WEBVTT
00:00:06.370 --> 00:00:08.370
DAVID SHIROKOFF: Hi, everyone.
00:00:08.370 --> 00:00:10.230
So for this problem,
we're just going
00:00:10.230 --> 00:00:14.180
to take a look at computing some
eigenvalues and eigenvectors
00:00:14.180 --> 00:00:16.059
of several matrices.
00:00:16.059 --> 00:00:19.300
And this is just a review
problem for exam number three.
00:00:19.300 --> 00:00:21.350
So specifically, we're
given a projection
00:00:21.350 --> 00:00:25.210
matrix which has the form
of a a transpose divided
00:00:25.210 --> 00:00:29.690
by a transpose a, where
a is the vector 3 and 4.
00:00:29.690 --> 00:00:32.820
The second problem is
for a rotation matrix
00:00:32.820 --> 00:00:39.920
Q, which is the numbers 0.6,
negative 0.8, 0.8, and 0.6.
00:00:39.920 --> 00:00:42.610
And then the third one is
for a reflection matrix
00:00:42.610 --> 00:00:45.280
which is 2P minus the identity.
00:00:45.280 --> 00:00:46.600
So I'll let you work these out.
00:00:46.600 --> 00:00:48.140
And then I'll come
back in a second,
00:00:48.140 --> 00:00:49.795
and I'll fill in my solutions.
00:01:02.380 --> 00:01:03.470
Hi, everyone.
00:01:03.470 --> 00:01:04.840
Welcome back.
00:01:04.840 --> 00:01:06.850
OK, so for the
first problem, we're
00:01:06.850 --> 00:01:10.040
given a matrix P, which
is a projection matrix.
00:01:10.040 --> 00:01:12.199
And from earlier
on in the course,
00:01:12.199 --> 00:01:14.740
we probably already know that
the eigenvalues of a projection
00:01:14.740 --> 00:01:17.720
matrix are either 0 or 1.
00:01:17.720 --> 00:01:20.160
And I'll just recall,
how do you know that?
00:01:20.160 --> 00:01:25.310
Well if x is an
eigenvector of P,
00:01:25.310 --> 00:01:29.180
then it satisfies the
equation P*x equals lambda*x.
00:01:29.180 --> 00:01:32.710
But for a projection matrix,
P squared is equal to P.
00:01:32.710 --> 00:01:37.700
So if P is a projection,
we have P squared equals P.
00:01:37.700 --> 00:01:45.530
And specifically, what this
means is P squared x is equal
00:01:45.530 --> 00:01:48.620
to lambda*x.
00:01:52.370 --> 00:01:59.940
So we have P acting on P
of x is equal to lambda*x.
00:01:59.940 --> 00:02:04.190
And on the left-hand side, P*
is going to give me a lambda*x.
00:02:04.190 --> 00:02:06.300
P* again will give
me a lambda x.
00:02:06.300 --> 00:02:12.130
So we get lambda squared
x equals lambda*x.
00:02:12.130 --> 00:02:14.310
And if I bring everything
to the left-hand side,
00:02:14.310 --> 00:02:20.420
I get lambda times lambda
minus 1 x equals 0.
00:02:20.420 --> 00:02:23.440
And because x is not a zero
vector, what that means is
00:02:23.440 --> 00:02:26.120
lambda has to be either 0 or 1.
00:02:26.120 --> 00:02:29.170
So this is just a quick
proof that the eigenvalue
00:02:29.170 --> 00:02:32.070
of a projection matrix
is either 0 or 1.
00:02:32.070 --> 00:02:33.650
So we already know
that P is going
00:02:33.650 --> 00:02:38.390
to have eigenvalues of 0 or 1.
00:02:38.390 --> 00:02:40.780
Now specifically,
how do I identify
00:02:40.780 --> 00:02:43.540
which eigenvectors
correspond to 0
00:02:43.540 --> 00:02:45.970
and which eigenvectors
correspond to 1?
00:02:45.970 --> 00:02:48.290
Well, in this case, P
has a specific form,
00:02:48.290 --> 00:02:52.030
which is a times a transpose
divided by a transpose a.
00:02:55.270 --> 00:02:59.300
So I'll just write out
explicitly what this is.
00:02:59.300 --> 00:03:03.640
So a transpose a, 1
divided by a transpose a
00:03:03.640 --> 00:03:10.550
is going to be 9 plus
16 on the denominator.
00:03:10.550 --> 00:03:17.920
Then we're going to have
3 and 4 and 3 and 4.
00:03:17.920 --> 00:03:19.980
Now when we have a
matrix of this form,
00:03:19.980 --> 00:03:22.790
it's always going to be
the case that the vector
00:03:22.790 --> 00:03:26.200
a is going to be an
eigenvector with eigenvalue 1.
00:03:30.100 --> 00:03:30.850
So let's check.
00:03:34.840 --> 00:03:38.510
What is P acting on a?
00:03:38.510 --> 00:03:50.190
Well, we end up with: the
matrix P is 1/25 [3; 4] [3, 4].
00:03:50.190 --> 00:03:59.120
This is the matrix P. And if
we acted on the vector [3; 4],
00:03:59.120 --> 00:04:02.460
notice how this piece right
here, we can multiply out.
00:04:02.460 --> 00:04:05.614
This is going to be a transpose,
and this is going to be a.
00:04:05.614 --> 00:04:07.280
And if we multiply
these two pieces out,
00:04:07.280 --> 00:04:14.630
we get 25, which is exactly
the denominator a transpose a.
00:04:14.630 --> 00:04:20.230
So at the end of the day, we get
[3, 4]; Because the 25 divides
00:04:20.230 --> 00:04:23.076
out with the 25.
00:04:23.076 --> 00:04:25.200
Now note that this is
exactly what we started with.
00:04:25.200 --> 00:04:28.260
This is exactly a.
00:04:28.260 --> 00:04:31.170
So note here that the
vector a corresponds
00:04:31.170 --> 00:04:32.340
to an eigenvalue of 1.
00:04:36.200 --> 00:04:45.780
Meanwhile, for an
eigenvalue of 0,
00:04:45.780 --> 00:04:48.290
well, it always turns
out to be the case
00:04:48.290 --> 00:04:51.116
that if I take any vector
perpendicular to a,
00:04:51.116 --> 00:04:54.810
P acting on that vector
is going to be 0.
00:04:54.810 --> 00:04:57.480
So what's a vector,
which I'll call b,
00:04:57.480 --> 00:04:59.980
that's perpendicular to a?
00:04:59.980 --> 00:05:01.962
Well, note that a is
just a two by two vector.
00:05:01.962 --> 00:05:04.420
So that means there's only
going to be one direction that's
00:05:04.420 --> 00:05:06.490
perpendicular to a.
00:05:06.490 --> 00:05:08.520
Now just by eyeballing
it, I can see
00:05:08.520 --> 00:05:11.260
that a vector that's going
to be perpendicular to a
00:05:11.260 --> 00:05:12.740
is negative 4 and 3.
00:05:16.550 --> 00:05:20.660
So let's quickly check
that this is an eigenvector
00:05:20.660 --> 00:05:22.860
of P with eigenvalue 0.
00:05:22.860 --> 00:05:25.600
So what we need to show is that
P acting on this vector, b,
00:05:25.600 --> 00:05:27.270
is 0.
00:05:27.270 --> 00:05:31.510
So P acting on b is
going to be 1/25.
00:05:34.430 --> 00:05:45.114
It's going to be [3; 4]
[3, 4], multiplied by [4, 3].
00:05:48.750 --> 00:05:55.730
And note how when I multiply
out this row on this column,
00:05:55.730 --> 00:05:59.220
I get negative 3 times
4 plus 3 times 4,
00:05:59.220 --> 00:06:02.910
which is going to be 0.
00:06:02.910 --> 00:06:03.860
OK?
00:06:03.860 --> 00:06:07.740
So this shows that this vector
b has an eigenvalue of 0
00:06:07.740 --> 00:06:10.700
because note that we
can write this as 0*b.
00:06:16.950 --> 00:06:17.450
OK.
00:06:17.450 --> 00:06:22.980
For the second part, Q,
what are the eigenvectors
00:06:22.980 --> 00:06:25.860
and eigenvalues
of this matrix, Q?
00:06:25.860 --> 00:06:28.740
Well, Q is a rotation matrix.
00:06:28.740 --> 00:06:37.860
So I'll just write out Q again,
0.6, negative 0.8, 0.8, 0.6.
00:06:37.860 --> 00:06:45.600
So note that we can identify the
diagonal elements with a cosine
00:06:45.600 --> 00:06:47.370
of some angle theta.
00:06:47.370 --> 00:06:50.010
And we can associate
the off-diagonal parts
00:06:50.010 --> 00:06:53.830
as sine theta and
negative sine theta.
00:06:53.830 --> 00:06:56.590
And the reason we can
do that is because 0.6
00:06:56.590 --> 00:07:01.030
squared plus 0.8 squared is 1.
00:07:01.030 --> 00:07:03.560
So this is a rotation matrix.
00:07:03.560 --> 00:07:05.660
Now, to work out
the eigenvalues,
00:07:05.660 --> 00:07:08.430
I take a look at the
characteristic equation.
00:07:08.430 --> 00:07:10.019
So this is going
to give me, if I
00:07:10.019 --> 00:07:11.810
take a look at the
characteristic equation,
00:07:11.810 --> 00:07:18.280
it's going to be 0.6
minus lambda, squared.
00:07:18.280 --> 00:07:22.020
Then we have minus times
0.8 times negative 0.8.
00:07:22.020 --> 00:07:27.150
So that's going to
be plus 0.8 squared.
00:07:27.150 --> 00:07:28.370
And we want this to be 0.
00:07:33.330 --> 00:07:37.820
So if I rewrite this, I get
lambda is 0.6 plus or minus
00:07:37.820 --> 00:07:42.740
0.8i, where i is the
imaginary number.
00:07:42.740 --> 00:07:45.450
So notice how the eigenvalues
come in complex conjugate
00:07:45.450 --> 00:07:46.450
pairs.
00:07:46.450 --> 00:07:50.880
And this is always the case
when we have a real matrix.
00:07:50.880 --> 00:07:55.210
So we can find, first off, just
the eigenvalue that corresponds
00:07:55.210 --> 00:07:57.820
to 0.6 plus 0.8i.
00:07:57.820 --> 00:07:59.770
And then at the end,
we'll be able to find
00:07:59.770 --> 00:08:02.540
the second eigenvector by just
taking the complex conjugate
00:08:02.540 --> 00:08:04.840
of the first one.
00:08:04.840 --> 00:08:10.510
So let's compute
Q minus lambda*I.
00:08:10.510 --> 00:08:13.420
And if we have this acting
on some eigenvector u,
00:08:13.420 --> 00:08:15.780
we want this to be 0.
00:08:15.780 --> 00:08:18.980
Now Q minus lambda*I
is going to be,
00:08:18.980 --> 00:08:25.560
for the case lambda
is 0.6 plus 0.8i,
00:08:25.560 --> 00:08:32.030
this is going to give me
a quantity of minus 0.8i,
00:08:32.030 --> 00:08:41.216
minus 0.8, 0.8, and minus 0.8i.
00:08:41.216 --> 00:08:42.799
And I'm going to
write down components
00:08:42.799 --> 00:08:46.151
of u, which are u_1 and u_2.
00:08:46.151 --> 00:08:50.010
And we want this to vanish.
00:08:50.010 --> 00:08:52.330
And we note that the second
row is a constant multiple
00:08:52.330 --> 00:08:53.710
of the first row.
00:08:53.710 --> 00:08:56.870
Specifically, if I multiplied
this first row through by i,
00:08:56.870 --> 00:08:59.965
we would get negative i
squared, which is just 1.
00:08:59.965 --> 00:09:01.840
And then the second part
would be negative i,
00:09:01.840 --> 00:09:04.710
so we would just get the
second row back, which is good.
00:09:07.770 --> 00:09:09.780
So we just need to
find u_1, u_2 that are
00:09:09.780 --> 00:09:13.030
orthogonal to this first row.
00:09:13.030 --> 00:09:17.650
And again, just by inspection,
I can pick 1 and negative i.
00:09:20.190 --> 00:09:25.750
So note that that would give
me negative 0.8i plus 0.8i,
00:09:25.750 --> 00:09:27.460
and this vanishes.
00:09:27.460 --> 00:09:31.520
So this is the eigenvector that
corresponds to the eigenvalue
00:09:31.520 --> 00:09:32.810
lambda 0.6 plus 0.8i.
00:09:37.590 --> 00:09:43.100
In the meantime, if I take
the second eigenvalue,
00:09:43.100 --> 00:09:49.690
which is negative 0.8i, I
can take u which is just
00:09:49.690 --> 00:09:54.780
the complex conjugate
of this u up here.
00:09:54.780 --> 00:09:57.780
So it'll be 1, plus i.
00:09:57.780 --> 00:10:00.610
So this concludes the
eigenvalues and eigenvectors
00:10:00.610 --> 00:10:04.261
of this matrix Q.
00:10:04.261 --> 00:10:04.760
OK.
00:10:04.760 --> 00:10:17.490
Now lastly, number three, we're
looking at a reflection matrix
00:10:17.490 --> 00:10:23.500
which has the form 2P minus
I, where P is the same matrix
00:10:23.500 --> 00:10:26.350
that we had in part one.
00:10:26.350 --> 00:10:28.690
Now at first glance,
it looks like we
00:10:28.690 --> 00:10:31.430
might have to diagonalize
this entire matrix.
00:10:31.430 --> 00:10:36.880
However, note that
by shifting 2P by I,
00:10:36.880 --> 00:10:38.810
we only shift the eigenvalues.
00:10:38.810 --> 00:10:41.210
And we don't actually
change the eigenvectors.
00:10:41.210 --> 00:10:44.790
So note that this matrix
R, which is 2P minus I,
00:10:44.790 --> 00:10:47.700
it's going to have the
same eigenvectors as P.
00:10:47.700 --> 00:10:50.560
It's just going to have
different eigenvalues.
00:10:50.560 --> 00:10:54.330
So first off, we're going
to have one eigenvector.
00:10:56.970 --> 00:10:59.990
So the first eigenvector
is going to be a.
00:10:59.990 --> 00:11:03.060
So we have one
eigenvector which is a.
00:11:03.060 --> 00:11:11.880
So we have one
eigenvector which is a.
00:11:11.880 --> 00:11:16.620
And note that for the
vector a, it corresponds
00:11:16.620 --> 00:11:19.500
to the eigenvalue of 1.
00:11:19.500 --> 00:11:23.340
So what eigenvalue does
this correspond to?
00:11:23.340 --> 00:11:25.950
This is going to give
me a lambda which
00:11:25.950 --> 00:11:31.150
is 2 times 1 minus 1.
00:11:31.150 --> 00:11:33.030
So it's 1.
00:11:33.030 --> 00:11:36.840
So note that a, the
vector a, not only
00:11:36.840 --> 00:11:40.290
has an eigenvalue of 1 for P,
but it has an eigenvalue of 1
00:11:40.290 --> 00:11:42.940
for R as well.
00:11:42.940 --> 00:11:53.380
The second case was b.
00:11:53.380 --> 00:11:57.710
And remember that b has
an eigenvalue of 0 for P.
00:11:57.710 --> 00:12:10.230
So when we act R acting on b,
we'll have 2 times 0 minus 1 b.
00:12:10.230 --> 00:12:14.040
So this is going to
give us negative b.
00:12:14.040 --> 00:12:16.960
So the eigenvalue for b
is going to be negative 1.
00:12:20.870 --> 00:12:21.370
OK.
00:12:21.370 --> 00:12:25.740
And this is actually a general
case for reflection matrices,
00:12:25.740 --> 00:12:28.030
is that they typically
have eigenvalues
00:12:28.030 --> 00:12:31.430
of plus 1 or negative 1.
00:12:31.430 --> 00:12:34.100
OK, so we've just taken a
look at several matrices
00:12:34.100 --> 00:12:36.010
that come up in practice.
00:12:36.010 --> 00:12:38.460
We've looked at projection
matrices, reflection matrices,
00:12:38.460 --> 00:12:40.290
and rotation matrices.
00:12:40.290 --> 00:12:42.600
And we've seen a little
bit of the properties
00:12:42.600 --> 00:12:44.940
of their eigenvalues
and eigenvectors.
00:12:44.940 --> 00:12:48.608
So I'll just conclude here,
and good luck on your test.