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.