, I\u0027ll leave you to pause and check that these are the calculations."},{"Start":"03:57.950 ","End":"04:04.310","Text":"Remember the solution space we called it w. The basis would be these,"},{"Start":"04:04.310 ","End":"04:06.050","Text":"but you got to get them in the right order."},{"Start":"04:06.050 ","End":"04:09.120","Text":"We want x, y, z, w."},{"Start":"04:09.120 ","End":"04:24.420","Text":"This row belongs"},{"Start":"04:24.420 ","End":"04:28.230","Text":"here and this 1 belongs here."},{"Start":"04:28.230 ","End":"04:31.715","Text":"That\u0027s right. You can see a 2.5 here from the y,"},{"Start":"04:31.715 ","End":"04:34.610","Text":"yeah, that\u0027s the right way round."},{"Start":"04:34.850 ","End":"04:41.920","Text":"The other dimension we have to do is count how many elements in the basis and that is 2."},{"Start":"04:41.920 ","End":"04:46.655","Text":"That\u0027s the first of 3."},{"Start":"04:46.655 ","End":"04:54.065","Text":"The second homogeneous system was this and here\u0027s the matrix."},{"Start":"04:54.065 ","End":"05:01.280","Text":"We want to subtract the first row from both the second and from the third."},{"Start":"05:01.280 ","End":"05:03.110","Text":"Then we get this."},{"Start":"05:03.110 ","End":"05:04.700","Text":"It\u0027s still not in echelon form."},{"Start":"05:04.700 ","End":"05:06.820","Text":"We need a 0 here."},{"Start":"05:06.820 ","End":"05:13.055","Text":"We subtract twice the second row from the third row, and we get this."},{"Start":"05:13.055 ","End":"05:16.295","Text":"Once again, we have a row of 0s."},{"Start":"05:16.295 ","End":"05:25.655","Text":"This means that once again we have just 2 equations and 4 unknowns."},{"Start":"05:25.655 ","End":"05:32.215","Text":"Again, we\u0027re going to use this method of the wandering 1s to each time let something,"},{"Start":"05:32.215 ","End":"05:36.315","Text":"1 of the free variables be 1 and the other 0."},{"Start":"05:36.315 ","End":"05:40.320","Text":"In this case, we have z and w. Just like before,"},{"Start":"05:40.320 ","End":"05:42.760","Text":"1 time I let w be 1 z is 0,"},{"Start":"05:42.760 ","End":"05:45.775","Text":"and then the other way around w is 0, z is 1."},{"Start":"05:45.775 ","End":"05:48.760","Text":"I first compute the y from the second equation,"},{"Start":"05:48.760 ","End":"05:51.990","Text":"and then I have all 3 of them,"},{"Start":"05:51.990 ","End":"05:54.570","Text":"y, z, and w to compute x from."},{"Start":"05:54.570 ","End":"05:55.870","Text":"This is what it comes out."},{"Start":"05:55.870 ","End":"05:58.345","Text":"Now I just have to get them in the proper order,"},{"Start":"05:58.345 ","End":"06:01.255","Text":"which gives us the basis for u."},{"Start":"06:01.255 ","End":"06:04.180","Text":"We want 1, 2, 0, 1."},{"Start":"06:04.180 ","End":"06:06.835","Text":"I guess, that 1 is this 1,"},{"Start":"06:06.835 ","End":"06:09.610","Text":"and this 1 is that 1,"},{"Start":"06:09.610 ","End":"06:12.190","Text":"and that\u0027s the basis for u."},{"Start":"06:12.190 ","End":"06:15.215","Text":"Again dimension is 2."},{"Start":"06:15.215 ","End":"06:19.230","Text":"Now, the third SLE homogeneous,"},{"Start":"06:19.230 ","End":"06:21.540","Text":"here is its matrix."},{"Start":"06:21.540 ","End":"06:23.260","Text":"Like I said, we don\u0027t need the 0s with"},{"Start":"06:23.260 ","End":"06:26.830","Text":"homogeneous and we want to bring it to echelon form."},{"Start":"06:26.830 ","End":"06:30.995","Text":"So I would subtract twice the first row from the second row."},{"Start":"06:30.995 ","End":"06:35.065","Text":"If we do that, what we end up with is this."},{"Start":"06:35.065 ","End":"06:37.375","Text":"There\u0027s a 0 row,"},{"Start":"06:37.375 ","End":"06:43.190","Text":"so there\u0027s only going to be 1 equation and 4 unknowns."},{"Start":"06:43.190 ","End":"06:45.985","Text":"This is the equation from the top row."},{"Start":"06:45.985 ","End":"06:48.790","Text":"Another 3 free variables, y, z,"},{"Start":"06:48.790 ","End":"06:53.320","Text":"and w. We\u0027re going to have 3 members of the basis."},{"Start":"06:53.320 ","End":"06:58.125","Text":"Each time we\u0027ll let 1 of these be 1 and the other 0."},{"Start":"06:58.125 ","End":"07:01.125","Text":"Here we are. I\u0027ve colored the 1."},{"Start":"07:01.125 ","End":"07:02.250","Text":"You have a 1, 0,"},{"Start":"07:02.250 ","End":"07:03.420","Text":"0, a 0, 1,"},{"Start":"07:03.420 ","End":"07:06.600","Text":"0, 0,0, 1 for w, z, and y."},{"Start":"07:06.600 ","End":"07:08.240","Text":"Once we have w, z and y,"},{"Start":"07:08.240 ","End":"07:13.560","Text":"we can plug in here and get the x and then gather these together."},{"Start":"07:13.560 ","End":"07:15.610","Text":"That gives us a basis."},{"Start":"07:15.610 ","End":"07:18.400","Text":"V was the name of the solution subspace for"},{"Start":"07:18.400 ","End":"07:23.095","Text":"this and so if we put them in the right order,"},{"Start":"07:23.095 ","End":"07:24.880","Text":"let\u0027s see minus 1, 0,"},{"Start":"07:24.880 ","End":"07:27.100","Text":"0, 1, that would be this 1."},{"Start":"07:27.100 ","End":"07:32.320","Text":"I guess the middle 1 would be this 1 and finally 1,"},{"Start":"07:32.320 ","End":"07:35.200","Text":"1, 0, 0, the last 1."},{"Start":"07:35.200 ","End":"07:37.870","Text":"The dimension is time is of course 3,"},{"Start":"07:37.870 ","End":"07:39.950","Text":"as we can count."},{"Start":"07:40.050 ","End":"07:45.115","Text":"Now on to part b, b1."},{"Start":"07:45.115 ","End":"07:52.240","Text":"Here again, the basis for u and v. If we want"},{"Start":"07:52.240 ","End":"07:58.690","Text":"to take the sum of 2 vector spaces,"},{"Start":"07:58.690 ","End":"08:03.400","Text":"if I want to compute a basis for u plus v,"},{"Start":"08:03.400 ","End":"08:08.860","Text":"all we do is throw in together the union of"},{"Start":"08:08.860 ","End":"08:18.265","Text":"these 2 bases and then just do row operations to cut it down to size."},{"Start":"08:18.265 ","End":"08:22.600","Text":"I\u0027ll show you. I put these 2 plus 3,"},{"Start":"08:22.600 ","End":"08:27.100","Text":"5 vectors in here for some reason I did it from right to left."},{"Start":"08:27.100 ","End":"08:30.230","Text":"This, then this, and then this."},{"Start":"08:33.060 ","End":"08:36.640","Text":"I even put the v\u0027s first."},{"Start":"08:36.640 ","End":"08:40.015","Text":"I guess. This 1 is here."},{"Start":"08:40.015 ","End":"08:41.275","Text":"This 1 is here."},{"Start":"08:41.275 ","End":"08:49.915","Text":"I don\u0027t know why I did v. These 3 from the v. These from u."},{"Start":"08:49.915 ","End":"08:52.270","Text":"That\u0027s not a u, that\u0027s the union."},{"Start":"08:52.270 ","End":"08:54.295","Text":"Maybe I\u0027ll erase it."},{"Start":"08:54.295 ","End":"08:56.185","Text":"I just get the order."},{"Start":"08:56.185 ","End":"08:58.540","Text":"Doesn\u0027t really matter that it first this,"},{"Start":"08:58.540 ","End":"09:01.090","Text":"this, this, that\u0027s how the order is."},{"Start":"09:01.090 ","End":"09:09.730","Text":"Now, something you know so well to do is to bring to row echelon form."},{"Start":"09:09.730 ","End":"09:12.400","Text":"I would just skim through that quickly."},{"Start":"09:12.400 ","End":"09:18.835","Text":"Add multiples of the top row to all the rest to make these 0."},{"Start":"09:18.835 ","End":"09:20.860","Text":"I\u0027ll spare you the details."},{"Start":"09:20.860 ","End":"09:23.125","Text":"This is what we get."},{"Start":"09:23.125 ","End":"09:27.355","Text":"Next. You want zeros in these places,"},{"Start":"09:27.355 ","End":"09:32.590","Text":"and this is what we get after we add multiples of the second row."},{"Start":"09:32.590 ","End":"09:35.170","Text":"If I could just subtract the second row from each of the third,"},{"Start":"09:35.170 ","End":"09:39.045","Text":"fourth and fifth and get this."},{"Start":"09:39.045 ","End":"09:43.020","Text":"This is the second."},{"Start":"09:43.020 ","End":"09:47.955","Text":"Can finally be brought to row echelon form by subtracting"},{"Start":"09:47.955 ","End":"09:52.910","Text":"the 3rd row"},{"Start":"09:52.910 ","End":"09:58.630","Text":"from the 4rth row."},{"Start":"09:58.630 ","End":"10:08.035","Text":"Anyway, we do have the echelon form and 2 rows of"},{"Start":"10:08.035 ","End":"10:14.050","Text":"zeros and the basis for the sum of"},{"Start":"10:14.050 ","End":"10:21.860","Text":"the vector subspaces u plus v is just what we got here without the zeros."},{"Start":"10:22.590 ","End":"10:27.040","Text":"Again, I don\u0027t know why I\u0027m doing it in the wrong order."},{"Start":"10:27.040 ","End":"10:28.540","Text":"This 1 1, 1,"},{"Start":"10:28.540 ","End":"10:30.730","Text":"0, 0 is here."},{"Start":"10:30.730 ","End":"10:33.025","Text":"Guess I\u0027m going in this direction."},{"Start":"10:33.025 ","End":"10:36.160","Text":"Then 0, 1, 1, 0 here,"},{"Start":"10:36.160 ","End":"10:37.900","Text":"and then 0, 0, minus 1,"},{"Start":"10:37.900 ","End":"10:41.395","Text":"1 is here, all accounted for."},{"Start":"10:41.395 ","End":"10:47.560","Text":"The dimension is 3 by counting the basis."},{"Start":"10:47.560 ","End":"10:52.270","Text":"We\u0027re still in the b, b2,"},{"Start":"10:52.270 ","End":"10:59.365","Text":"which are asked for- I just exposed at all and no point doing your row at a time."},{"Start":"10:59.365 ","End":"11:06.775","Text":"After the dimension of u plus v, the subspace,"},{"Start":"11:06.775 ","End":"11:14.845","Text":"there\u0027s a formula that this is equal to the dimension of each 1 of them added,"},{"Start":"11:14.845 ","End":"11:19.105","Text":"and then we subtract the dimension of the intersection."},{"Start":"11:19.105 ","End":"11:21.235","Text":"Now since we know this,"},{"Start":"11:21.235 ","End":"11:24.565","Text":"this and this, we can use it to find what this is."},{"Start":"11:24.565 ","End":"11:26.680","Text":"This is 3, this was 2."},{"Start":"11:26.680 ","End":"11:31.795","Text":"This was 3 from the previous exercises."},{"Start":"11:31.795 ","End":"11:35.695","Text":"We have 3 equals 2 plus 3 minus something."},{"Start":"11:35.695 ","End":"11:37.765","Text":"That something is 2."},{"Start":"11:37.765 ","End":"11:45.040","Text":"That\u0027s the answer for the dimension of the intersection subspace."},{"Start":"11:45.040 ","End":"11:51.760","Text":"Now, for part c, we want the intersection of the solution sets."},{"Start":"11:51.760 ","End":"11:57.730","Text":"We want to u intersection with v. And that was the 2 systems of equations."},{"Start":"11:57.730 ","End":"11:59.560","Text":"We call them 2 and 3."},{"Start":"11:59.560 ","End":"12:03.580","Text":"So what we do is if you want something to belong to both,"},{"Start":"12:03.580 ","End":"12:09.620","Text":"it has to satisfy the equations of 2 and of 3."},{"Start":"12:09.810 ","End":"12:18.025","Text":"Here we are. The first 3 equations belong to u,"},{"Start":"12:18.025 ","End":"12:23.965","Text":"that\u0027s part 2, and the second 2 belong to the system number 3."},{"Start":"12:23.965 ","End":"12:26.935","Text":"We take them all together and that will give us the end,"},{"Start":"12:26.935 ","End":"12:29.660","Text":"the intersection of both."},{"Start":"12:30.240 ","End":"12:33.940","Text":"Now, here\u0027s the matrix corresponding to this."},{"Start":"12:33.940 ","End":"12:35.860","Text":"We just take the coefficients,"},{"Start":"12:35.860 ","End":"12:39.805","Text":"and we\u0027re going to do row operations on this."},{"Start":"12:39.805 ","End":"12:44.335","Text":"First step was to get all zeros here."},{"Start":"12:44.335 ","End":"12:51.475","Text":"Just subtracted this top row from these 3 and twice the top row from this,"},{"Start":"12:51.475 ","End":"12:53.755","Text":"and we get this."},{"Start":"12:53.755 ","End":"13:03.055","Text":"Then, after we subtract twice this second row from the third row, we get this."},{"Start":"13:03.055 ","End":"13:09.985","Text":"Notice that there\u0027s only 2 rows and the rest of it is 0 s."},{"Start":"13:09.985 ","End":"13:15.529","Text":"If I go back to the equation form,"},{"Start":"13:15.660 ","End":"13:24.790","Text":"here it is and w and z are free variables and x and y depend on these,"},{"Start":"13:24.790 ","End":"13:26.470","Text":"and we use back substitution,"},{"Start":"13:26.470 ","End":"13:29.980","Text":"and we\u0027ll use the wandering 1s method again."},{"Start":"13:29.980 ","End":"13:33.490","Text":"Here\u0027s where w is 1 and z is 0."},{"Start":"13:33.490 ","End":"13:34.990","Text":"First we compute y,"},{"Start":"13:34.990 ","End":"13:37.510","Text":"and then we use that to compute x."},{"Start":"13:37.510 ","End":"13:40.480","Text":"Similarly, the other way around w is 0,"},{"Start":"13:40.480 ","End":"13:45.610","Text":"z is 1 and that gives us"},{"Start":"13:45.610 ","End":"13:49.120","Text":"a basis for u intersection with"},{"Start":"13:49.120 ","End":"13:53.170","Text":"v as being these 2 vectors once you have them in the right order."},{"Start":"13:53.170 ","End":"13:55.195","Text":"Let\u0027s see, 1, 2, 0, 1."},{"Start":"13:55.195 ","End":"13:58.730","Text":"That\u0027s this 1 here."},{"Start":"13:59.010 ","End":"14:01.225","Text":"This 1, let\u0027s see."},{"Start":"14:01.225 ","End":"14:03.280","Text":"Minus 2, minus 1, 1, 0,"},{"Start":"14:03.280 ","End":"14:07.460","Text":"that would be this 1 here."},{"Start":"14:07.830 ","End":"14:10.660","Text":"I\u0027m just repeating what we saw earlier,"},{"Start":"14:10.660 ","End":"14:13.690","Text":"that the dimension of u intersection v really is 2."},{"Start":"14:13.690 ","End":"14:16.975","Text":"We got it previously from a computation."},{"Start":"14:16.975 ","End":"14:20.660","Text":"That\u0027s it for this exercise."}],"ID":26727},{"Watched":false,"Name":"Exercise 2","Duration":"2m 32s","ChapterTopicVideoID":25915,"CourseChapterTopicPlaylistID":246318,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.610","Text":"In this exercise, we\u0027re looking at a subspace of R^4,"},{"Start":"00:05.610 ","End":"00:07.800","Text":"4-dimensional real space."},{"Start":"00:07.800 ","End":"00:11.610","Text":"We want all the vectors a, b, c, d,"},{"Start":"00:11.610 ","End":"00:14.310","Text":"which satisfy these 2 conditions,"},{"Start":"00:14.310 ","End":"00:19.320","Text":"a equals c and b equals d. I\u0027ll give you an example."},{"Start":"00:19.320 ","End":"00:25.245","Text":"Let\u0027s take 1,4,1,4,"},{"Start":"00:25.245 ","End":"00:28.785","Text":"a equals c, the first and the third are the same,"},{"Start":"00:28.785 ","End":"00:32.865","Text":"b equals d means the second and the fourth are the same."},{"Start":"00:32.865 ","End":"00:39.710","Text":"So that\u0027s a subset anyway,"},{"Start":"00:39.710 ","End":"00:41.180","Text":"we\u0027ll see that it is a subspace."},{"Start":"00:41.180 ","End":"00:48.425","Text":"We want to find a basis for this subspace and its dimension."},{"Start":"00:48.425 ","End":"00:56.090","Text":"Now notice that I can rewrite these 2 conditions as a system of linear equations,"},{"Start":"00:56.090 ","End":"00:59.120","Text":"2 equations and 4 unknowns, a b, c, d."},{"Start":"00:59.120 ","End":"01:01.940","Text":"A equals c, then a minus c is 0,"},{"Start":"01:01.940 ","End":"01:04.430","Text":"b equal d, b minus d is 0."},{"Start":"01:04.430 ","End":"01:10.415","Text":"Arrange them nicely and notice that it already is in echelon form."},{"Start":"01:10.415 ","End":"01:12.890","Text":"You just look for the free variables,"},{"Start":"01:12.890 ","End":"01:14.210","Text":"those would be c and d,"},{"Start":"01:14.210 ","End":"01:16.340","Text":"and a and b depend on them."},{"Start":"01:16.340 ","End":"01:18.230","Text":"To get the basis,"},{"Start":"01:18.230 ","End":"01:22.320","Text":"we use what I call the wandering ones method."},{"Start":"01:22.340 ","End":"01:27.345","Text":"Once we let d be 1 and c be 0."},{"Start":"01:27.345 ","End":"01:33.010","Text":"Then the other way around we take d as the 0 and c as 1."},{"Start":"01:33.230 ","End":"01:36.050","Text":"From this equation, we compute b,"},{"Start":"01:36.050 ","End":"01:37.740","Text":"from this 1 we compute a."},{"Start":"01:37.740 ","End":"01:41.960","Text":"Well, because a equals c and b equals d, that\u0027s pretty easy."},{"Start":"01:41.960 ","End":"01:44.135","Text":"If d is 0, b is 0."},{"Start":"01:44.135 ","End":"01:46.280","Text":"Whatever c is a is."},{"Start":"01:46.280 ","End":"01:50.320","Text":"We\u0027ve got these solutions."},{"Start":"01:50.320 ","End":"01:57.615","Text":"These give us the basis for the solution space u."},{"Start":"01:57.615 ","End":"02:00.540","Text":"The first 1 gives us 0, 1,"},{"Start":"02:00.540 ","End":"02:03.060","Text":"0, 1, which is this."},{"Start":"02:03.060 ","End":"02:04.260","Text":"Second 1 gives us 1,"},{"Start":"02:04.260 ","End":"02:06.495","Text":"0, 1, 0, which is this."},{"Start":"02:06.495 ","End":"02:11.399","Text":"Certainly, they satisfy the condition that a equals c and b equals d."},{"Start":"02:11.399 ","End":"02:19.595","Text":"These 2 form a basis for all solutions to this condition."},{"Start":"02:19.595 ","End":"02:23.150","Text":"They all linear combination of these 2."},{"Start":"02:23.150 ","End":"02:26.000","Text":"The dimension of u, of course,"},{"Start":"02:26.000 ","End":"02:33.270","Text":"just by counting the number of elements in the basis is 2, and we\u0027re done."}],"ID":26725},{"Watched":false,"Name":"Exercise 3","Duration":"2m 11s","ChapterTopicVideoID":25916,"CourseChapterTopicPlaylistID":246318,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.740 ","End":"00:07.410","Text":"We start out with a vector space R^4,"},{"Start":"00:07.410 ","End":"00:11.835","Text":"and we let U be the subset of R^4."},{"Start":"00:11.835 ","End":"00:14.265","Text":"All those a, b, c, d,"},{"Start":"00:14.265 ","End":"00:17.610","Text":"such that these 2 conditions are satisfied,"},{"Start":"00:17.610 ","End":"00:19.875","Text":"that c is the sum of a and b,"},{"Start":"00:19.875 ","End":"00:24.660","Text":"and d is the sum of b and c. We want to"},{"Start":"00:24.660 ","End":"00:30.675","Text":"find a basis for this subspace."},{"Start":"00:30.675 ","End":"00:33.725","Text":"We\u0027ll show that it is a subspace much as the subset,"},{"Start":"00:33.725 ","End":"00:35.480","Text":"and if we have the basis,"},{"Start":"00:35.480 ","End":"00:38.400","Text":"then also the dimension."},{"Start":"00:38.600 ","End":"00:45.970","Text":"What we can do is write these 2 equations in this form,"},{"Start":"00:45.970 ","End":"00:49.520","Text":"and then we can see that it\u0027s a system of linear equations,"},{"Start":"00:49.520 ","End":"00:53.450","Text":"but it\u0027s homogeneous because of the 0s on the right."},{"Start":"00:53.450 ","End":"01:00.220","Text":"Since it\u0027s homogeneous, the solution space is a subspace."},{"Start":"01:00.220 ","End":"01:03.035","Text":"What we have to do now is solve it."},{"Start":"01:03.035 ","End":"01:06.725","Text":"It already is in echelon form, and clearly,"},{"Start":"01:06.725 ","End":"01:09.545","Text":"c and d are the free variables,"},{"Start":"01:09.545 ","End":"01:14.149","Text":"but a and b are dependent on those."},{"Start":"01:14.170 ","End":"01:17.420","Text":"We\u0027re going to use the method we just call it in this course,"},{"Start":"01:17.420 ","End":"01:20.360","Text":"wandering ones, it\u0027s not a general name."},{"Start":"01:20.360 ","End":"01:22.280","Text":"Where from the free variables,"},{"Start":"01:22.280 ","End":"01:23.330","Text":"in this case, d and c,"},{"Start":"01:23.330 ","End":"01:26.495","Text":"each time we let 1 of them be 1 and the rest 0,"},{"Start":"01:26.495 ","End":"01:32.325","Text":"so first we\u0027ll let d equal 1 and c be 0 and then the other way around."},{"Start":"01:32.325 ","End":"01:33.930","Text":"Once we have c and d,"},{"Start":"01:33.930 ","End":"01:35.640","Text":"we can compute b from here,"},{"Start":"01:35.640 ","End":"01:38.800","Text":"and once we have b, we can compute a."},{"Start":"01:39.470 ","End":"01:42.950","Text":"We collect these in the right order, a, b,"},{"Start":"01:42.950 ","End":"01:46.470","Text":"c, d. This is minus 1, 1,"},{"Start":"01:46.470 ","End":"01:49.395","Text":"0, 1, that\u0027s this vector,"},{"Start":"01:49.395 ","End":"01:51.620","Text":"and this one gives us 2 minus 1,"},{"Start":"01:51.620 ","End":"01:53.615","Text":"1, 0, that\u0027s here."},{"Start":"01:53.615 ","End":"02:00.385","Text":"These 2 vectors will give us a basis of u,"},{"Start":"02:00.385 ","End":"02:05.270","Text":"and by counting the number of elements in the set, that\u0027s 1,"},{"Start":"02:05.270 ","End":"02:11.280","Text":"2, that\u0027s the dimension of U. We\u0027re done."}],"ID":26726}],"Thumbnail":null,"ID":246318},{"Name":"Subspaces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"2m 26s","ChapterTopicVideoID":25918,"CourseChapterTopicPlaylistID":246319,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.390","Text":"In this exercise, we\u0027re considering the space M_2 over R,"},{"Start":"00:06.390 ","End":"00:12.870","Text":"which means 2 by 2 square matrices of real numbers."},{"Start":"00:12.870 ","End":"00:20.435","Text":"The subspace U is the span of these 3 matrices."},{"Start":"00:20.435 ","End":"00:28.650","Text":"Our task is to find the basis for this subspace and its dimension."},{"Start":"00:28.780 ","End":"00:32.600","Text":"We\u0027re going to use matrices,"},{"Start":"00:32.600 ","End":"00:34.730","Text":"but we can\u0027t have a matrix of matrices."},{"Start":"00:34.730 ","End":"00:38.285","Text":"We have to squash these, flatten them out."},{"Start":"00:38.285 ","End":"00:41.630","Text":"Well, remember the snake method or we go like this, like this,"},{"Start":"00:41.630 ","End":"00:46.065","Text":"like so, like so, like so."},{"Start":"00:46.065 ","End":"00:49.080","Text":"Then this 1 is like a flat vector,"},{"Start":"00:49.080 ","End":"00:51.525","Text":"1, 1, minus 1, 2."},{"Start":"00:51.525 ","End":"00:52.800","Text":"This 1 is 4, 1,"},{"Start":"00:52.800 ","End":"00:54.600","Text":"minus 1, 1."},{"Start":"00:54.600 ","End":"00:57.300","Text":"Similarly here, 2 minus 1,"},{"Start":"00:57.300 ","End":"00:59.100","Text":"1, 3, that\u0027s here."},{"Start":"00:59.100 ","End":"01:02.270","Text":"These are now in a matrix and we\u0027re going to"},{"Start":"01:02.270 ","End":"01:06.240","Text":"do row operations to bring it to echelon form."},{"Start":"01:06.240 ","End":"01:12.760","Text":"Well, first off, let\u0027s 0 out the rest of this column below the 1."},{"Start":"01:12.760 ","End":"01:19.780","Text":"We\u0027ll subtract 4 times the first column from the second and twice from the last."},{"Start":"01:19.780 ","End":"01:21.550","Text":"You know all this stuff."},{"Start":"01:21.550 ","End":"01:25.200","Text":"After we do that, we get this,"},{"Start":"01:25.200 ","End":"01:27.300","Text":"and now still not echelon,"},{"Start":"01:27.300 ","End":"01:29.100","Text":"we want a 0 here,"},{"Start":"01:29.100 ","End":"01:31.245","Text":"but look it is the same row."},{"Start":"01:31.245 ","End":"01:34.750","Text":"On any event we\u0027re going to subtract the second row"},{"Start":"01:34.750 ","End":"01:38.395","Text":"from the third row in row notation like this,"},{"Start":"01:38.395 ","End":"01:45.730","Text":"and we end up, there we are all 0s in the last row."},{"Start":"01:45.730 ","End":"01:50.690","Text":"That means that we only have 2 vectors."},{"Start":"01:50.690 ","End":"01:57.420","Text":"Now, we have to unsnake them and flatten them out like this 1,"},{"Start":"01:57.420 ","End":"01:59.760","Text":"1, minus 1, 2, I just place them in order,"},{"Start":"01:59.760 ","End":"02:01.200","Text":"1, 1,"},{"Start":"02:01.200 ","End":"02:03.390","Text":"minus 1, 2, and this 1."},{"Start":"02:03.390 ","End":"02:06.840","Text":"If I had blank here I put 0 minus 3,"},{"Start":"02:06.840 ","End":"02:10.500","Text":"then I\u0027d write the 3, then I\u0027d write the minus 7."},{"Start":"02:10.500 ","End":"02:17.400","Text":"The basis for the subspace U is these 2 matrices."},{"Start":"02:17.400 ","End":"02:19.905","Text":"Counting there are 2 elements in it,"},{"Start":"02:19.905 ","End":"02:27.010","Text":"2 matrices, which means that the dimension is 2 and we\u0027re done."}],"ID":31490},{"Watched":false,"Name":"Exercise 2","Duration":"2m 35s","ChapterTopicVideoID":25919,"CourseChapterTopicPlaylistID":246319,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.600","Text":"In this exercise, we\u0027re dealing with the space P_3 over R,"},{"Start":"00:07.600 ","End":"00:17.130","Text":"the space of polynomials of degree up to 3 over the real numbers."},{"Start":"00:17.130 ","End":"00:22.995","Text":"We want the subspace which is spanned by these 3 polynomials."},{"Start":"00:22.995 ","End":"00:28.500","Text":"Span, of course, meaning the set of all linear combinations of these."},{"Start":"00:28.500 ","End":"00:31.075","Text":"Then we have to find the basis and the dimension."},{"Start":"00:31.075 ","End":"00:33.170","Text":"Now, we want to use matrices."},{"Start":"00:33.170 ","End":"00:35.390","Text":"With matrices, we don\u0027t have polynomials,"},{"Start":"00:35.390 ","End":"00:40.875","Text":"we have to look upon this to somehow make these into regular vectors."},{"Start":"00:40.875 ","End":"00:46.410","Text":"Now, a typical member of P_3 over"},{"Start":"00:46.410 ","End":"00:52.740","Text":"R is a plus bx plus cx squared plus dx cubed."},{"Start":"00:52.740 ","End":"00:58.410","Text":"We could represent it by the vector a, b, c,"},{"Start":"00:58.410 ","End":"01:02.510","Text":"d. Then if we do that here,"},{"Start":"01:02.510 ","End":"01:04.730","Text":"then here we\u0027ll have vector 1,"},{"Start":"01:04.730 ","End":"01:07.835","Text":"1 minus 1, 2."},{"Start":"01:07.835 ","End":"01:10.915","Text":"Now, similarly for the other 2."},{"Start":"01:10.915 ","End":"01:14.360","Text":"So this next 1 would be 4, 1 minus 1,"},{"Start":"01:14.360 ","End":"01:15.785","Text":"1, which is here,"},{"Start":"01:15.785 ","End":"01:18.575","Text":"2 minus 1, 1 minus 3."},{"Start":"01:18.575 ","End":"01:22.315","Text":"Of course, we have to make sure that they\u0027re in the correct order."},{"Start":"01:22.315 ","End":"01:25.310","Text":"If there\u0027s a missing term in it\u0027s 0 and so on."},{"Start":"01:25.310 ","End":"01:27.320","Text":"But yeah, this is the matrix,"},{"Start":"01:27.320 ","End":"01:32.620","Text":"and we are going to do row operations to bring it to echelon form."},{"Start":"01:32.620 ","End":"01:37.070","Text":"We subtract 4 times this row from this row and twice this row from this row,"},{"Start":"01:37.070 ","End":"01:38.300","Text":"and that\u0027s what it says here,"},{"Start":"01:38.300 ","End":"01:39.995","Text":"and we end up with this."},{"Start":"01:39.995 ","End":"01:42.200","Text":"There are 0s here."},{"Start":"01:42.200 ","End":"01:46.320","Text":"Then we want to get 0 here also,"},{"Start":"01:46.320 ","End":"01:49.760","Text":"so just subtract this row from this row,"},{"Start":"01:49.760 ","End":"01:53.185","Text":"which in notation is like this."},{"Start":"01:53.185 ","End":"01:55.830","Text":"That gives us 0 here."},{"Start":"01:55.830 ","End":"02:00.114","Text":"Now, we see that we have just 2 non 0 rows,"},{"Start":"02:00.114 ","End":"02:02.650","Text":"but we have to convert these to polynomials,"},{"Start":"02:02.650 ","End":"02:04.675","Text":"the opposite of what we did before."},{"Start":"02:04.675 ","End":"02:11.920","Text":"This will be 1 plus x minus x squared plus 2x cubed."},{"Start":"02:11.920 ","End":"02:15.580","Text":"Similarly, for the second row,"},{"Start":"02:15.580 ","End":"02:17.455","Text":"which will be 0,"},{"Start":"02:17.455 ","End":"02:22.870","Text":"it\u0027ll be minus 3x plus 3x squared minus 7x cubed."},{"Start":"02:22.870 ","End":"02:25.175","Text":"That\u0027s the basis,"},{"Start":"02:25.175 ","End":"02:27.365","Text":"and the dimension of course, is 2,"},{"Start":"02:27.365 ","End":"02:30.130","Text":"just by counting how many elements in the basis,"},{"Start":"02:30.130 ","End":"02:31.945","Text":"the number of rows here,"},{"Start":"02:31.945 ","End":"02:35.450","Text":"which are not 0. We\u0027re done."}],"ID":31491},{"Watched":false,"Name":"Direct sum of subspaces","Duration":"11m 30s","ChapterTopicVideoID":27899,"CourseChapterTopicPlaylistID":246319,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:09.000","Text":"Welcome back. Today we have a video on direct sums, so let\u0027s proceed."},{"Start":"00:09.000 ","End":"00:16.740","Text":"Let\u0027s R_3(x) be the space of polynomials in x of degree less than or equal to 3."},{"Start":"00:16.740 ","End":"00:19.395","Text":"Now consider a subset U,"},{"Start":"00:19.395 ","End":"00:22.215","Text":"which is equal to the polynomial p(x),"},{"Start":"00:22.215 ","End":"00:26.110","Text":"which is a_3 x^3 plus a_ 2"},{"Start":"00:26.690 ","End":"00:34.155","Text":"x^2 such that p(0) is equal to 0 and p(1) is equal to 0 as well."},{"Start":"00:34.155 ","End":"00:38.090","Text":"We also need to consider a subspace V,"},{"Start":"00:38.090 ","End":"00:41.480","Text":"which is equal to the span of 1 and x^2,"},{"Start":"00:41.480 ","End":"00:45.545","Text":"which is a well-defined subspace"},{"Start":"00:45.545 ","End":"00:51.180","Text":"of R_3(x) and we need to prove that our R_3(x) is equal to"},{"Start":"00:51.180 ","End":"00:58.505","Text":"the direct sum of U and V. Notice here that we have this circle"},{"Start":"00:58.505 ","End":"01:02.015","Text":"around the plus sign and that indicates that we are"},{"Start":"01:02.015 ","End":"01:06.260","Text":"dealing with a direct sum here and not a normal sum."},{"Start":"01:06.260 ","End":"01:12.105","Text":"Let\u0027s start by defining what a direct sum is."},{"Start":"01:12.105 ","End":"01:19.860","Text":"As it\u0027s defined, if R_3(x) is equal to the direct sum of U and V,"},{"Start":"01:19.860 ","End":"01:22.950","Text":"then we have 2 conditions that need to be met."},{"Start":"01:22.950 ","End":"01:27.430","Text":"The first of these conditions is that R_3(x)"},{"Start":"01:27.430 ","End":"01:32.340","Text":"is equal to U plus V. These are just the sets and"},{"Start":"01:32.340 ","End":"01:36.860","Text":"the subspaces added together but the second condition tells us that"},{"Start":"01:36.860 ","End":"01:42.845","Text":"the intersection of this set in this subspace must be equal to 0."},{"Start":"01:42.845 ","End":"01:47.105","Text":"What that means is when we take the intersection of these 2 sets,"},{"Start":"01:47.105 ","End":"01:51.950","Text":"we can\u0027t have some polynomial with x^3 and x^2 in it,"},{"Start":"01:51.950 ","End":"01:56.430","Text":"we must only have 0 as the overlap."},{"Start":"01:56.430 ","End":"02:01.100","Text":"The other piece of information that this question tells us is that"},{"Start":"02:01.100 ","End":"02:06.670","Text":"p(0) is equal to 0 and p(1) is equal to 0,"},{"Start":"02:06.670 ","End":"02:10.750","Text":"so let\u0027s see what that gives us."},{"Start":"02:10.850 ","End":"02:16.040","Text":"P(0) equals 0 is quite easy because if we subzero"},{"Start":"02:16.040 ","End":"02:20.780","Text":"into p(x) then all these terms with x then just disappear."},{"Start":"02:20.780 ","End":"02:25.115","Text":"What that tells us is that the only remaining term is this a_naught,"},{"Start":"02:25.115 ","End":"02:30.465","Text":"so that tells us that a_naught is equal to 0."},{"Start":"02:30.465 ","End":"02:33.885","Text":"Now, if we sub in 1 into p(x),"},{"Start":"02:33.885 ","End":"02:41.325","Text":"then this part will tell us that a _3 times 1^3 plus"},{"Start":"02:41.325 ","End":"02:49.065","Text":"a_ 2 times 1^2 plus a_1 times x plus a_naught,"},{"Start":"02:49.065 ","End":"02:50.490","Text":"which we know is 0 now,"},{"Start":"02:50.490 ","End":"02:54.675","Text":"so we can just say plus 0=0."},{"Start":"02:54.675 ","End":"02:59.400","Text":"Then obviously all these 1^3 and 1^2 is just equal to 1."},{"Start":"02:59.400 ","End":"03:03.570","Text":"In here, this should be a 1 as well."},{"Start":"03:03.570 ","End":"03:08.835","Text":"Then that just tells us that a_3 plus"},{"Start":"03:08.835 ","End":"03:14.295","Text":"a_2 plus a_1 is equal to 0."},{"Start":"03:14.295 ","End":"03:18.750","Text":"Here we have 2 constraints on the"},{"Start":"03:18.750 ","End":"03:24.140","Text":"a_i\u0027s based on the information that is given to us in the question."},{"Start":"03:24.140 ","End":"03:28.985","Text":"Now let\u0027s go about proving these 2 constraints which we defined earlier,"},{"Start":"03:28.985 ","End":"03:35.975","Text":"which are necessary for our 3(x) to be the direct sum of U and V. Now,"},{"Start":"03:35.975 ","End":"03:43.805","Text":"if we know that a_naught is equal to 0 and a_ 3 plus a_2 plus a_1 is equal to 0,"},{"Start":"03:43.805 ","End":"03:48.095","Text":"then we can re-express U in a simpler way."},{"Start":"03:48.095 ","End":"03:50.080","Text":"Let\u0027s do that now."},{"Start":"03:50.080 ","End":"03:53.400","Text":"Here is our simplified version of U,"},{"Start":"03:53.400 ","End":"03:58.610","Text":"so you\u0027ll notice there\u0027s now no a_naught term because we addressed the a_naught with"},{"Start":"03:58.610 ","End":"04:05.955","Text":"0 and you\u0027ll see that I\u0027ve replaced a_1 here with minus a_2 minus a_3."},{"Start":"04:05.955 ","End":"04:12.600","Text":"All I did here was I brought this a_2 and a_3 to the other side,"},{"Start":"04:12.600 ","End":"04:18.585","Text":"so that would tell us that a_1 is equal to minus a_2"},{"Start":"04:18.585 ","End":"04:25.265","Text":"plus a_3 and then here we\u0027ve just distributed the minus sign inside."},{"Start":"04:25.265 ","End":"04:31.000","Text":"Let\u0027s go about proving these 2 constraints that we saw earlier."},{"Start":"04:31.000 ","End":"04:35.180","Text":"Just for ease, it might be useful to re-express V."},{"Start":"04:35.180 ","End":"04:40.445","Text":"Here we\u0027re told that V is the span of 1 and x^2."},{"Start":"04:40.445 ","End":"04:46.410","Text":"What that means is we can write V as being equal to say,"},{"Start":"04:46.410 ","End":"04:52.570","Text":"b_naught plus b_2 x^2."},{"Start":"04:52.570 ","End":"04:57.260","Text":"Because this encompasses all the polynomials of degree"},{"Start":"04:57.260 ","End":"05:02.520","Text":"less than or equal to 2 but we are remember excluding the linear term,"},{"Start":"05:02.520 ","End":"05:07.155","Text":"so the term that involves just coefficients of x."},{"Start":"05:07.155 ","End":"05:11.520","Text":"Let\u0027s prove these constraints now."},{"Start":"05:11.520 ","End":"05:15.935","Text":"Maybe we\u0027ll start by proving constraint 2,"},{"Start":"05:15.935 ","End":"05:17.945","Text":"because that might be easier."},{"Start":"05:17.945 ","End":"05:19.945","Text":"Let\u0027s do that now."},{"Start":"05:19.945 ","End":"05:24.120","Text":"At our intersection between U and V,"},{"Start":"05:24.120 ","End":"05:26.338","Text":"because these are both polynomials,"},{"Start":"05:26.338 ","End":"05:27.980","Text":"if we have an intersection,"},{"Start":"05:27.980 ","End":"05:34.755","Text":"then it means that the coefficients of each U and V must be equal to each other."},{"Start":"05:34.755 ","End":"05:36.630","Text":"If we look at, say,"},{"Start":"05:36.630 ","End":"05:39.930","Text":"the x^3 terms of each,"},{"Start":"05:39.930 ","End":"05:46.895","Text":"this would imply that a_3 is equal to 0 because we don\u0027t have a cubic term in V,"},{"Start":"05:46.895 ","End":"05:52.405","Text":"so this tells us that a_3=0."},{"Start":"05:52.405 ","End":"05:55.910","Text":"If we look at the squared terms,"},{"Start":"05:55.910 ","End":"06:01.870","Text":"then this will tell us that a_2=b_2."},{"Start":"06:05.960 ","End":"06:09.605","Text":"If we look at the linear terms,"},{"Start":"06:09.605 ","End":"06:11.390","Text":"then this tells us that"},{"Start":"06:11.390 ","End":"06:18.485","Text":"minus a_2 minus a_3 is equal to 0 because we have no linear terms belonging to V,"},{"Start":"06:18.485 ","End":"06:26.565","Text":"so we have minus a_2 minus a_3 is equal to 0."},{"Start":"06:26.565 ","End":"06:34.103","Text":"That tells us that a_2 is equal to minus a_3,"},{"Start":"06:34.103 ","End":"06:38.810","Text":"but we\u0027ve already established that a_3 is equal to 0,"},{"Start":"06:38.810 ","End":"06:45.390","Text":"so then that tells us that a_2 must also be equal to 0,"},{"Start":"06:45.390 ","End":"06:49.095","Text":"which then tells us that in the x^2 term,"},{"Start":"06:49.095 ","End":"06:52.395","Text":"that a_2 is equal to b_2,"},{"Start":"06:52.395 ","End":"06:54.450","Text":"which is equal to 0."},{"Start":"06:54.450 ","End":"06:56.450","Text":"Then finally, we just need to"},{"Start":"06:56.450 ","End":"07:05.460","Text":"compare the constant terms."},{"Start":"07:05.460 ","End":"07:11.085","Text":"We have b_naught here and we have nothing over here,"},{"Start":"07:11.085 ","End":"07:17.220","Text":"so then that will tell us that b_naught is equal to 0."},{"Start":"07:17.220 ","End":"07:21.200","Text":"In essence, at our intersection,"},{"Start":"07:21.200 ","End":"07:23.810","Text":"all the coefficients are equal to 0,"},{"Start":"07:23.810 ","End":"07:28.250","Text":"so U intersected with V=0,"},{"Start":"07:28.250 ","End":"07:32.195","Text":"and therefore, we have satisfied our second constraint."},{"Start":"07:32.195 ","End":"07:36.080","Text":"Now all we need to show to prove the property of"},{"Start":"07:36.080 ","End":"07:41.845","Text":"direct sums that U plus V is equal to R_3(x),"},{"Start":"07:41.845 ","End":"07:47.375","Text":"so the polynomial space of x with degree less than or equal to 3."},{"Start":"07:47.375 ","End":"07:49.235","Text":"Let\u0027s do that now."},{"Start":"07:49.235 ","End":"07:55.835","Text":"We\u0027ve written up our second condition that we need for the property of direct sum,"},{"Start":"07:55.835 ","End":"07:57.770","Text":"which is R_3(x) is equal to"},{"Start":"07:57.770 ","End":"08:04.100","Text":"U plus V. What we\u0027ve done here is we\u0027ve just written down what"},{"Start":"08:04.100 ","End":"08:08.195","Text":"U plus V is equal to and essentially just"},{"Start":"08:08.195 ","End":"08:13.115","Text":"combined the coefficients from each to give us this result."},{"Start":"08:13.115 ","End":"08:17.255","Text":"We have U plus V is equal to a_3(x^3),"},{"Start":"08:17.255 ","End":"08:19.000","Text":"which comes from this term,"},{"Start":"08:19.000 ","End":"08:21.645","Text":"plus a_2 plus b_2 x^2,"},{"Start":"08:21.645 ","End":"08:23.925","Text":"so we have the a_2 from U,"},{"Start":"08:23.925 ","End":"08:27.840","Text":"and the b_2 from V and then we have this plus,"},{"Start":"08:27.840 ","End":"08:30.810","Text":"minus a_2 minus a_3 x,"},{"Start":"08:30.810 ","End":"08:36.335","Text":"so that\u0027s from the U and then we have plus this constant term b_naught,"},{"Start":"08:36.335 ","End":"08:43.270","Text":"which is from V. If we need to show that u plus V is equal to R_3(x),"},{"Start":"08:43.270 ","End":"08:48.020","Text":"then it\u0027s useful to just expand as to what R_3(x) is equal to."},{"Start":"08:48.020 ","End":"08:55.520","Text":"Remember, R_3(x) is just the span of polynomials of degree less than or equal to 3."},{"Start":"08:55.520 ","End":"09:05.930","Text":"We can write this in general as C_3x^3 plus c_2 x^2 plus c_1x plus c_naught."},{"Start":"09:05.930 ","End":"09:09.620","Text":"If we want to make these 2 things equal to each other,"},{"Start":"09:09.620 ","End":"09:14.300","Text":"then we can once again just compare the coefficients."},{"Start":"09:14.300 ","End":"09:20.835","Text":"Let\u0027s be sensible here and take a_ 3 is equal to c_3,"},{"Start":"09:20.835 ","End":"09:26.055","Text":"and then a_2 plus b_2 is equal to c_2."},{"Start":"09:26.055 ","End":"09:31.140","Text":"Minus a_2 minus a_3 is equal to c_1,"},{"Start":"09:31.140 ","End":"09:37.970","Text":"which we will just write minus a_2 plus a_3 is equal to c_1."},{"Start":"09:37.970 ","End":"09:42.920","Text":"Then finally, b_naught is equal to c_naught."},{"Start":"09:42.920 ","End":"09:46.880","Text":"The question is, does this polynomial on"},{"Start":"09:46.880 ","End":"09:54.925","Text":"the left hand side span all possible polynomials of degree less than or equal to 3?"},{"Start":"09:54.925 ","End":"09:57.270","Text":"Well, we can see that,"},{"Start":"09:57.270 ","End":"10:00.065","Text":"if we let a_3 vary,"},{"Start":"10:00.065 ","End":"10:04.325","Text":"then that means that we can choose whatever c_3 that we want."},{"Start":"10:04.325 ","End":"10:08.165","Text":"We\u0027re happy with this part."},{"Start":"10:08.165 ","End":"10:10.805","Text":"b_naught is equal to c_naught."},{"Start":"10:10.805 ","End":"10:17.103","Text":"If we just let b_naught vary than we can vary c_naught as we please as well,"},{"Start":"10:17.103 ","End":"10:19.675","Text":"so the constant term we\u0027re happy with."},{"Start":"10:19.675 ","End":"10:23.655","Text":"Now, a_2 plus b_2 is equal to c_2."},{"Start":"10:23.655 ","End":"10:26.814","Text":"We haven\u0027t let a_2 vary yet,"},{"Start":"10:26.814 ","End":"10:31.125","Text":"so if we let a_2 vary and therefore b_2 vary,"},{"Start":"10:31.125 ","End":"10:33.715","Text":"then we can have whatever c_2 we want."},{"Start":"10:33.715 ","End":"10:40.310","Text":"Then the only coefficient that has dependency is this c_1,"},{"Start":"10:40.310 ","End":"10:45.575","Text":"because we\u0027ve already let a_2 and a_3 vary."},{"Start":"10:45.575 ","End":"10:51.380","Text":"But then if we start by assigning a particular value to this constant term,"},{"Start":"10:51.380 ","End":"10:55.250","Text":"then we can just adjust all the other coefficients to make sure"},{"Start":"10:55.250 ","End":"10:59.615","Text":"that we get the desired polynomial that we are looking for."},{"Start":"10:59.615 ","End":"11:05.795","Text":"In essence, this is the only one that doesn\u0027t depend or has dependency,"},{"Start":"11:05.795 ","End":"11:10.655","Text":"but because all the other ones we can range freely then we can span"},{"Start":"11:10.655 ","End":"11:16.495","Text":"R_3 as required and therefore we have met the second condition."},{"Start":"11:16.495 ","End":"11:26.845","Text":"We have proven that R_3(x) is equal to the direct sum of U plus V as required."},{"Start":"11:26.845 ","End":"11:31.770","Text":"Thank you very much and see you in the next video."}],"ID":31492},{"Watched":false,"Name":"Sum of Subspaces - Part 1","Duration":"8m 18s","ChapterTopicVideoID":28180,"CourseChapterTopicPlaylistID":246319,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.205","Text":"Today we\u0027re going to be looking at a linear algebra problem"},{"Start":"00:05.205 ","End":"00:12.190","Text":"and this particular question is going to be broken down into 3 videos."},{"Start":"00:12.190 ","End":"00:18.150","Text":"Each video is going to be answering each part that we have given here."},{"Start":"00:18.150 ","End":"00:20.520","Text":"This is our question."},{"Start":"00:20.520 ","End":"00:27.650","Text":"Let W be a finite dimensional vector space and we are also letting A and"},{"Start":"00:27.650 ","End":"00:34.850","Text":"B be subspaces of W. In the first video or this first part,"},{"Start":"00:34.850 ","End":"00:37.790","Text":"we have to prove that the set U,"},{"Start":"00:37.790 ","End":"00:40.820","Text":"which is equal to this set A plus B,"},{"Start":"00:40.820 ","End":"00:42.200","Text":"such that A is in A,"},{"Start":"00:42.200 ","End":"00:46.510","Text":"B is in B is a subspace of W."},{"Start":"00:46.510 ","End":"00:52.835","Text":"Remember what we need to consider in order for something to be a subspace."},{"Start":"00:52.835 ","End":"01:01.250","Text":"There are 3 axioms that we need to satisfy so we will write this down more formally."},{"Start":"01:01.250 ","End":"01:02.940","Text":"Let\u0027s just discuss them."},{"Start":"01:02.940 ","End":"01:08.780","Text":"The first axiom, remember is that 0 will belong in the set U."},{"Start":"01:08.780 ","End":"01:13.465","Text":"The second axiom is that it\u0027s closed under addition,"},{"Start":"01:13.465 ","End":"01:19.850","Text":"and the third axiom is that the set U is closed under scalar multiplication."},{"Start":"01:19.850 ","End":"01:22.825","Text":"Let\u0027s write that down a bit more formally."},{"Start":"01:22.825 ","End":"01:26.060","Text":"We\u0027ve written down the conditions that need to be"},{"Start":"01:26.060 ","End":"01:30.245","Text":"satisfied in order for the set U to be a subspace."},{"Start":"01:30.245 ","End":"01:37.400","Text":"But one thing that we should also consider is that if A and B are also subspaces of W,"},{"Start":"01:37.400 ","End":"01:43.945","Text":"so these subspaces themselves must also satisfy these 3 conditions."},{"Start":"01:43.945 ","End":"01:50.010","Text":"That fact in itself is going to help us to answer this question here."},{"Start":"01:50.010 ","End":"01:54.980","Text":"Let\u0027s go about proving each of these 1 by 1."},{"Start":"01:54.980 ","End":"02:01.330","Text":"We\u0027ll start with this first condition that 0 belongs to this set U."},{"Start":"02:01.330 ","End":"02:11.300","Text":"We can start by saying since A and B are themselves subspaces of W,"},{"Start":"02:11.300 ","End":"02:16.370","Text":"then we know that 0 must belong to"},{"Start":"02:16.370 ","End":"02:22.850","Text":"the set A and 0 must also belong to the set B."},{"Start":"02:22.850 ","End":"02:25.205","Text":"If we take say,"},{"Start":"02:25.205 ","End":"02:29.300","Text":"little a which is an element of the set big A."},{"Start":"02:29.300 ","End":"02:34.580","Text":"Take little a is equal to 0,"},{"Start":"02:34.580 ","End":"02:37.865","Text":"and also take little b,"},{"Start":"02:37.865 ","End":"02:40.280","Text":"which is an element of the set big B,"},{"Start":"02:40.280 ","End":"02:42.590","Text":"which is also equal to 0,"},{"Start":"02:42.590 ","End":"02:46.220","Text":"then we know that if we add these 2 things together,"},{"Start":"02:46.220 ","End":"02:47.960","Text":"so a plus b,"},{"Start":"02:47.960 ","End":"02:50.840","Text":"then that\u0027s just equal to 0 plus 0,"},{"Start":"02:50.840 ","End":"02:55.310","Text":"which is an element in U and what is 0 plus 0?"},{"Start":"02:55.310 ","End":"02:57.110","Text":"That\u0027s obviously just 0."},{"Start":"02:57.110 ","End":"03:04.095","Text":"We can therefore say that 0 belongs in our set U."},{"Start":"03:04.095 ","End":"03:09.615","Text":"We have indeed proven this first condition here."},{"Start":"03:09.615 ","End":"03:17.400","Text":"Now, let\u0027s look about seeing if our set U is closed under addition."},{"Start":"03:17.450 ","End":"03:19.670","Text":"First we might ask,"},{"Start":"03:19.670 ","End":"03:25.775","Text":"what does it mean for our set u to be closed under addition?"},{"Start":"03:25.775 ","End":"03:28.820","Text":"Well, that means that if we have 2 elements,"},{"Start":"03:28.820 ","End":"03:31.430","Text":"say u_1 and u_2,"},{"Start":"03:31.430 ","End":"03:33.170","Text":"and this is little u here,"},{"Start":"03:33.170 ","End":"03:36.470","Text":"that are elements of the set big U."},{"Start":"03:36.470 ","End":"03:41.645","Text":"Then if it\u0027s closed under addition then this means that u_1"},{"Start":"03:41.645 ","End":"03:47.825","Text":"plus u_2 is also an element of the set big U."},{"Start":"03:47.825 ","End":"03:52.175","Text":"How do we show that this is closed under addition?"},{"Start":"03:52.175 ","End":"03:58.260","Text":"Well, if u_1 is a member of the set U,"},{"Start":"03:58.260 ","End":"04:00.620","Text":"well, remember how you was defined."},{"Start":"04:00.620 ","End":"04:05.880","Text":"Big U was just equal to the set a plus b,"},{"Start":"04:05.880 ","End":"04:09.855","Text":"where a is in A and b is in B."},{"Start":"04:09.855 ","End":"04:11.795","Text":"We can write a general term,"},{"Start":"04:11.795 ","End":"04:15.070","Text":"u_1 in U is being equal to say,"},{"Start":"04:15.070 ","End":"04:17.880","Text":"a_1 plus b_1,"},{"Start":"04:17.880 ","End":"04:24.825","Text":"where a_1 is in A and b_1 is in B."},{"Start":"04:24.825 ","End":"04:28.780","Text":"Then we can do the same for the elements u_2."},{"Start":"04:28.780 ","End":"04:33.180","Text":"We can say that that\u0027s equal to a_2 plus b_2,"},{"Start":"04:33.180 ","End":"04:40.075","Text":"where again, a_2 is in A and b_2 is in B."},{"Start":"04:40.075 ","End":"04:44.440","Text":"Well then, because remember we said that we\u0027re going to use"},{"Start":"04:44.440 ","End":"04:48.775","Text":"the fact that the sets a and b themselves are sub-spaces,"},{"Start":"04:48.775 ","End":"04:52.630","Text":"then this means that A is closed under addition."},{"Start":"04:52.630 ","End":"04:58.310","Text":"That means that a_1 plus a_2 belongs to the set A."},{"Start":"04:58.310 ","End":"05:03.675","Text":"We can just call this, say a_3."},{"Start":"05:03.675 ","End":"05:08.145","Text":"We can also say that because b is itself a subspace,"},{"Start":"05:08.145 ","End":"05:15.185","Text":"that B is closed under addition so b_1 plus b_2 also belongs to the set B,"},{"Start":"05:15.185 ","End":"05:18.305","Text":"and we can call this b_3."},{"Start":"05:18.305 ","End":"05:21.830","Text":"Then essentially what we\u0027re doing here is we\u0027re just"},{"Start":"05:21.830 ","End":"05:25.370","Text":"adding again 2 elements of A and B so we"},{"Start":"05:25.370 ","End":"05:32.275","Text":"can say that a_3 plus b_3 will belong to the set U."},{"Start":"05:32.275 ","End":"05:36.680","Text":"Because remember, a_3 belongs to A and b_3 belongs to B,"},{"Start":"05:36.680 ","End":"05:38.945","Text":"and this is how U is defined."},{"Start":"05:38.945 ","End":"05:44.285","Text":"Again, we can just then say with certainty that our set,"},{"Start":"05:44.285 ","End":"05:46.925","Text":"U is closed under addition."},{"Start":"05:46.925 ","End":"05:50.060","Text":"Now we\u0027re going to play very similar trick in"},{"Start":"05:50.060 ","End":"05:54.190","Text":"showing that it\u0027s also closed under multiplication."},{"Start":"05:54.190 ","End":"05:56.550","Text":"Let\u0027s do that now."},{"Start":"05:56.550 ","End":"05:59.730","Text":"What does it mean for U,"},{"Start":"05:59.730 ","End":"06:02.255","Text":"our set to be closed under multiplication?"},{"Start":"06:02.255 ","End":"06:08.555","Text":"Well, it means if we haven\u0027t elements little u of our set big U,"},{"Start":"06:08.555 ","End":"06:11.270","Text":"then if it\u0027s closed under multiplication,"},{"Start":"06:11.270 ","End":"06:18.155","Text":"then it means that Lambda times U is also a member of the set big U,"},{"Start":"06:18.155 ","End":"06:23.420","Text":"where Lambda here is just some scalar or some real number."},{"Start":"06:23.420 ","End":"06:28.990","Text":"Maybe Lambda is equal to 1-1.5, something like that."},{"Start":"06:28.990 ","End":"06:32.405","Text":"How do we show that this is true?"},{"Start":"06:32.405 ","End":"06:35.600","Text":"Well, remember what, how U is defined."},{"Start":"06:35.600 ","End":"06:41.055","Text":"Well, U is just defined as a plus b."},{"Start":"06:41.055 ","End":"06:45.740","Text":"That means that Lambda times u is equal to"},{"Start":"06:45.740 ","End":"06:51.980","Text":"Lambda times a plus b and then if we expand this,"},{"Start":"06:51.980 ","End":"06:57.375","Text":"then we get Lambda a plus Lambda b."},{"Start":"06:57.375 ","End":"07:01.985","Text":"What we need to show here is does this term,"},{"Start":"07:01.985 ","End":"07:06.720","Text":"Lambda a plus Lambda b belong to the set big U?"},{"Start":"07:06.720 ","End":"07:14.465","Text":"Now remember that the sets A and B themselves are closed under multiplication."},{"Start":"07:14.465 ","End":"07:16.535","Text":"We can say that this term here,"},{"Start":"07:16.535 ","End":"07:19.940","Text":"Lambda A belongs to the set A,"},{"Start":"07:19.940 ","End":"07:25.130","Text":"and this term here Lambda b belongs to the set B."},{"Start":"07:25.130 ","End":"07:29.060","Text":"In essence, we can just write this as a term,"},{"Start":"07:29.060 ","End":"07:34.530","Text":"maybe let\u0027s call it a_1 and this is a term b_1."},{"Start":"07:34.530 ","End":"07:38.295","Text":"Because a_1 belongs to A and b_1 belongs to B,"},{"Start":"07:38.295 ","End":"07:41.650","Text":"then we can say that a_1 plus b_1,"},{"Start":"07:41.650 ","End":"07:44.585","Text":"which is equal to Lambda U,"},{"Start":"07:44.585 ","End":"07:46.850","Text":"because remember, everything is the same."},{"Start":"07:46.850 ","End":"07:50.855","Text":"We\u0027ve just been changing things around and how they\u0027re expressed."},{"Start":"07:50.855 ","End":"07:53.760","Text":"But we can say that a_1 plus b_1,"},{"Start":"07:53.760 ","End":"07:58.935","Text":"which is equal to Lambda U must also belong to the set B."},{"Start":"07:58.935 ","End":"08:02.195","Text":"Essentially, what we\u0027ve done here is we have"},{"Start":"08:02.195 ","End":"08:05.920","Text":"shown that the 3 conditions of subspaces hold,"},{"Start":"08:05.920 ","End":"08:10.415","Text":"and therefore we have shown that this sets big U is in fact"},{"Start":"08:10.415 ","End":"08:15.500","Text":"a subspace as we were required to in the question."},{"Start":"08:15.500 ","End":"08:19.529","Text":"I will see you in the next part."}],"ID":31493},{"Watched":false,"Name":"Sum of Subspaces - Part 2","Duration":"4m 38s","ChapterTopicVideoID":28181,"CourseChapterTopicPlaylistID":246319,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:05.110","Text":"We\u0027re returning back to our 3 part series on"},{"Start":"00:05.110 ","End":"00:09.550","Text":"this linear algebra problem that we looked up before."},{"Start":"00:09.550 ","End":"00:11.845","Text":"In the previous video,"},{"Start":"00:11.845 ","End":"00:14.395","Text":"we showed that this set U,"},{"Start":"00:14.395 ","End":"00:17.275","Text":"which is equal to a plus b,"},{"Start":"00:17.275 ","End":"00:21.370","Text":"where a is an element of A and b is an element of"},{"Start":"00:21.370 ","End":"00:26.605","Text":"B is a subspace of W. In this video today,"},{"Start":"00:26.605 ","End":"00:31.300","Text":"we\u0027re going to prove that this subspace that we just found is"},{"Start":"00:31.300 ","End":"00:37.295","Text":"the smallest possible subspace that contains both A and B."},{"Start":"00:37.295 ","End":"00:43.285","Text":"How are we going to do this is by a proof by contradiction."},{"Start":"00:43.285 ","End":"00:52.575","Text":"How this starts is that we start with a statement that we consider to be true,"},{"Start":"00:52.575 ","End":"00:56.375","Text":"and then we\u0027re going to show that this statement is in fact"},{"Start":"00:56.375 ","End":"01:01.445","Text":"untrue by a contradiction that we will meet later on."},{"Start":"01:01.445 ","End":"01:08.595","Text":"The statement that we consider to be true is we\u0027re going to let W"},{"Start":"01:08.595 ","End":"01:16.005","Text":"be a smaller subspace than a plus b that contains both A and B."},{"Start":"01:16.005 ","End":"01:21.110","Text":"Remember this goes against what we\u0027re being asked for in the proof because remember,"},{"Start":"01:21.110 ","End":"01:26.525","Text":"we\u0027re being asked to show that this set U is in fact the smallest subspace."},{"Start":"01:26.525 ","End":"01:30.980","Text":"We\u0027re saying that there exists a set W that is, in fact,"},{"Start":"01:30.980 ","End":"01:34.355","Text":"a smallest subspace than our set U,"},{"Start":"01:34.355 ","End":"01:36.215","Text":"which we\u0027ve called a plus b,"},{"Start":"01:36.215 ","End":"01:38.840","Text":"that contains both A and B."},{"Start":"01:38.840 ","End":"01:42.560","Text":"This is the start of our proof."},{"Start":"01:42.560 ","End":"01:45.475","Text":"Now, how do we proceed?"},{"Start":"01:45.475 ","End":"01:49.895","Text":"Since W is itself a subspace,"},{"Start":"01:49.895 ","End":"01:51.530","Text":"let\u0027s write this down."},{"Start":"01:51.530 ","End":"01:59.135","Text":"Since W is itself a subspace,"},{"Start":"01:59.135 ","End":"02:04.955","Text":"then remember the conditions that needed to be satisfied earlier."},{"Start":"02:04.955 ","End":"02:11.075","Text":"Well, that meant that W had to be or has to be closed under addition."},{"Start":"02:11.075 ","End":"02:13.605","Text":"If we have elements,"},{"Start":"02:13.605 ","End":"02:21.210","Text":"if the set big A and the set big B both belong to the set W, well,"},{"Start":"02:21.210 ","End":"02:24.935","Text":"then that means that W must be closed under addition"},{"Start":"02:24.935 ","End":"02:29.690","Text":"for all the elements in A and all the elements in B."},{"Start":"02:29.690 ","End":"02:38.670","Text":"This means that for all a in A and for all b in B."},{"Start":"02:38.670 ","End":"02:43.150","Text":"For all elements little a in the set big A,"},{"Start":"02:43.150 ","End":"02:47.420","Text":"and for all elements little b in the set big B,"},{"Start":"02:47.420 ","End":"02:50.915","Text":"then because W is closed under addition,"},{"Start":"02:50.915 ","End":"02:54.410","Text":"then this means that a plus b,"},{"Start":"02:54.410 ","End":"02:56.360","Text":"which is itself an element,"},{"Start":"02:56.360 ","End":"03:00.920","Text":"must be an element in W. Essentially,"},{"Start":"03:00.920 ","End":"03:04.580","Text":"what this is saying is that the set,"},{"Start":"03:04.580 ","End":"03:08.990","Text":"all the sets A plus B are themselves"},{"Start":"03:08.990 ","End":"03:14.270","Text":"a subspace of W. But remember what we said at the start."},{"Start":"03:14.270 ","End":"03:18.515","Text":"We said that W was a smaller subspace,"},{"Start":"03:18.515 ","End":"03:20.015","Text":"than A plus B."},{"Start":"03:20.015 ","End":"03:24.945","Text":"Essentially what this means here is the W"},{"Start":"03:24.945 ","End":"03:30.485","Text":"is a smaller or a subset or subspace of A and B."},{"Start":"03:30.485 ","End":"03:34.100","Text":"But here we\u0027re saying that A plus B is smaller than"},{"Start":"03:34.100 ","End":"03:39.875","Text":"W. You can see that already we have arrived at our contradiction."},{"Start":"03:39.875 ","End":"03:43.625","Text":"We can conclude the proof here."},{"Start":"03:43.625 ","End":"03:46.805","Text":"But just so we can make it a bit more solid,"},{"Start":"03:46.805 ","End":"03:50.210","Text":"we can just write this down formally."},{"Start":"03:50.210 ","End":"03:55.720","Text":"This contradicts our assumption"},{"Start":"03:55.720 ","End":"04:01.035","Text":"that W is smaller than A plus B."},{"Start":"04:01.035 ","End":"04:11.400","Text":"Let\u0027s write that. This contradicts our assumption that W is smaller than A plus B."},{"Start":"04:11.400 ","End":"04:14.900","Text":"Essentially, we have used the facts of"},{"Start":"04:14.900 ","End":"04:19.385","Text":"closure under addition to give us our final proof."},{"Start":"04:19.385 ","End":"04:21.410","Text":"Then how we can finish this,"},{"Start":"04:21.410 ","End":"04:23.600","Text":"you can write Q,"},{"Start":"04:23.600 ","End":"04:26.870","Text":"E, D, or you could do A."},{"Start":"04:26.870 ","End":"04:30.365","Text":"Sometimes people do a square box and they fill it in."},{"Start":"04:30.365 ","End":"04:32.825","Text":"That is the Part 2 done,"},{"Start":"04:32.825 ","End":"04:38.910","Text":"and we\u0027ll see you in the final video of this 3 part series."}],"ID":31494},{"Watched":false,"Name":"Sum of Subspaces - Part 3","Duration":"5m 4s","ChapterTopicVideoID":28182,"CourseChapterTopicPlaylistID":246319,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.020","Text":"Welcome back. This is the final video"},{"Start":"00:04.020 ","End":"00:08.505","Text":"in our 3 part series of this linear algebra problem."},{"Start":"00:08.505 ","End":"00:11.085","Text":"Just to recap, so far,"},{"Start":"00:11.085 ","End":"00:16.710","Text":"we have proved that this set U is in fact a subspace of W. We have"},{"Start":"00:16.710 ","End":"00:24.450","Text":"shown that this is the smallest possible subspace that contains both these sets A and B,"},{"Start":"00:24.450 ","End":"00:28.350","Text":"which remember we\u0027re both subspaces of W initially."},{"Start":"00:28.350 ","End":"00:30.135","Text":"Now in this final video,"},{"Start":"00:30.135 ","End":"00:32.895","Text":"which is probably going to be quite quick,"},{"Start":"00:32.895 ","End":"00:37.865","Text":"we\u0027re going to give examples subspace is A and B in which"},{"Start":"00:37.865 ","End":"00:44.935","Text":"the set U or the subspace U is equal to the direct sum of A and B."},{"Start":"00:44.935 ","End":"00:48.085","Text":"How are we going to start is,"},{"Start":"00:48.085 ","End":"00:52.850","Text":"it\u0027d be useful if we understood exactly and formally what it"},{"Start":"00:52.850 ","End":"00:58.645","Text":"means for 2 things to be a direct sum of another set."},{"Start":"00:58.645 ","End":"01:03.455","Text":"Let\u0027s write down the formal definition for that now."},{"Start":"01:03.455 ","End":"01:08.015","Text":"Here is our definition for direct sums."},{"Start":"01:08.015 ","End":"01:12.505","Text":"If A and B are direct sums of U,"},{"Start":"01:12.505 ","End":"01:18.450","Text":"then that means that U is equal to the set A plus the set B,"},{"Start":"01:18.450 ","End":"01:25.395","Text":"but the intersection of the 2 sets A and B is the empty set."},{"Start":"01:25.395 ","End":"01:31.970","Text":"In other words, there are no elements that belong to both A and B."},{"Start":"01:31.970 ","End":"01:35.900","Text":"The overlap is just equal to the empty set."},{"Start":"01:35.900 ","End":"01:38.255","Text":"That doesn\u0027t mean that 0 belongs to both."},{"Start":"01:38.255 ","End":"01:43.450","Text":"It means that there are exactly 0 elements belonging to both."},{"Start":"01:43.450 ","End":"01:51.200","Text":"For an example, let\u0027s just write this one down and then we can discuss why this is true."},{"Start":"01:51.200 ","End":"01:57.915","Text":"If we let U equal R^2,"},{"Start":"01:57.915 ","End":"02:02.700","Text":"so this is just coordinate pairs."},{"Start":"02:02.700 ","End":"02:08.690","Text":"You might have seen this before as maybe the sets (x,"},{"Start":"02:08.690 ","End":"02:15.440","Text":"y), where x and y are just real valued numbers."},{"Start":"02:15.440 ","End":"02:17.780","Text":"If we let U equal R^2,"},{"Start":"02:17.780 ","End":"02:25.785","Text":"and then we take A as being equal to the span of (1,0)."},{"Start":"02:25.785 ","End":"02:36.660","Text":"Sorry, so we take A is equal to the span of (1,0) and B is equal to the span of (0,1)."},{"Start":"02:36.660 ","End":"02:43.580","Text":"Then this does indeed satisfy our conditions for a direct sum."},{"Start":"02:43.580 ","End":"02:46.565","Text":"Let\u0027s see how it applies to 1,"},{"Start":"02:46.565 ","End":"02:49.480","Text":"so A plus B."},{"Start":"02:49.480 ","End":"02:51.015","Text":"Let\u0027s just do that here."},{"Start":"02:51.015 ","End":"02:55.605","Text":"A plus B is equal to the span"},{"Start":"02:55.605 ","End":"03:02.400","Text":"of (0,1) plus the span of (1,0)."},{"Start":"03:02.400 ","End":"03:07.695","Text":"This is just equal to the span of (1,1)."},{"Start":"03:07.695 ","End":"03:12.220","Text":"That\u0027s another way of just writing R^2."},{"Start":"03:13.160 ","End":"03:17.840","Text":"We\u0027ve shown that A plus B is equal to R^2,"},{"Start":"03:17.840 ","End":"03:20.240","Text":"which remember, is equal to U,"},{"Start":"03:20.240 ","End":"03:22.280","Text":"because that\u0027s what we define."},{"Start":"03:22.280 ","End":"03:26.830","Text":"We\u0027re happy with this first condition here."},{"Start":"03:26.830 ","End":"03:30.770","Text":"Then the only other one that we need to satisfy is that"},{"Start":"03:30.770 ","End":"03:35.045","Text":"the intersection of A and B is the empty set."},{"Start":"03:35.045 ","End":"03:40.070","Text":"The intersection of A and B is the intersection of"},{"Start":"03:40.070 ","End":"03:47.085","Text":"span of (1,0) with the span of (0,1)."},{"Start":"03:47.085 ","End":"03:56.415","Text":"Clearly the intersection of these things is just equal to the empty set."},{"Start":"03:56.415 ","End":"04:00.750","Text":"Because another way you can look at the span of (0,1),"},{"Start":"04:00.750 ","End":"04:04.605","Text":"so span of (0,1)."},{"Start":"04:04.605 ","End":"04:06.480","Text":"This is just the set,"},{"Start":"04:06.480 ","End":"04:10.485","Text":"we can say Alpha of (0,1),"},{"Start":"04:10.485 ","End":"04:16.520","Text":"where Alpha belongs to the real numbers or is just some scalar and"},{"Start":"04:16.520 ","End":"04:22.755","Text":"then span of (1,0) is just equal to,"},{"Start":"04:22.755 ","End":"04:26.280","Text":"let\u0027s just say Beta of (1,0),"},{"Start":"04:26.280 ","End":"04:30.559","Text":"where Beta is just some scalar."},{"Start":"04:30.559 ","End":"04:37.070","Text":"Well, the intersection of these 2 things has to be the empty set because there\u0027s"},{"Start":"04:37.070 ","End":"04:41.030","Text":"no elements here or there\u0027s no quantity that regardless"},{"Start":"04:41.030 ","End":"04:45.260","Text":"of our choice for Alpha and Beta that when we take the intersection,"},{"Start":"04:45.260 ","End":"04:47.685","Text":"these 2 things can be equal."},{"Start":"04:47.685 ","End":"04:56.700","Text":"That is the justification for our choices of U, A, and B."},{"Start":"04:56.700 ","End":"05:04.440","Text":"That concludes this 3 part video series on this particular linear algebra question."}],"ID":31495},{"Watched":false,"Name":"The Dimension Formula","Duration":"3m 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example and then also explain what I mean by an ordered basis."},{"Start":"00:12.675 ","End":"00:18.490","Text":"We\u0027ll consider the vectors base R_3, 3-dimensional real space."},{"Start":"00:19.010 ","End":"00:21.960","Text":"We\u0027ll take the basis B,"},{"Start":"00:21.960 ","End":"00:26.730","Text":"and I\u0027ll just ask you to take my word for it that this is indeed a basis."},{"Start":"00:26.730 ","End":"00:29.905","Text":"I don\u0027t want to spend time verifying that."},{"Start":"00:29.905 ","End":"00:32.780","Text":"But all you would have to do would be to show that these"},{"Start":"00:32.780 ","End":"00:35.150","Text":"3 are linearly independent because the 3 of them,"},{"Start":"00:35.150 ","End":"00:37.425","Text":"anyway, take my word for it,"},{"Start":"00:37.425 ","End":"00:39.270","Text":"this is a basis."},{"Start":"00:39.270 ","End":"00:45.165","Text":"Now, in this section makes a difference, the order."},{"Start":"00:45.165 ","End":"00:47.025","Text":"The basis is in a specific order."},{"Start":"00:47.025 ","End":"00:50.710","Text":"This is the 1st, this is the 2nd, this is the 3rd."},{"Start":"00:51.260 ","End":"00:55.820","Text":"It\u0027s not quite a set because in a set the order doesn\u0027t make a difference,"},{"Start":"00:55.820 ","End":"00:57.155","Text":"but here it does."},{"Start":"00:57.155 ","End":"00:59.705","Text":"That\u0027s as far as ordered goes."},{"Start":"00:59.705 ","End":"01:05.195","Text":"Now, because it\u0027s a basis of R_3 of course,"},{"Start":"01:05.195 ","End":"01:12.445","Text":"then any vector in R_3 can be written as a linear combination of these 3,"},{"Start":"01:12.445 ","End":"01:15.350","Text":"and in only 1 way."},{"Start":"01:15.350 ","End":"01:19.670","Text":"It\u0027ll be something times this plus something times this plus something times this."},{"Start":"01:19.670 ","End":"01:22.640","Text":"I\u0027ll take just some random example."},{"Start":"01:22.640 ","End":"01:27.005","Text":"Let\u0027s take the vector 2, 8, 12 in R_3."},{"Start":"01:27.005 ","End":"01:35.580","Text":"You could check that this is equal to minus 2 times the 1st element of the basis,"},{"Start":"01:35.580 ","End":"01:37.740","Text":"and then plus 4 times the 2nd,"},{"Start":"01:37.740 ","End":"01:40.080","Text":"plus 10 times the 3rd."},{"Start":"01:40.080 ","End":"01:42.390","Text":"How I got to this minus 2,"},{"Start":"01:42.390 ","End":"01:45.320","Text":"4,10 is not important at this moment."},{"Start":"01:45.320 ","End":"01:47.600","Text":"In fact, I might have even cooked it up backwards."},{"Start":"01:47.600 ","End":"01:49.700","Text":"I might\u0027ve started with this and given you"},{"Start":"01:49.700 ","End":"01:55.770","Text":"this just to verify that this is so with computation."},{"Start":"01:57.080 ","End":"02:00.235","Text":"Now that we have these numbers,"},{"Start":"02:00.235 ","End":"02:05.125","Text":"these coefficients of the basis,"},{"Start":"02:05.125 ","End":"02:12.220","Text":"we say that the coordinate vector of the vector u, which is 2, 8,"},{"Start":"02:12.220 ","End":"02:16.000","Text":"12 with respect to the basis B,"},{"Start":"02:16.000 ","End":"02:18.030","Text":"it\u0027s got to be an ordered basis,"},{"Start":"02:18.030 ","End":"02:25.940","Text":"is we write it as u in a square brackets with the name of the basis here."},{"Start":"02:25.940 ","End":"02:30.795","Text":"These 3 components are these 3 scalars from here,"},{"Start":"02:30.795 ","End":"02:34.320","Text":"and that\u0027s what gives us the coordinate vector."},{"Start":"02:34.320 ","End":"02:35.920","Text":"Remember, the basis,"},{"Start":"02:35.920 ","End":"02:41.034","Text":"you have to keep the order of the 3 basis elements."},{"Start":"02:41.034 ","End":"02:46.185","Text":"If we didn\u0027t give the name u,"},{"Start":"02:46.185 ","End":"02:49.670","Text":"we could just write it like this to put the vector itself"},{"Start":"02:49.670 ","End":"02:52.970","Text":"in the square brackets to indicate the name of the basis,"},{"Start":"02:52.970 ","End":"02:55.745","Text":"and this is the coordinate vector."},{"Start":"02:55.745 ","End":"02:59.275","Text":"Here\u0027s the 2nd example."},{"Start":"02:59.275 ","End":"03:04.080","Text":"Another random vector 7, 2, 7."},{"Start":"03:04.080 ","End":"03:07.230","Text":"Check that this holds."},{"Start":"03:07.230 ","End":"03:10.050","Text":"It doesn\u0027t matter where I got the 2,"},{"Start":"03:10.050 ","End":"03:11.985","Text":"5, 0 from."},{"Start":"03:11.985 ","End":"03:14.235","Text":"Later, we\u0027ll learn how to find them,"},{"Start":"03:14.235 ","End":"03:15.920","Text":"but I might have just cooked it up by"},{"Start":"03:15.920 ","End":"03:19.115","Text":"starting from the right-hand side and getting to the left."},{"Start":"03:19.115 ","End":"03:21.770","Text":"Anyway, just verify this."},{"Start":"03:21.770 ","End":"03:24.125","Text":"When you\u0027ve checked that,"},{"Start":"03:24.125 ","End":"03:33.290","Text":"then we\u0027ll be able to say that the coordinate vector of this vector 7,"},{"Start":"03:33.290 ","End":"03:39.750","Text":"2, 7, v is the vector 2, 5, 0,"},{"Start":"03:39.750 ","End":"03:41.850","Text":"which I get from these 3 here."},{"Start":"03:41.850 ","End":"03:48.855","Text":"Again, we put it in a square brackets with the name of the ordered basis here,"},{"Start":"03:48.855 ","End":"03:53.070","Text":"and if we haven\u0027t given this a name v,"},{"Start":"03:53.070 ","End":"03:57.890","Text":"we can just write like so."},{"Start":"03:57.890 ","End":"04:01.110","Text":"Our previous example was R_3."},{"Start":"04:01.110 ","End":"04:03.735","Text":"Now let\u0027s look at another space."},{"Start":"04:03.735 ","End":"04:08.205","Text":"The space of 2 by 2 matrices over the real numbers,"},{"Start":"04:08.205 ","End":"04:12.130","Text":"the short, just M_2 by 2."},{"Start":"04:12.220 ","End":"04:16.370","Text":"I\u0027ll ask you to take my word for it that this is a basis."},{"Start":"04:16.370 ","End":"04:19.190","Text":"At least you can see it has the right number of elements because 2"},{"Start":"04:19.190 ","End":"04:22.505","Text":"times 2 is 4 and there are 4 elements here."},{"Start":"04:22.505 ","End":"04:25.130","Text":"If you are going to check, you would just need to"},{"Start":"04:25.130 ","End":"04:27.380","Text":"check that they are linearly independent anyway."},{"Start":"04:27.380 ","End":"04:29.570","Text":"We\u0027ll take it on trust."},{"Start":"04:29.570 ","End":"04:33.095","Text":"Since it\u0027s a basis, it also spans,"},{"Start":"04:33.095 ","End":"04:38.030","Text":"which means that any matrix in"},{"Start":"04:38.030 ","End":"04:44.990","Text":"this space can be written as a linear combination of these 4 matrices,"},{"Start":"04:44.990 ","End":"04:48.275","Text":"and as I said the order is important."},{"Start":"04:48.275 ","End":"04:54.155","Text":"Let\u0027s take some random example from this space,"},{"Start":"04:54.155 ","End":"04:55.610","Text":"call it A,"},{"Start":"04:55.610 ","End":"04:59.640","Text":"and it\u0027s minus 2, 4,13, 28."},{"Start":"04:59.810 ","End":"05:05.420","Text":"I\u0027d like you to check that this matrix A can be written as"},{"Start":"05:05.420 ","End":"05:11.000","Text":"this linear combination of the 4 basis elements."},{"Start":"05:11.000 ","End":"05:13.370","Text":"Do the computation, verify it."},{"Start":"05:13.370 ","End":"05:17.090","Text":"It doesn\u0027t matter how I got to these numbers,"},{"Start":"05:17.090 ","End":"05:18.590","Text":"we\u0027ll see that later."},{"Start":"05:18.590 ","End":"05:20.330","Text":"There is a way of finding them,"},{"Start":"05:20.330 ","End":"05:24.760","Text":"but I might have just cooked it up from right to left."},{"Start":"05:24.760 ","End":"05:29.470","Text":"Anyway, once you\u0027ve seen that this is so,"},{"Start":"05:29.980 ","End":"05:34.925","Text":"then we could say that the coordinate vector of"},{"Start":"05:34.925 ","End":"05:41.000","Text":"this matrix A relative to the basis B is given by,"},{"Start":"05:41.000 ","End":"05:45.240","Text":"and we just put the name of the,"},{"Start":"05:45.550 ","End":"05:50.610","Text":"it\u0027s a vector and it\u0027s also a matrix, in square brackets."},{"Start":"05:50.610 ","End":"05:54.735","Text":"We put the B as a subscript as before,"},{"Start":"05:54.735 ","End":"05:58.490","Text":"and these numbers are the coordinates."},{"Start":"05:58.490 ","End":"06:01.715","Text":"If we didn\u0027t give the matrix a name,"},{"Start":"06:01.715 ","End":"06:04.745","Text":"we could also just write it like this."},{"Start":"06:04.745 ","End":"06:07.535","Text":"The coordinates of this matrix,"},{"Start":"06:07.535 ","End":"06:12.310","Text":"the coordinate vector with respect to the base is this."},{"Start":"06:12.310 ","End":"06:16.985","Text":"Now it\u0027s time to discuss how we can actually compute"},{"Start":"06:16.985 ","End":"06:23.930","Text":"a coordinate vector of a given vector relative to a given basis."},{"Start":"06:23.930 ","End":"06:27.255","Text":"As usual, we\u0027ll do it with examples."},{"Start":"06:27.255 ","End":"06:35.380","Text":"I\u0027ll use the example we had earlier for this ordered basis of R_3."},{"Start":"06:35.380 ","End":"06:40.130","Text":"Go back and check and see this is for the example we had."},{"Start":"06:40.130 ","End":"06:42.720","Text":"We actually had 2 examples,"},{"Start":"06:42.720 ","End":"06:44.170","Text":"remember we had u,"},{"Start":"06:44.170 ","End":"06:47.395","Text":"which was this, and we had v, which was this."},{"Start":"06:47.395 ","End":"06:51.550","Text":"In each case, I pulled these numbers out of a hat,"},{"Start":"06:51.550 ","End":"06:53.740","Text":"so to speak, the minus 2, 4, 10."},{"Start":"06:53.740 ","End":"06:55.615","Text":"I just asked you to verify,"},{"Start":"06:55.615 ","End":"06:56.964","Text":"and once we verify,"},{"Start":"06:56.964 ","End":"07:01.750","Text":"then we said the coordinate vector of"},{"Start":"07:01.750 ","End":"07:08.450","Text":"u with respect to the basis B is this from these numbers,"},{"Start":"07:08.450 ","End":"07:13.990","Text":"and we said that the coordinate vector of this"},{"Start":"07:13.990 ","End":"07:20.970","Text":"v with respect to this same ordered basis was 2, 5, 0."},{"Start":"07:20.970 ","End":"07:22.470","Text":"All these we had,"},{"Start":"07:22.470 ","End":"07:23.810","Text":"but the question is,"},{"Start":"07:23.810 ","End":"07:29.360","Text":"how would we compute these numbers if I didn\u0027t just produce them?"},{"Start":"07:29.360 ","End":"07:31.520","Text":"The idea is,"},{"Start":"07:31.520 ","End":"07:32.540","Text":"instead of taking,"},{"Start":"07:32.540 ","End":"07:33.650","Text":"say, 2, 8,"},{"Start":"07:33.650 ","End":"07:35.810","Text":"12 or 7, 2, 7,"},{"Start":"07:35.810 ","End":"07:42.555","Text":"is to take a general vector like x, y, z."},{"Start":"07:42.555 ","End":"07:46.910","Text":"If I do the computation for a general x, y, z,"},{"Start":"07:46.910 ","End":"07:51.110","Text":"then I can afterwards plug in whatever I want,"},{"Start":"07:51.110 ","End":"07:52.700","Text":"2, 8, 12, 7,"},{"Start":"07:52.700 ","End":"07:55.055","Text":"2, 7, or anything else."},{"Start":"07:55.055 ","End":"07:57.440","Text":"Here\u0027s our vector, x, y, z,"},{"Start":"07:57.440 ","End":"08:05.195","Text":"give it a name w. You want to find it as the linear combination of the basis vectors."},{"Start":"08:05.195 ","End":"08:10.580","Text":"These scalars a, b, c are the unknowns."},{"Start":"08:10.580 ","End":"08:12.890","Text":"In other words, we want to find a, b,"},{"Start":"08:12.890 ","End":"08:14.870","Text":"and c in terms of x,"},{"Start":"08:14.870 ","End":"08:16.010","Text":"y, z as if x,"},{"Start":"08:16.010 ","End":"08:18.190","Text":"y, z were known."},{"Start":"08:18.190 ","End":"08:22.595","Text":"I multiply the scalars and then I do the addition,"},{"Start":"08:22.595 ","End":"08:24.894","Text":"then we get this."},{"Start":"08:24.894 ","End":"08:32.370","Text":"This gives us a system of 3 linear equations and 3 unknowns,"},{"Start":"08:32.370 ","End":"08:36.700","Text":"a, b, c. We\u0027ll do it using matrices."},{"Start":"08:36.700 ","End":"08:41.760","Text":"This is the corresponding augmented matrix."},{"Start":"08:41.760 ","End":"08:46.550","Text":"What we\u0027re going to do is not just bring it into row echelon form."},{"Start":"08:46.550 ","End":"08:53.500","Text":"Let\u0027s make this part the identity matrix and then it will be easier."},{"Start":"08:53.500 ","End":"08:55.460","Text":"This has 1 way of doing it."},{"Start":"08:55.460 ","End":"09:02.450","Text":"I\u0027ll start out by subtracting this top row from the other 2."},{"Start":"09:02.450 ","End":"09:04.490","Text":"In row notation,"},{"Start":"09:04.490 ","End":"09:06.185","Text":"this is what I mean."},{"Start":"09:06.185 ","End":"09:10.130","Text":"If we do that, we get this."},{"Start":"09:10.130 ","End":"09:14.135","Text":"This is already in row echelon form, but we\u0027re continuing."},{"Start":"09:14.135 ","End":"09:18.455","Text":"Let\u0027s subtract this row from this row."},{"Start":"09:18.455 ","End":"09:20.270","Text":"This is the notation,"},{"Start":"09:20.270 ","End":"09:23.840","Text":"and then we get this."},{"Start":"09:23.840 ","End":"09:29.040","Text":"The next thing is to add this row to this row,"},{"Start":"09:29.040 ","End":"09:31.320","Text":"and that will make this 0."},{"Start":"09:31.320 ","End":"09:40.120","Text":"Then finally, we get this and all we have to do is multiply the middle row by minus 1."},{"Start":"09:40.120 ","End":"09:43.505","Text":"We have the identity matrix here,"},{"Start":"09:43.505 ","End":"09:48.830","Text":"which means that we have exactly what a,"},{"Start":"09:48.830 ","End":"09:52.330","Text":"b, and c are in terms of x, y, z."},{"Start":"09:52.330 ","End":"09:55.685","Text":"If we go back to where we came from,"},{"Start":"09:55.685 ","End":"10:01.580","Text":"we had that w was a times this vector plus b times this,"},{"Start":"10:01.580 ","End":"10:03.110","Text":"plus c times this."},{"Start":"10:03.110 ","End":"10:05.240","Text":"I just replaced a, b,"},{"Start":"10:05.240 ","End":"10:09.000","Text":"and c with what they are from here."},{"Start":"10:09.080 ","End":"10:11.600","Text":"We have this formula,"},{"Start":"10:11.600 ","End":"10:13.159","Text":"and now we can convert"},{"Start":"10:13.159 ","End":"10:22.810","Text":"any vector into its coordinate vector."},{"Start":"10:22.810 ","End":"10:26.505","Text":"For instance, let\u0027s take 1, 2, 3."},{"Start":"10:26.505 ","End":"10:29.095","Text":"I just follow this recipe."},{"Start":"10:29.095 ","End":"10:33.500","Text":"These 3 basis vectors stay and all I do is this computation."},{"Start":"10:33.500 ","End":"10:35.825","Text":"So x plus y minus z,"},{"Start":"10:35.825 ","End":"10:40.020","Text":"1 plus 2 minus 3, and so on."},{"Start":"10:40.800 ","End":"10:43.090","Text":"Here we get 0,"},{"Start":"10:43.090 ","End":"10:45.250","Text":"here 1, and here 2,"},{"Start":"10:45.250 ","End":"10:49.870","Text":"which means that the coordinate vector of this 1, 2,"},{"Start":"10:49.870 ","End":"10:54.775","Text":"3 with respect to that basis is 0, 1, 2."},{"Start":"10:54.775 ","End":"10:57.595","Text":"These are the 0, 1, 2 from here."},{"Start":"10:57.595 ","End":"11:00.220","Text":"One more example, take 4,"},{"Start":"11:00.220 ","End":"11:02.200","Text":"5, 6 this time."},{"Start":"11:02.200 ","End":"11:04.990","Text":"Same setup we put in 4,"},{"Start":"11:04.990 ","End":"11:07.180","Text":"5, 6 for x, y, z."},{"Start":"11:07.180 ","End":"11:13.390","Text":"Compute these 3 scalar coefficients."},{"Start":"11:13.390 ","End":"11:18.430","Text":"Make these into a vector and that\u0027s the coordinate vector of 4,"},{"Start":"11:18.430 ","End":"11:21.175","Text":"5, 6 with respect to the basis B."},{"Start":"11:21.175 ","End":"11:25.165","Text":"In general, we can say that our w,"},{"Start":"11:25.165 ","End":"11:28.810","Text":"which is the arbitrary vector x, y, z"},{"Start":"11:28.810 ","End":"11:33.145","Text":"has this expression here as"},{"Start":"11:33.145 ","End":"11:39.610","Text":"the coordinate vector with respect to the bases."},{"Start":"11:39.610 ","End":"11:44.125","Text":"Now another example, the last one was from R3."},{"Start":"11:44.125 ","End":"11:50.365","Text":"This time we\u0027re going to take the example from the space M_2x2,"},{"Start":"11:50.365 ","End":"11:52.450","Text":"which you write like this."},{"Start":"11:52.450 ","End":"11:58.735","Text":"Just look back and recall that we had this as a basis for that space,"},{"Start":"11:58.735 ","End":"12:02.005","Text":"these 4 elements matrices."},{"Start":"12:02.005 ","End":"12:08.125","Text":"Again, I\u0027m just reminding you that we saw that this equation holds."},{"Start":"12:08.125 ","End":"12:11.830","Text":"This gives us this matrix in terms of"},{"Start":"12:11.830 ","End":"12:16.975","Text":"a linear combination of the basis matrices or vectors."},{"Start":"12:16.975 ","End":"12:20.110","Text":"These numbers minus 2, 4, 3, 2"},{"Start":"12:20.110 ","End":"12:30.440","Text":"gave us the coordinate vector of this matrix relative to the ordered basis B up here."},{"Start":"12:30.930 ","End":"12:34.390","Text":"What I haven\u0027t shown you is how you would go"},{"Start":"12:34.390 ","End":"12:37.030","Text":"about computing them if I just gave you this."},{"Start":"12:37.030 ","End":"12:39.130","Text":"I pulled them out of a hat."},{"Start":"12:39.130 ","End":"12:42.685","Text":"It turns out that there is a methodology."},{"Start":"12:42.685 ","End":"12:51.100","Text":"The idea is to just take 1 computation for a general matrix, x, y, z, t."},{"Start":"12:51.100 ","End":"12:53.635","Text":"If I computed it for this,"},{"Start":"12:53.635 ","End":"12:56.710","Text":"then afterwards I could substitute x equals minus 2,"},{"Start":"12:56.710 ","End":"12:58.750","Text":"y equals 4, and so on,"},{"Start":"12:58.750 ","End":"13:02.480","Text":"or any other matrix I please."},{"Start":"13:02.610 ","End":"13:05.710","Text":"We take this x, y, z,"},{"Start":"13:05.710 ","End":"13:10.495","Text":"t matrix as a given and we want to find a,"},{"Start":"13:10.495 ","End":"13:14.410","Text":"b, c, and d such that this holds."},{"Start":"13:14.410 ","End":"13:16.870","Text":"In other words, we want to compute a, b,"},{"Start":"13:16.870 ","End":"13:20.170","Text":"c, and d in terms of x, y, z,"},{"Start":"13:20.170 ","End":"13:27.615","Text":"and t. Moving through this quickly multiply out the scalars in each of these 4."},{"Start":"13:27.615 ","End":"13:31.890","Text":"Then we do the addition for each place like the upper-left,"},{"Start":"13:31.890 ","End":"13:34.050","Text":"we add them all together and so on."},{"Start":"13:34.050 ","End":"13:37.010","Text":"Then we get this."},{"Start":"13:37.010 ","End":"13:46.120","Text":"From this, we can get 4 equations like x is equal to a,"},{"Start":"13:46.120 ","End":"13:47.620","Text":"only I just wrote it backwards."},{"Start":"13:47.620 ","End":"13:48.970","Text":"a equals x and so on,"},{"Start":"13:48.970 ","End":"13:51.325","Text":"y equals 2a plus 2b."},{"Start":"13:51.325 ","End":"13:53.905","Text":"I wrote the x, y, z, t on the right."},{"Start":"13:53.905 ","End":"14:00.340","Text":"Notice that I took these 4 in a particular order, x, y, z,"},{"Start":"14:00.340 ","End":"14:08.245","Text":"t. We want to be consistent as I want to show you something about this in a moment."},{"Start":"14:08.245 ","End":"14:15.880","Text":"The augmented matrix we get from this system of linear equations is this."},{"Start":"14:15.880 ","End":"14:24.580","Text":"There\u0027s a shortcut way of getting to this almost immediately once we have the basis."},{"Start":"14:24.580 ","End":"14:30.760","Text":"From the basis, we can get these columns if we just follow the same order,"},{"Start":"14:30.760 ","End":"14:32.620","Text":"this snake, so to speak,"},{"Start":"14:32.620 ","End":"14:36.640","Text":"1, 2, 3, 4 gives us 1, 2, 3,"},{"Start":"14:36.640 ","End":"14:43.360","Text":"4, 0 2, 4, 4, 0, 2, 4, 4,"},{"Start":"14:43.360 ","End":"14:46.810","Text":"0, 0, 1, 4, 0, 0,"},{"Start":"14:46.810 ","End":"14:50.290","Text":"1, 4, and 0, 0,"},{"Start":"14:50.290 ","End":"14:55.765","Text":"0, 4, 0, 0, 0, 4."},{"Start":"14:55.765 ","End":"15:02.200","Text":"Now we want to solve this equation in the matrix form."},{"Start":"15:02.200 ","End":"15:06.295","Text":"One way of solving it is to bring this into row echelon form."},{"Start":"15:06.295 ","End":"15:10.690","Text":"But actually, we can also continue doing row operations and until we"},{"Start":"15:10.690 ","End":"15:16.595","Text":"get the identity matrix here, and then the answer comes out quicker."},{"Start":"15:16.595 ","End":"15:19.155","Text":"I\u0027ll go through this quickly."},{"Start":"15:19.155 ","End":"15:24.345","Text":"We get this by subtracting multiples of the first row from the others,"},{"Start":"15:24.345 ","End":"15:27.405","Text":"twice this, from this, and so on."},{"Start":"15:27.405 ","End":"15:31.655","Text":"Not to forget to do it also to the right-hand column."},{"Start":"15:31.655 ","End":"15:33.745","Text":"That\u0027s the first step."},{"Start":"15:33.745 ","End":"15:38.995","Text":"Then from here, we can subtract twice this row from each of these."},{"Start":"15:38.995 ","End":"15:44.630","Text":"Now that gives us zeros here and this is what we have left."},{"Start":"15:44.760 ","End":"15:51.730","Text":"Then we subtract the third row 4 times from the fourth row."},{"Start":"15:51.730 ","End":"15:54.010","Text":"That gives us this."},{"Start":"15:54.010 ","End":"15:57.880","Text":"Notice that this is actually in diagonal form already."},{"Start":"15:57.880 ","End":"16:01.900","Text":"All I have to do is divide this by 2 and this by 4."},{"Start":"16:01.900 ","End":"16:04.240","Text":"I\u0027m leaving you to check the computations."},{"Start":"16:04.240 ","End":"16:05.755","Text":"This is what we get."},{"Start":"16:05.755 ","End":"16:09.475","Text":"From here, we can actually get everything."},{"Start":"16:09.475 ","End":"16:12.100","Text":"We get a, b, c,"},{"Start":"16:12.100 ","End":"16:20.455","Text":"and d as what\u0027s in this right column, these 4 expressions."},{"Start":"16:20.455 ","End":"16:23.650","Text":"Now if you look back as to what a, b, c,"},{"Start":"16:23.650 ","End":"16:25.210","Text":"and d where they were,"},{"Start":"16:25.210 ","End":"16:26.890","Text":"the numbers here a, b,"},{"Start":"16:26.890 ","End":"16:33.100","Text":"c, and d. Such that this general, the x, y, z,"},{"Start":"16:33.100 ","End":"16:38.620","Text":"t is equal to the linear combination of the basis vectors with a, b, c,"},{"Start":"16:38.620 ","End":"16:44.350","Text":"d. These 4 quantities in green,"},{"Start":"16:44.350 ","End":"16:50.215","Text":"a, b, c, and d are the components of the coordinate vectors."},{"Start":"16:50.215 ","End":"16:52.240","Text":"If I just put them here,"},{"Start":"16:52.240 ","End":"17:00.504","Text":"I get that the coordinate vector of this matrix V with respect to basis B is this."},{"Start":"17:00.504 ","End":"17:03.700","Text":"I\u0027m returning to our example that we had before."},{"Start":"17:03.700 ","End":"17:07.360","Text":"In order to verify this double-check,"},{"Start":"17:07.360 ","End":"17:11.290","Text":"we had minus 2, 4, 13, 28."},{"Start":"17:11.290 ","End":"17:14.080","Text":"Now if I do it according to this computation,"},{"Start":"17:14.080 ","End":"17:15.850","Text":"I see x here."},{"Start":"17:15.850 ","End":"17:17.785","Text":"I put minus 2."},{"Start":"17:17.785 ","End":"17:20.500","Text":"Instead of y I put 4 instead of z,"},{"Start":"17:20.500 ","End":"17:23.004","Text":"I put 13, instead of t I put 28,"},{"Start":"17:23.004 ","End":"17:25.165","Text":"and all of this green."},{"Start":"17:25.165 ","End":"17:31.090","Text":"If we do that and do the computation,"},{"Start":"17:31.090 ","End":"17:32.320","Text":"minus 2 is minus 2."},{"Start":"17:32.320 ","End":"17:33.790","Text":"This thing, if you compute,"},{"Start":"17:33.790 ","End":"17:35.365","Text":"it comes out to be 4."},{"Start":"17:35.365 ","End":"17:38.695","Text":"This, I don\u0027t know 13 minus 8,"},{"Start":"17:38.695 ","End":"17:41.530","Text":"minus 2 is 3, and so on."},{"Start":"17:41.530 ","End":"17:44.230","Text":"We get these 4 numbers."},{"Start":"17:44.230 ","End":"17:52.660","Text":"The coordinate vector of this matrix with respect to the basis that we had is minus 2,"},{"Start":"17:52.660 ","End":"17:54.610","Text":"4, 3, 2."},{"Start":"17:54.610 ","End":"17:56.800","Text":"I\u0027m just noting that\u0027s what we got earlier,"},{"Start":"17:56.800 ","End":"18:03.050","Text":"so that\u0027s a double-check and we are done."}],"ID":26730},{"Watched":false,"Name":"Exercise 1 Part a","Duration":"3m 50s","ChapterTopicVideoID":25921,"CourseChapterTopicPlaylistID":246320,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.330","Text":"This exercise is the first of a set of 3 related exercises."},{"Start":"00:06.330 ","End":"00:09.945","Text":"In all of them, it involves"},{"Start":"00:09.945 ","End":"00:16.890","Text":"this space polynomials of degree up to 2 with real coefficients."},{"Start":"00:16.890 ","End":"00:19.350","Text":"Just like these where a,"},{"Start":"00:19.350 ","End":"00:20.700","Text":"b, and c are real numbers,"},{"Start":"00:20.700 ","End":"00:23.940","Text":"we\u0027re given 2 bases,"},{"Start":"00:23.940 ","End":"00:28.780","Text":"we just accept it that these are bases."},{"Start":"00:30.050 ","End":"00:33.000","Text":"Well, in this particular exercise,"},{"Start":"00:33.000 ","End":"00:35.340","Text":"we don\u0027t use B2, we just use B1."},{"Start":"00:35.340 ","End":"00:41.655","Text":"Our task is to take this general element, a polynomial,"},{"Start":"00:41.655 ","End":"00:50.550","Text":"and write it\u0027s coordinate vector relative to the basis B1."},{"Start":"00:50.550 ","End":"00:54.880","Text":"In the next exercise, we\u0027ll do a relative to B2."},{"Start":"00:55.370 ","End":"01:01.685","Text":"What we do, we can\u0027t just use polynomials directly in a matrix."},{"Start":"01:01.685 ","End":"01:06.320","Text":"We convert them to vectors."},{"Start":"01:06.320 ","End":"01:10.310","Text":"This would be vector a,"},{"Start":"01:10.310 ","End":"01:14.090","Text":"b, c. For example,"},{"Start":"01:14.090 ","End":"01:23.100","Text":"1 plus x would be 1,1,0 and this would be 011."},{"Start":"01:23.600 ","End":"01:26.370","Text":"If we do all this translation,"},{"Start":"01:26.370 ","End":"01:28.920","Text":"it\u0027s as if this space was 3,"},{"Start":"01:28.920 ","End":"01:32.010","Text":"just like 3D vectors."},{"Start":"01:32.010 ","End":"01:38.330","Text":"Our basis B1 becomes this and our polynomial is just the vector V,"},{"Start":"01:38.330 ","End":"01:41.880","Text":"we\u0027ll call it, which is abc."},{"Start":"01:41.920 ","End":"01:47.225","Text":"If we then put them in a matrix,"},{"Start":"01:47.225 ","End":"01:49.940","Text":"we\u0027ll use our usual technique."},{"Start":"01:49.940 ","End":"01:52.999","Text":"We put them in as column vectors."},{"Start":"01:52.999 ","End":"01:56.435","Text":"Notice 110 is this,"},{"Start":"01:56.435 ","End":"01:58.965","Text":"this 1 here is this 1 here,"},{"Start":"01:58.965 ","End":"02:04.530","Text":"and the next 1 010011 and then we put a,"},{"Start":"02:04.530 ","End":"02:08.380","Text":"b, c with a separator, an augmented matrix."},{"Start":"02:08.380 ","End":"02:12.455","Text":"We\u0027re going to do row operations,"},{"Start":"02:12.455 ","End":"02:15.470","Text":"not just to bring it to row echelon form,"},{"Start":"02:15.470 ","End":"02:21.270","Text":"but until we get the identity matrix."},{"Start":"02:21.370 ","End":"02:25.070","Text":"First step, subtract the first row from"},{"Start":"02:25.070 ","End":"02:31.685","Text":"the second row on the left and the right of the separator and this is what we get."},{"Start":"02:31.685 ","End":"02:35.660","Text":"Next. I mean, this is an echelon form,"},{"Start":"02:35.660 ","End":"02:37.400","Text":"but like I said, we\u0027re continuing."},{"Start":"02:37.400 ","End":"02:41.330","Text":"We\u0027ll subtract the third row from"},{"Start":"02:41.330 ","End":"02:46.475","Text":"the second row and now we have the identity matrix here."},{"Start":"02:46.475 ","End":"02:53.150","Text":"What\u0027s important to know these entries to the right of the separator."},{"Start":"02:53.150 ","End":"02:57.920","Text":"These tell us how to write our general a, b,"},{"Start":"02:57.920 ","End":"03:03.720","Text":"c in terms of the 3 basis vectors."},{"Start":"03:04.280 ","End":"03:09.770","Text":"A is here, b minus a minus c here and c here."},{"Start":"03:09.770 ","End":"03:15.305","Text":"You could check, you could actually expand this and simplify it and you should get this."},{"Start":"03:15.305 ","End":"03:23.900","Text":"The coordinate vector would be this here with respect to the basis B1."},{"Start":"03:23.900 ","End":"03:31.010","Text":"What this means is that if we have any polynomial of degree 2 or less,"},{"Start":"03:31.010 ","End":"03:40.310","Text":"then we can write it as a linear combination of these 3 basis vectors from B1,"},{"Start":"03:40.310 ","End":"03:45.000","Text":"1 plus x, x and x plus x squared as follows."},{"Start":"03:46.360 ","End":"03:50.520","Text":"That concludes part 1 of 3."}],"ID":26731},{"Watched":false,"Name":"Exercise 1 Part b","Duration":"3m 30s","ChapterTopicVideoID":25922,"CourseChapterTopicPlaylistID":246320,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.530","Text":"This exercise is part 2 of a 3-part set."},{"Start":"00:05.870 ","End":"00:08.970","Text":"In the previous part,"},{"Start":"00:08.970 ","End":"00:16.870","Text":"we considered the vector space of polynomials up to degree 2 over the real numbers."},{"Start":"00:17.240 ","End":"00:24.300","Text":"We took a general polynomial which would be a plus bx plus cx"},{"Start":"00:24.300 ","End":"00:30.655","Text":"squared and wrote its coordinate vector in terms of the basis B_1."},{"Start":"00:30.655 ","End":"00:33.725","Text":"In this exercise, you want to do the same thing,"},{"Start":"00:33.725 ","End":"00:37.865","Text":"but relative to basis B_2."},{"Start":"00:37.865 ","End":"00:41.210","Text":"We won\u0027t be using B_1 in this part 2."},{"Start":"00:41.210 ","End":"00:43.535","Text":"We used it in part 1."},{"Start":"00:43.535 ","End":"00:46.980","Text":"Just here for completeness."},{"Start":"00:47.180 ","End":"00:52.430","Text":"As before, the technique is to write"},{"Start":"00:52.430 ","End":"00:58.370","Text":"a polynomial as if it was a 3D vector like a,"},{"Start":"00:58.370 ","End":"01:03.110","Text":"b, c. If we do this conversion,"},{"Start":"01:03.110 ","End":"01:07.700","Text":"then our basis B_2 becomes this,"},{"Start":"01:07.700 ","End":"01:13.170","Text":"like 1 plus x squared is 1 plus 0x plus 1x squared."},{"Start":"01:13.170 ","End":"01:15.000","Text":"That\u0027s 1, 0, 1."},{"Start":"01:15.000 ","End":"01:16.710","Text":"X plus x squared is 0, 1,"},{"Start":"01:16.710 ","End":"01:19.125","Text":"1, x squared is 0, 0, 1."},{"Start":"01:19.125 ","End":"01:22.280","Text":"Our general polynomial will be vector a,"},{"Start":"01:22.280 ","End":"01:25.460","Text":"b, c, like here."},{"Start":"01:26.320 ","End":"01:33.110","Text":"What we do is we put"},{"Start":"01:33.110 ","End":"01:39.200","Text":"the members of the basis into a matrix."},{"Start":"01:39.200 ","End":"01:41.420","Text":"These become column vectors."},{"Start":"01:41.420 ","End":"01:42.530","Text":"For example, this 1,"},{"Start":"01:42.530 ","End":"01:44.540","Text":"0, 1 is this 1,"},{"Start":"01:44.540 ","End":"01:46.250","Text":"0, 1, and the 0, 1,"},{"Start":"01:46.250 ","End":"01:48.875","Text":"1 is this, and so on."},{"Start":"01:48.875 ","End":"01:51.500","Text":"We put it into the augmented matrix."},{"Start":"01:51.500 ","End":"01:53.940","Text":"Here we put the a, b,"},{"Start":"01:53.940 ","End":"01:58.600","Text":"c. Next we\u0027re going to do row operations,"},{"Start":"01:58.600 ","End":"02:01.720","Text":"but not just to bring it to row echelon form,"},{"Start":"02:01.720 ","End":"02:06.310","Text":"but we\u0027ll go all the way until the part on the left,"},{"Start":"02:06.310 ","End":"02:09.910","Text":"the restricted part is the identity matrix."},{"Start":"02:09.910 ","End":"02:16.880","Text":"We can start off by subtracting the first row from the last row to get this."},{"Start":"02:17.540 ","End":"02:23.095","Text":"Next we subtract the second row from the last row and get this."},{"Start":"02:23.095 ","End":"02:27.540","Text":"Lo and behold, this is the identity matrix."},{"Start":"02:27.540 ","End":"02:32.645","Text":"What we do now is we look at these entries here."},{"Start":"02:32.645 ","End":"02:37.060","Text":"What these tell us is how to write a general vector a,"},{"Start":"02:37.060 ","End":"02:40.975","Text":"b, c in terms of the 3 basis vectors."},{"Start":"02:40.975 ","End":"02:44.180","Text":"These 3 become the coefficients."},{"Start":"02:44.180 ","End":"02:45.380","Text":"This a is this a,"},{"Start":"02:45.380 ","End":"02:49.500","Text":"this b is this b, c minus a minus b."},{"Start":"02:50.270 ","End":"02:56.870","Text":"The coordinate vector with respect to basis B_2 is a,"},{"Start":"02:56.870 ","End":"03:00.490","Text":"b, c minus a minus b."},{"Start":"03:00.490 ","End":"03:05.020","Text":"What this means, returning to polynomials,"},{"Start":"03:05.020 ","End":"03:09.465","Text":"is that only polynomial from P_2, in other words,"},{"Start":"03:09.465 ","End":"03:11.835","Text":"up to degree 2 a quadratic,"},{"Start":"03:11.835 ","End":"03:19.180","Text":"can be broken up in terms of these 3 basis vectors."},{"Start":"03:20.900 ","End":"03:25.560","Text":"That\u0027s basically it for part 2."},{"Start":"03:25.560 ","End":"03:26.600","Text":"In the next clip,"},{"Start":"03:26.600 ","End":"03:30.030","Text":"we\u0027ll do the last part of the 3."}],"ID":26732},{"Watched":false,"Name":"Exercise 1 Part c","Duration":"2m 56s","ChapterTopicVideoID":25923,"CourseChapterTopicPlaylistID":246320,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.785","Text":"This exercise is the third and last in a 3-part set of exercises that are related,"},{"Start":"00:07.785 ","End":"00:14.340","Text":"all involving quadratic polynomials with real coefficients."},{"Start":"00:14.340 ","End":"00:16.485","Text":"In part 1,"},{"Start":"00:16.485 ","End":"00:20.910","Text":"we dealt with the coordinate vector with respect to B_1,"},{"Start":"00:20.910 ","End":"00:23.280","Text":"and part 2 we did B_2."},{"Start":"00:23.280 ","End":"00:27.210","Text":"Each of these are basis for the space."},{"Start":"00:27.210 ","End":"00:34.605","Text":"In this part, we have to compute the change of basis matrix from B_1 to B_2."},{"Start":"00:34.605 ","End":"00:40.255","Text":"We\u0027re going to follow the recipe for this is the definite set of steps 3 of them."},{"Start":"00:40.255 ","End":"00:49.255","Text":"In step 1, we want to compute the coordinate vector relative to the old basis."},{"Start":"00:49.255 ","End":"00:55.520","Text":"We did this in part 1 and we also wrote the polynomials as vectors."},{"Start":"00:55.520 ","End":"01:02.205","Text":"Those are the 3 vectors from B_1 in this form,"},{"Start":"01:02.205 ","End":"01:08.260","Text":"and these coefficients are the components of the coordinate vector."},{"Start":"01:08.260 ","End":"01:11.710","Text":"Using this, in step 2,"},{"Start":"01:11.710 ","End":"01:19.880","Text":"we\u0027re going to write each vector in the new basis in terms of the old basis."},{"Start":"01:19.880 ","End":"01:23.040","Text":"When we say from B_1 to B_2,"},{"Start":"01:23.040 ","End":"01:27.520","Text":"it means that B_1 is the old and B_2 is the new."},{"Start":"01:27.620 ","End":"01:35.535","Text":"We\u0027re going to write each of these 3 according to this formula."},{"Start":"01:35.535 ","End":"01:38.355","Text":"Let me see."},{"Start":"01:38.355 ","End":"01:41.130","Text":"I copied this down here,"},{"Start":"01:41.130 ","End":"01:43.905","Text":"now we can scroll a bit more."},{"Start":"01:43.905 ","End":"01:46.625","Text":"I gave you all 3 of them at once."},{"Start":"01:46.625 ","End":"01:48.230","Text":"Here\u0027s the 1, 0,"},{"Start":"01:48.230 ","End":"01:49.700","Text":"1, here\u0027s the 0,"},{"Start":"01:49.700 ","End":"01:51.260","Text":"1, 1, here\u0027s the 0,"},{"Start":"01:51.260 ","End":"01:53.240","Text":"0, 1, that\u0027s each of these."},{"Start":"01:53.240 ","End":"01:56.730","Text":"For each of them, I apply this formula."},{"Start":"01:57.200 ","End":"02:00.925","Text":"Let\u0027s take, for example, the first 1."},{"Start":"02:00.925 ","End":"02:07.245","Text":"This is my a, b and c. Just a is 1,"},{"Start":"02:07.245 ","End":"02:14.865","Text":"and then I want b minus a minus c. It 0 minus 1 minus 1 is minus 2,"},{"Start":"02:14.865 ","End":"02:16.785","Text":"and then c is 1,"},{"Start":"02:16.785 ","End":"02:19.665","Text":"similarly with the other 2."},{"Start":"02:19.665 ","End":"02:29.240","Text":"That was step 2. The final step is to take all these numbers that I didn\u0027t color,"},{"Start":"02:29.240 ","End":"02:35.240","Text":"these coefficients, and put them into a matrix but transposed."},{"Start":"02:35.240 ","End":"02:38.330","Text":"Notice, here we have 1 minus 2,"},{"Start":"02:38.330 ","End":"02:41.405","Text":"1, that becomes the first column,"},{"Start":"02:41.405 ","End":"02:45.210","Text":"0, 0, 1 is the second column,"},{"Start":"02:45.210 ","End":"02:46.440","Text":"0 minus 1,"},{"Start":"02:46.440 ","End":"02:48.450","Text":"1 is the third column."},{"Start":"02:48.450 ","End":"02:51.630","Text":"The transpose, don\u0027t forget that."},{"Start":"02:51.630 ","End":"02:54.000","Text":"That\u0027s the answer,"},{"Start":"02:54.000 ","End":"02:56.620","Text":"and we are done."}],"ID":26733},{"Watched":false,"Name":"Exercise 2 Part a","Duration":"6m 25s","ChapterTopicVideoID":25924,"CourseChapterTopicPlaylistID":246320,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:08.055","Text":"In this exercise, we\u0027re given 2 different bases for the same space,"},{"Start":"00:08.055 ","End":"00:13.755","Text":"M_2 over R. This means square 2 by 2 matrices,"},{"Start":"00:13.755 ","End":"00:16.650","Text":"just like you see here."},{"Start":"00:16.650 ","End":"00:22.390","Text":"This is the first of a 3 part set of exercises."},{"Start":"00:22.390 ","End":"00:28.790","Text":"In fact, we won\u0027t be using this base here in this exercise."},{"Start":"00:28.790 ","End":"00:30.965","Text":"It will be used for the next exercise."},{"Start":"00:30.965 ","End":"00:33.680","Text":"In this 1, we\u0027ll be using base B,"},{"Start":"00:33.680 ","End":"00:40.910","Text":"which is these 4 matrices and you could check that they are a base."},{"Start":"00:40.910 ","End":"00:43.580","Text":"You just take my word for it."},{"Start":"00:43.580 ","End":"00:47.600","Text":"I hope that this is a base for this space."},{"Start":"00:47.600 ","End":"00:51.185","Text":"Now, our task is to find the coordinate vector."},{"Start":"00:51.185 ","End":"00:52.835","Text":"Just to be specific,"},{"Start":"00:52.835 ","End":"00:55.910","Text":"I took a general matrix called it X,"},{"Start":"00:55.910 ","End":"00:57.380","Text":"which is x, y, z,"},{"Start":"00:57.380 ","End":"01:01.325","Text":"t and I want to find its coordinate vector."},{"Start":"01:01.325 ","End":"01:02.990","Text":"It\u0027s always relative to a base,"},{"Start":"01:02.990 ","End":"01:09.050","Text":"so relative to B. I\u0027ll let you know already now that next exercise will be"},{"Start":"01:09.050 ","End":"01:15.170","Text":"the same thing but relative to E. Now in order to proceed with our usual methods,"},{"Start":"01:15.170 ","End":"01:18.185","Text":"we have to flatten things out."},{"Start":"01:18.185 ","End":"01:21.290","Text":"Matrices have to become vectors and"},{"Start":"01:21.290 ","End":"01:25.100","Text":"we flatten them out using what I call the snake method."},{"Start":"01:25.100 ","End":"01:27.905","Text":"We take them in order 1, 1, 0, 0,"},{"Start":"01:27.905 ","End":"01:31.620","Text":"this will be vector 0,"},{"Start":"01:31.620 ","End":"01:34.215","Text":"1, 1, 0, and so on."},{"Start":"01:34.215 ","End":"01:41.660","Text":"What we do then is put these into a matrix,"},{"Start":"01:41.660 ","End":"01:43.970","Text":"as well as this x, y, z,"},{"Start":"01:43.970 ","End":"01:49.940","Text":"t. Remember for this kind of exercise, we use columns."},{"Start":"01:49.940 ","End":"01:51.680","Text":"This 1, 1, 0,"},{"Start":"01:51.680 ","End":"01:53.770","Text":"0 is the first column,"},{"Start":"01:53.770 ","End":"01:55.590","Text":"this 1, 0, 1, 1,"},{"Start":"01:55.590 ","End":"01:57.335","Text":"0 is the second column,"},{"Start":"01:57.335 ","End":"01:58.835","Text":"third column, fourth column,"},{"Start":"01:58.835 ","End":"02:03.410","Text":"and the last column we put our variable 1."},{"Start":"02:03.410 ","End":"02:05.270","Text":"It\u0027s an augmented matrix,"},{"Start":"02:05.270 ","End":"02:08.720","Text":"I guess I should have put a vertical separator."},{"Start":"02:08.720 ","End":"02:10.840","Text":"We don\u0027t really need it."},{"Start":"02:10.840 ","End":"02:17.960","Text":"Our task is to get this matrix not into row echelon form,"},{"Start":"02:17.960 ","End":"02:25.050","Text":"but to keep going until we get the identity matrix to the left of the separator."},{"Start":"02:25.610 ","End":"02:33.395","Text":"Here I got the rest of the first column to be 0 under the 1 by subtracting"},{"Start":"02:33.395 ","End":"02:42.110","Text":"the first row from the second row and you can also see here this y became y minus x."},{"Start":"02:42.110 ","End":"02:52.020","Text":"Let\u0027s keep going. Let\u0027s subtract the second row from the third row and then we have this."},{"Start":"02:52.020 ","End":"02:56.135","Text":"We\u0027re almost there, we just have to get rid of that 1 there."},{"Start":"02:56.135 ","End":"03:03.040","Text":"What we\u0027ll do is subtract the third row from the fourth row."},{"Start":"03:03.040 ","End":"03:10.755","Text":"1 minus 1 is 0 and t minus this is this."},{"Start":"03:10.755 ","End":"03:16.570","Text":"That\u0027s the first part of the process."},{"Start":"03:18.260 ","End":"03:21.600","Text":"Then I colored them when you aren\u0027t looking."},{"Start":"03:21.600 ","End":"03:25.200","Text":"These are now the components,"},{"Start":"03:25.200 ","End":"03:28.350","Text":"so coordinates or coefficients in"},{"Start":"03:28.350 ","End":"03:34.370","Text":"the coordinate vector which I wrote over here so"},{"Start":"03:34.370 ","End":"03:40.910","Text":"that this x would be this x and this y minus x is here."},{"Start":"03:40.910 ","End":"03:43.555","Text":"This is over here."},{"Start":"03:43.555 ","End":"03:47.225","Text":"This 1 here, you can\u0027t go wrong, I\u0027ve colored it."},{"Start":"03:47.225 ","End":"03:51.520","Text":"Now let me explain what it means for this to be a coordinate vector."},{"Start":"03:51.520 ","End":"03:55.880","Text":"You\u0027ve studied it, but I want to explain it again."},{"Start":"03:56.600 ","End":"04:06.390","Text":"These 4 quantities expressions will tell us how to write any 2 by 2 matrix,"},{"Start":"04:06.390 ","End":"04:10.600","Text":"which is also a vector as a linear combination of"},{"Start":"04:10.600 ","End":"04:19.615","Text":"the 4 bases matrices or vectors, whatever."},{"Start":"04:19.615 ","End":"04:24.415","Text":"I\u0027ll show you. It means that this vector"},{"Start":"04:24.415 ","End":"04:30.100","Text":"here is the linear combination of these as follows."},{"Start":"04:30.100 ","End":"04:31.920","Text":"We take x,"},{"Start":"04:31.920 ","End":"04:33.970","Text":"I\u0027m not going to color them again."},{"Start":"04:33.970 ","End":"04:36.040","Text":"X times the first 1,"},{"Start":"04:36.040 ","End":"04:38.860","Text":"y minus x times the second."},{"Start":"04:38.860 ","End":"04:43.950","Text":"Then z minus y plus x here times this 1,"},{"Start":"04:43.950 ","End":"04:48.220","Text":"and this times the last 1."},{"Start":"04:48.260 ","End":"04:51.420","Text":"You know what? Wasn\u0027t planning to,"},{"Start":"04:51.420 ","End":"04:53.979","Text":"let\u0027s do a numerical example."},{"Start":"04:54.200 ","End":"04:58.170","Text":"What should I take? Let\u0027s see."},{"Start":"04:58.170 ","End":"05:00.030","Text":"Let\u0027s just take, I don\u0027t know, 1, 2,"},{"Start":"05:00.030 ","End":"05:03.240","Text":"3, 4 and see what we get."},{"Start":"05:03.240 ","End":"05:07.880","Text":"X is this would be 1 times,"},{"Start":"05:07.880 ","End":"05:11.110","Text":"and then we would write the first 1, 1,"},{"Start":"05:11.110 ","End":"05:14.660","Text":"1, 0, 0 and then y minus x."},{"Start":"05:14.660 ","End":"05:16.700","Text":"This is y and this is x,"},{"Start":"05:16.700 ","End":"05:18.620","Text":"so it\u0027s 2 minus 1."},{"Start":"05:18.620 ","End":"05:25.020","Text":"It\u0027s 1 times the next 1 which is 0, 1, 1, 0."},{"Start":"05:25.020 ","End":"05:30.275","Text":"Then z minus y plus x."},{"Start":"05:30.275 ","End":"05:34.200","Text":"This is z minus this plus this,"},{"Start":"05:34.200 ","End":"05:36.360","Text":"3 minus 2 plus 1."},{"Start":"05:36.360 ","End":"05:41.740","Text":"I make it 2 times this 1."},{"Start":"05:41.780 ","End":"05:47.280","Text":"Finally, t minus z plus y minus x."},{"Start":"05:47.280 ","End":"05:52.660","Text":"Let\u0027s see. This minus this plus this minus this,"},{"Start":"05:53.360 ","End":"05:59.745","Text":"I think make it 2 also times 0,"},{"Start":"05:59.745 ","End":"06:02.245","Text":"0, 0, 1."},{"Start":"06:02.245 ","End":"06:04.100","Text":"Might\u0027ve made a mistake here,"},{"Start":"06:04.100 ","End":"06:07.655","Text":"but if you multiply this out and add, it should work."},{"Start":"06:07.655 ","End":"06:10.025","Text":"That will check the top left."},{"Start":"06:10.025 ","End":"06:15.000","Text":"1 plus 0 plus twice 0 plus twice 0 is 1."},{"Start":"06:15.000 ","End":"06:19.020","Text":"That\u0027s okay. Hopefully the rest are okay."},{"Start":"06:19.020 ","End":"06:25.270","Text":"That\u0027s the idea and that concludes part 2 of the 3 part set."}],"ID":26734},{"Watched":false,"Name":"Exercise 2 Part b","Duration":"1m 44s","ChapterTopicVideoID":25925,"CourseChapterTopicPlaylistID":246320,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.845","Text":"This exercise is part 2 in a 3-part set,"},{"Start":"00:04.845 ","End":"00:08.610","Text":"and it\u0027s the same as the previous exercise."},{"Start":"00:08.610 ","End":"00:16.230","Text":"We are considering the space of 2 by 2 matrices with real numbers."},{"Start":"00:16.230 ","End":"00:18.840","Text":"In the previous exercise,"},{"Start":"00:18.840 ","End":"00:23.595","Text":"we dealt with basis B."},{"Start":"00:23.595 ","End":"00:29.025","Text":"This time we\u0027re going with the basis E. If you look at it closely,"},{"Start":"00:29.025 ","End":"00:34.590","Text":"this is actually the standard basis of M2."},{"Start":"00:34.590 ","End":"00:42.510","Text":"The 1 just keeps going in order from here to here to here to here."},{"Start":"00:42.510 ","End":"00:44.770","Text":"E is also for elementary,"},{"Start":"00:44.770 ","End":"00:46.420","Text":"elementary or standard basis,"},{"Start":"00:46.420 ","End":"00:48.835","Text":"they\u0027re both good names for it."},{"Start":"00:48.835 ","End":"00:55.150","Text":"This time we\u0027re computing the coordinate vector of a general vector called X,"},{"Start":"00:55.150 ","End":"01:01.940","Text":"Y, Z, T relative to the standard basis."},{"Start":"01:02.760 ","End":"01:05.830","Text":"Now because it\u0027s the standard basis,"},{"Start":"01:05.830 ","End":"01:09.894","Text":"our work is very easy because the whole point about the standard basis"},{"Start":"01:09.894 ","End":"01:14.919","Text":"is that when we have a vector or in this case matrix,"},{"Start":"01:14.919 ","End":"01:20.450","Text":"then these numbers will be just the coordinates in the right order."},{"Start":"01:20.670 ","End":"01:27.840","Text":"I can write this as X times the first 1 plus Y times a second,"},{"Start":"01:27.840 ","End":"01:28.910","Text":"Z times the third,"},{"Start":"01:28.910 ","End":"01:30.365","Text":"and T times the fourth."},{"Start":"01:30.365 ","End":"01:32.870","Text":"And these coefficients are X, Y,"},{"Start":"01:32.870 ","End":"01:38.905","Text":"Z and T. The coordinate vector is just this,"},{"Start":"01:38.905 ","End":"01:40.140","Text":"X, Y, Z,"},{"Start":"01:40.140 ","End":"01:42.380","Text":"T. That was easy."},{"Start":"01:42.380 ","End":"01:44.820","Text":"We\u0027re done with part 2."}],"ID":26735},{"Watched":false,"Name":"Exercise 2 Part c","Duration":"4m 19s","ChapterTopicVideoID":25926,"CourseChapterTopicPlaylistID":246320,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.690","Text":"This exercise is the third and last in a 3-part set of exercises."},{"Start":"00:06.690 ","End":"00:14.100","Text":"To remind you, we are in this base of 2 by 2 square matrices with real numbers,"},{"Start":"00:14.100 ","End":"00:16.590","Text":"and we had 2 different bases."},{"Start":"00:16.590 ","End":"00:24.015","Text":"We had this basis and this 1 which happens to be the standard or elementary basis."},{"Start":"00:24.015 ","End":"00:27.120","Text":"This time our task is to find the change of"},{"Start":"00:27.120 ","End":"00:32.415","Text":"basis matrix from B to E. This will be our old basis,"},{"Start":"00:32.415 ","End":"00:34.465","Text":"this will be the new basis."},{"Start":"00:34.465 ","End":"00:37.470","Text":"There\u0027s a recipe for how to do this in 3 steps,"},{"Start":"00:37.470 ","End":"00:41.550","Text":"but we can\u0027t work with these matrices"},{"Start":"00:41.550 ","End":"00:46.475","Text":"which are also vectors as is we have to flatten them out."},{"Start":"00:46.475 ","End":"00:51.650","Text":"Now referring to what I call the snake method where we just take them in order 1,"},{"Start":"00:51.650 ","End":"00:54.025","Text":"0, 0, 0."},{"Start":"00:54.025 ","End":"00:56.970","Text":"Here, for example, this would be 1,"},{"Start":"00:56.970 ","End":"00:58.345","Text":"1, 0, 0."},{"Start":"00:58.345 ","End":"01:07.500","Text":"Each of these matrices can be flattened out by taking it but this is the usual order,"},{"Start":"01:07.500 ","End":"01:08.880","Text":"we take, top-left."},{"Start":"01:08.880 ","End":"01:14.250","Text":"I mean, we go from left to right on the rows and then we go down the rows."},{"Start":"01:14.810 ","End":"01:17.040","Text":"Let\u0027s just start."},{"Start":"01:17.040 ","End":"01:18.980","Text":"It\u0027s a 3-step recipe."},{"Start":"01:18.980 ","End":"01:27.995","Text":"In part 1, we have to compute the coordinate vector relative to the old basis B."},{"Start":"01:27.995 ","End":"01:34.315","Text":"But we did this in part 1 of this 3-part exercise,"},{"Start":"01:34.315 ","End":"01:37.065","Text":"and I copied the result."},{"Start":"01:37.065 ","End":"01:40.760","Text":"Now in step 2, instead of the general x,"},{"Start":"01:40.760 ","End":"01:42.590","Text":"y, z, t,"},{"Start":"01:42.590 ","End":"01:46.340","Text":"we\u0027re going to use each vector in"},{"Start":"01:46.340 ","End":"01:53.330","Text":"the new basis which is the standard basis in terms of the old basis."},{"Start":"01:53.330 ","End":"01:55.730","Text":"Well, you\u0027ll see what I mean."},{"Start":"01:55.730 ","End":"01:59.880","Text":"Here again, on the 2 basis,"},{"Start":"01:59.880 ","End":"02:01.110","Text":"they disappeared off screen."},{"Start":"02:01.110 ","End":"02:03.705","Text":"Here they are again, and I want to name them."},{"Start":"02:03.705 ","End":"02:06.570","Text":"The ones in the old basis,"},{"Start":"02:06.570 ","End":"02:08.130","Text":"I want to call them u_1,"},{"Start":"02:08.130 ","End":"02:10.920","Text":"u_2, u_3, and u_4."},{"Start":"02:10.920 ","End":"02:18.860","Text":"Earlier we saw that the coordinate vector with respect to the old basis was this,"},{"Start":"02:18.860 ","End":"02:24.800","Text":"and what that means is that the general vector x, y, z,"},{"Start":"02:24.800 ","End":"02:28.730","Text":"t can be written using these as coefficients to get"},{"Start":"02:28.730 ","End":"02:34.410","Text":"a linear combination of the old basis u_1,"},{"Start":"02:34.410 ","End":"02:36.000","Text":"u_2, u_3, u_4."},{"Start":"02:36.000 ","End":"02:37.440","Text":"This is the expression."},{"Start":"02:37.440 ","End":"02:38.610","Text":"This 1 times u_1,"},{"Start":"02:38.610 ","End":"02:40.500","Text":"this 1 times u_2, and so on."},{"Start":"02:40.500 ","End":"02:42.130","Text":"The idea is now,"},{"Start":"02:42.130 ","End":"02:48.490","Text":"instead of taking this general vector to take each of these in turn,"},{"Start":"02:48.490 ","End":"02:53.510","Text":"and so it\u0027s just a matter of replacing x, y, z,"},{"Start":"02:53.510 ","End":"02:58.475","Text":"t successively by each of these 4 which I\u0027m doing here,"},{"Start":"02:58.475 ","End":"03:01.900","Text":"and then interpreting x,"},{"Start":"03:01.900 ","End":"03:04.670","Text":"and I could do it across or vertically, doesn\u0027t matter."},{"Start":"03:04.670 ","End":"03:06.200","Text":"I\u0027ll just take x and each of them."},{"Start":"03:06.200 ","End":"03:07.280","Text":"This would be the 1,"},{"Start":"03:07.280 ","End":"03:08.930","Text":"0, 0, 0."},{"Start":"03:08.930 ","End":"03:11.510","Text":"Then I need y minus x."},{"Start":"03:11.510 ","End":"03:13.730","Text":"Here I need 0 minus 1,"},{"Start":"03:13.730 ","End":"03:15.630","Text":"and here I need 1 minus 0."},{"Start":"03:15.630 ","End":"03:16.860","Text":"Then 0 minus 0,"},{"Start":"03:16.860 ","End":"03:19.185","Text":"and 0 minus 0 gives me the second column."},{"Start":"03:19.185 ","End":"03:22.745","Text":"Here I need z minus y plus x."},{"Start":"03:22.745 ","End":"03:28.150","Text":"Anyway, you can do these computations and this is what we get."},{"Start":"03:28.150 ","End":"03:33.515","Text":"The important numbers are the ones that I have colored here in blue,"},{"Start":"03:33.515 ","End":"03:41.305","Text":"and these numbers almost give me the change of basis matrix."},{"Start":"03:41.305 ","End":"03:46.985","Text":"The almost is that you might forget we have to transpose,"},{"Start":"03:46.985 ","End":"03:49.920","Text":"and that is our step 3."},{"Start":"03:49.920 ","End":"03:55.270","Text":"We just take these and exchange rows and columns, transpose."},{"Start":"03:55.270 ","End":"03:58.090","Text":"The first column will be 1 minus 1,"},{"Start":"03:58.090 ","End":"04:00.505","Text":"1 minus 1, and that\u0027s this."},{"Start":"04:00.505 ","End":"04:04.045","Text":"The second row becomes the second column,"},{"Start":"04:04.045 ","End":"04:06.835","Text":"and so on, and we write it,"},{"Start":"04:06.835 ","End":"04:10.315","Text":"the change of basis matrix M from"},{"Start":"04:10.315 ","End":"04:19.330","Text":"the basis B to the basis E, and that\u0027s it."}],"ID":26736}],"Thumbnail":null,"ID":246320}]

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1.1

1