1 00:00:07,110 --> 00:00:08,540 PROFESSOR: Hi guys. 2 00:00:08,540 --> 00:00:10,700 Today, we are going to play around 3 00:00:10,700 --> 00:00:14,040 with the basics of eigenvalues and eigenvectors. 4 00:00:14,040 --> 00:00:16,940 We're going to do the following problem, 5 00:00:16,940 --> 00:00:20,450 we're given this invertible matrix A, 6 00:00:20,450 --> 00:00:24,350 and we'll find the eigenvalues and eigenvectors 7 00:00:24,350 --> 00:00:33,630 not of A, but of A squared and A inverse minus the identity. 8 00:00:33,630 --> 00:00:40,250 So, this problem might seem daunting at first, squaring a 3 9 00:00:40,250 --> 00:00:44,690 by 3 matrix, or taking an inverse of a 3 by 3 matrix 10 00:00:44,690 --> 00:00:49,880 is a fairly computationally intensive task, 11 00:00:49,880 --> 00:00:52,530 but if you've seen Professor Strang's lecture 12 00:00:52,530 --> 00:00:54,540 on eigenvalues and eigenvectors you 13 00:00:54,540 --> 00:00:55,910 shouldn't be all too worried. 14 00:00:55,910 --> 00:00:59,440 So I'll give you a few moments to think 15 00:00:59,440 --> 00:01:02,650 of your own line of attack and then you'll see mine. 16 00:01:11,120 --> 00:01:19,160 Hi again, OK, so the observation that makes our life really easy 17 00:01:19,160 --> 00:01:20,480 is the following one. 18 00:01:20,480 --> 00:01:28,170 So say v is an eigenvector with associated eigenvalue lambda 19 00:01:28,170 --> 00:01:39,070 to the matrix A. Then, if we hit v with A squared, well, 20 00:01:39,070 --> 00:01:45,621 this we can write it as A times A*v, but A*v is lambda*v, 21 00:01:45,621 --> 00:01:46,120 right? 22 00:01:46,120 --> 00:01:50,150 So we have A*lambda*v. 23 00:01:50,150 --> 00:01:55,965 Lambda is a scalar, so we can move it in front and get 24 00:01:55,965 --> 00:02:03,670 lambda*A*v, and lambda*A*v is, when we plug in A*v, lambda*v, 25 00:02:03,670 --> 00:02:10,235 is just lambda squared v. So, what we've find out is that 26 00:02:10,235 --> 00:02:14,370 if v is an eigenvector for A then it's also an eigenvector 27 00:02:14,370 --> 00:02:15,450 for A squared. 28 00:02:15,450 --> 00:02:19,840 Just that the eigenvalue is the eigenvalue squared. 29 00:02:19,840 --> 00:02:28,070 Similarly, if we hit A inverse-- if you hit v with A inverse. 30 00:02:28,070 --> 00:02:36,760 So in this case we can write v as A*v over lambda, 31 00:02:36,760 --> 00:02:41,150 given that of course, lambda is non-zero. 32 00:02:41,150 --> 00:02:45,870 But the eigenvalues of an invertible matrix 33 00:02:45,870 --> 00:02:50,065 are always non-zero, which is an exercise 34 00:02:50,065 --> 00:02:52,560 you should do yourselves. 35 00:02:52,560 --> 00:03:00,210 So if we just, then, take out the A 36 00:03:00,210 --> 00:03:06,830 and combine it with A inverse, this is the identity, 37 00:03:06,830 --> 00:03:11,210 and so we get 1 over lambda v. So v 38 00:03:11,210 --> 00:03:14,020 is also an eigenvector for a inverse, 39 00:03:14,020 --> 00:03:19,060 with eigenvalue the reciprocal of lambda. 40 00:03:19,060 --> 00:03:25,350 OK, and from here, of course, A inverse minus the identity 41 00:03:25,350 --> 00:03:38,640 is lambda inverse minus 1 v, so the eigenvalue 42 00:03:38,640 --> 00:03:44,920 of A inverse minus the identity is 1 over lambda minus 1. 43 00:03:44,920 --> 00:03:50,120 OK, so, what we've figured out is: 44 00:03:50,120 --> 00:03:53,340 we just need to find the eigenvalues and eigenvectors 45 00:03:53,340 --> 00:03:58,350 of A and then we have a way of finding what the eigenvalues 46 00:03:58,350 --> 00:04:03,740 and eigenvectors of A squared and A inverse 47 00:04:03,740 --> 00:04:05,290 minus the identity will be. 48 00:04:05,290 --> 00:04:11,570 OK so, how do we find the eigenvalues? 49 00:04:11,570 --> 00:04:17,089 Well, what does it mean for lambda 50 00:04:17,089 --> 00:04:19,180 to be an eigenvalue of A? 51 00:04:19,180 --> 00:04:22,355 It means that the matrix A minus lambda the identity 52 00:04:22,355 --> 00:04:28,500 is singular, which is precisely the case when 53 00:04:28,500 --> 00:04:35,220 its determinant is 0, OK? 54 00:04:35,220 --> 00:04:49,760 So we need to solve the following equation: 55 00:04:49,760 --> 00:05:06,230 1 minus lambda, 2, 3; 0, 1 minus lambda, -2; and 0, 1, 56 00:05:06,230 --> 00:05:08,420 4 minus lambda. 57 00:05:08,420 --> 00:05:17,050 OK, it's fairly obvious which column we should 58 00:05:17,050 --> 00:05:20,190 use to expand this determinant. 59 00:05:20,190 --> 00:05:22,490 We should use the first column, because we 60 00:05:22,490 --> 00:05:26,800 have only one non-zero entry, and so this 61 00:05:26,800 --> 00:05:36,000 is equal to 1 minus lambda times the determinant of the two 62 00:05:36,000 --> 00:05:45,050 by two matrix 1 minus lambda, -2; 1, 4 63 00:05:45,050 --> 00:05:51,090 minus lambda, which is, I'm going to do the computation up 64 00:05:51,090 --> 00:05:51,590 here. 65 00:05:57,740 --> 00:06:08,260 1 minus lambda, lambda squared minus 5 lambda plus 6. 66 00:06:08,260 --> 00:06:11,580 Which is a fairly familiar quadratic, 67 00:06:11,580 --> 00:06:18,140 and we can write it as the product of linear factors, 68 00:06:18,140 --> 00:06:23,230 as lambda minus 2, lambda minus three. 69 00:06:23,230 --> 00:06:33,000 So the three eigenvalues of A are 1, 2, and 3. 70 00:06:33,000 --> 00:06:37,990 OK so, first half of our problem is done, 71 00:06:37,990 --> 00:06:41,830 now we just need to find what the eigenvectors associated 72 00:06:41,830 --> 00:06:44,740 with each of these eigenvalues are. 73 00:06:44,740 --> 00:06:46,690 How we do that? 74 00:06:46,690 --> 00:06:51,562 Well, let's see. 75 00:06:51,562 --> 00:06:54,960 Let's figure out what the eigenvector associated 76 00:06:54,960 --> 00:06:56,380 with lambda equals 1 is. 77 00:06:59,260 --> 00:07:01,610 So, we know that the eigenvector needs 78 00:07:01,610 --> 00:07:05,990 to be in the null space of A minus lambda 79 00:07:05,990 --> 00:07:13,270 the identity, so A minus the identity, v, 80 00:07:13,270 --> 00:07:37,740 so-- write this out-- it's, 0, 0, 3, 2, 3, 0, -2, 0, 1. 81 00:07:37,740 --> 00:07:40,130 And we see that the first column is 0, 82 00:07:40,130 --> 00:07:52,630 so the first variable will be our free variable 83 00:07:52,630 --> 00:07:54,840 if we want to solve this linear system of equations. 84 00:07:57,360 --> 00:08:01,970 And you can just set it to 1 and it's not 85 00:08:01,970 --> 00:08:09,950 hard to see that the other two entries should be 0. 86 00:08:09,950 --> 00:08:14,930 So we can do the same procedure with the other two eigenvalues 87 00:08:14,930 --> 00:08:24,540 and we'll get an eigenvector for each eigenvalue. 88 00:08:24,540 --> 00:08:27,280 And in the end-- let me go back here. 89 00:08:30,100 --> 00:08:36,010 So I'm going to put our results in a little table. 90 00:08:36,010 --> 00:08:46,830 So A squared, inverse minus the identity, so the first row 91 00:08:46,830 --> 00:08:48,560 will be eigenvalues. 92 00:08:54,920 --> 00:08:59,800 So it's going to be: if lambda is an eigenvalue for A, 93 00:08:59,800 --> 00:09:02,150 then we saw that lambda squared will 94 00:09:02,150 --> 00:09:08,250 be the eigenvalue for A squared and lambda inverse minus 1 95 00:09:08,250 --> 00:09:11,940 will be the eigenvalue for A inverse minus the identity. 96 00:09:11,940 --> 00:09:19,630 And the eigenvectors will be the same. 97 00:09:19,630 --> 00:09:21,483 OK, we're done.