WEBVTT
00:00:06.109 --> 00:00:07.900
PROFESSOR: Today, we're
going to be solving
00:00:07.900 --> 00:00:10.110
a problem from a final exam.
00:00:10.110 --> 00:00:11.360
And here it is.
00:00:11.360 --> 00:00:16.800
It's about a matrix A,
[1, 0, 1; 0, 1, 1; 1, 1, 0].
00:00:16.800 --> 00:00:21.320
And we know that this matrix
has two eigenvalues, 1 and 2.
00:00:21.320 --> 00:00:23.910
And we also know that
if we do elimination,
00:00:23.910 --> 00:00:27.130
the first two pivots
will be 1 and 1.
00:00:27.130 --> 00:00:31.020
And here are two questions
about this matrix.
00:00:31.020 --> 00:00:35.270
The first one is find lambda_3
and d_3, the third eigenvalue
00:00:35.270 --> 00:00:36.870
and the third pivot.
00:00:36.870 --> 00:00:41.780
And the second one asks you,
what is the smallest a_(3, 3)--
00:00:41.780 --> 00:00:45.310
so if you can change this entry,
what is the smallest number
00:00:45.310 --> 00:00:48.320
that you can put there that
will make the matrix A positive
00:00:48.320 --> 00:00:49.660
semidefinite?
00:00:49.660 --> 00:00:55.820
And also, if instead of changing
that entry, you do A plus c*I,
00:00:55.820 --> 00:00:58.850
what is the smallest number
c that will make that matrix,
00:00:58.850 --> 00:01:02.160
A plus c*I, positive
semidefinite?
00:01:02.160 --> 00:01:03.910
There's also a third
part to the question,
00:01:03.910 --> 00:01:05.370
but we'll get to that later.
00:01:05.370 --> 00:01:08.024
Why don't you hit pause and
work on these two parts?
00:01:08.024 --> 00:01:10.690
And when you're ready, come back
and I'll show you how I did it.
00:01:18.920 --> 00:01:19.860
Hi.
00:01:19.860 --> 00:01:24.610
I hope you managed to do parts
A and B. Let's work on it
00:01:24.610 --> 00:01:26.710
together.
00:01:26.710 --> 00:01:33.055
Part A. Well, we want to know
what the third eigenvalue is.
00:01:33.055 --> 00:01:35.080
And you know what
the first two are.
00:01:35.080 --> 00:01:38.510
What else do you know about
eigenvalues and the matrix?
00:01:38.510 --> 00:01:41.720
You know that the sum of all
the eigenvalues of the matrix
00:01:41.720 --> 00:01:44.260
is equal to the
trace of the matrix.
00:01:44.260 --> 00:01:49.570
So lambda_1 plus
lambda_2 plus lambda_3
00:01:49.570 --> 00:01:51.730
is equal to the
trace of the matrix.
00:01:51.730 --> 00:01:56.750
In this case, you have
1 plus 2 plus lambda_3
00:01:56.750 --> 00:01:58.180
equals to the trace.
00:01:58.180 --> 00:02:01.285
The trace is the sum of
the diagonal entries.
00:02:01.285 --> 00:02:03.290
So, come over here.
00:02:03.290 --> 00:02:05.720
The trace is 1 plus 1 plus 0.
00:02:05.720 --> 00:02:07.860
The trace is equal to 2.
00:02:07.860 --> 00:02:12.620
So we have 3 plus
lambda_3 is equal to 2.
00:02:12.620 --> 00:02:19.800
So lambda_3 is equal to minus 1.
00:02:19.800 --> 00:02:23.800
On to the third pivot.
00:02:23.800 --> 00:02:25.890
We don't really want
to do elimination.
00:02:25.890 --> 00:02:27.190
That would take too long.
00:02:27.190 --> 00:02:30.750
So there must be some
trick that we can use.
00:02:30.750 --> 00:02:33.400
Well, we have the
first two pivots,
00:02:33.400 --> 00:02:35.570
and we want to know the third.
00:02:35.570 --> 00:02:37.880
Remember, when you do
elimination steps, that
00:02:37.880 --> 00:02:40.380
does not change the
determinant of the matrix.
00:02:40.380 --> 00:02:43.410
And you're left with
an upper triangular.
00:02:43.410 --> 00:02:47.820
So the determinant
of that matrix
00:02:47.820 --> 00:02:50.310
will be d_1 times d_2 times d_3.
00:02:50.310 --> 00:02:52.800
And it will still be
equal to the determinant
00:02:52.800 --> 00:02:58.240
of A. I guess there's a small
caveat that I should point out.
00:02:58.240 --> 00:03:01.140
The pivots are not always
the diagonal entries.
00:03:01.140 --> 00:03:04.960
It might be that one of the
diagonal entries will be 0.
00:03:04.960 --> 00:03:07.180
That happens if the
matrix is singular.
00:03:07.180 --> 00:03:11.820
But here, all my three
eigenvalues are non-zero.
00:03:11.820 --> 00:03:14.270
They are 1, 2, and -1.
00:03:14.270 --> 00:03:15.330
So that won't happen.
00:03:15.330 --> 00:03:17.840
So this is actually
possible to do.
00:03:17.840 --> 00:03:19.770
The product of the
three pivots will
00:03:19.770 --> 00:03:23.060
be equal to the determinant
of A. And the determinant of A
00:03:23.060 --> 00:03:28.920
is the product of the
eigenvalues, 1 times 2 times
00:03:28.920 --> 00:03:29.840
-1.
00:03:29.840 --> 00:03:33.400
So it's equal to -2.
00:03:33.400 --> 00:03:39.060
1 times 1 times
d_33 is equal to -2.
00:03:39.060 --> 00:03:41.880
Here is your third pivot, d_3.
00:03:41.880 --> 00:03:46.460
That finishes part A. Is
that the result that you got?
00:03:46.460 --> 00:03:54.020
Let's do part B. What is the
smallest a_(3,3) entry that
00:03:54.020 --> 00:03:57.520
would make the matrix
positive semidefinite?
00:03:57.520 --> 00:04:02.290
Well, first of all, note that
A is not positive semidefinite
00:04:02.290 --> 00:04:03.080
yet.
00:04:03.080 --> 00:04:07.280
The eigenvalues
are 1, 2, and -1.
00:04:07.280 --> 00:04:10.500
-1 is negative, so the matrix
is not positive definite and not
00:04:10.500 --> 00:04:12.410
even positive semidefinite.
00:04:12.410 --> 00:04:15.740
Positive semidefinite means
that all the eigenvalues
00:04:15.740 --> 00:04:18.529
will be either positive or 0.
00:04:18.529 --> 00:04:20.850
That is, non-negative.
00:04:20.850 --> 00:04:30.430
So our matrix will be
1, 0, 1; 0, 1, 1; 1, 1,
00:04:30.430 --> 00:04:34.660
and we're allowed to
change this third entry.
00:04:34.660 --> 00:04:40.410
How do we figure out if this
matrix is positive semidefinite
00:04:40.410 --> 00:04:42.440
or not?
00:04:42.440 --> 00:04:44.700
Well, I was talking
about the eigenvalues.
00:04:44.700 --> 00:04:50.100
But maybe the easiest way is
to do the determinant test.
00:04:50.100 --> 00:04:52.800
The determinant of
the small one by one
00:04:52.800 --> 00:04:56.720
left uppermost matrix is 1.
00:04:56.720 --> 00:05:00.380
The determinant of the two
by two upper leftmost matrix
00:05:00.380 --> 00:05:05.950
is 1 times 1 minus 0
times 0, 1, also positive.
00:05:05.950 --> 00:05:09.410
So we need to check that the
determinant of the whole matrix
00:05:09.410 --> 00:05:11.760
will also be non-negative.
00:05:11.760 --> 00:05:14.870
So what is the determinant
of this matrix?
00:05:14.870 --> 00:05:17.490
It is equal to the
three by three matrix.
00:05:17.490 --> 00:05:19.870
So do you know how
to do it quickly?
00:05:19.870 --> 00:05:23.810
There's this way that only
works for three by three and not
00:05:23.810 --> 00:05:24.940
for bigger.
00:05:24.940 --> 00:05:31.150
Which is, the determinant will
be 1 times 1 times a_(3,3) plus
00:05:31.150 --> 00:05:33.495
0 times 1 times 1.
00:05:33.495 --> 00:05:34.990
That's 0.
00:05:34.990 --> 00:05:39.700
Plus 1 times 0 times
1 That's 0 again.
00:05:39.700 --> 00:05:48.560
Minus 1 times 1 times 1 minus
1 times 1 times 1 minus a_(3,3)
00:05:48.560 --> 00:05:49.720
times 0 times 0.
00:05:49.720 --> 00:05:51.320
That's 0.
00:05:51.320 --> 00:05:52.850
So this is the determinant.
00:05:52.850 --> 00:05:55.110
It's a_(3,3) minus 2.
00:05:55.110 --> 00:05:58.350
And I want it to be
greater than or equal to 0.
00:05:58.350 --> 00:06:00.090
This will guarantee
that the matrix
00:06:00.090 --> 00:06:01.990
is positive semidefinite.
00:06:01.990 --> 00:06:07.540
So a_(3,3) must be bigger
than or equal to 2.
00:06:07.540 --> 00:06:11.820
The smallest value for a_(3,3)
that will make the matrix
00:06:11.820 --> 00:06:16.900
positive semidefinite is 2.
00:06:16.900 --> 00:06:19.030
There's another part
to the question still,
00:06:19.030 --> 00:06:23.900
which is what is the smallest
c that will make the matrix A
00:06:23.900 --> 00:06:26.210
plus c*I positive semidefinite?
00:06:29.520 --> 00:06:31.620
How should we do this?
00:06:31.620 --> 00:06:38.060
The quickest way is to
do the eigenvalue test.
00:06:38.060 --> 00:06:50.570
A has eigenvalues 1, 2, and -1.
00:06:50.570 --> 00:06:58.620
So A plus c*I has
eigenvalues-- well,
00:06:58.620 --> 00:07:01.190
you're just adding
c*I to the matrix.
00:07:01.190 --> 00:07:05.520
And in this particular
case, you should know by now
00:07:05.520 --> 00:07:07.550
that that keeps the
eigenvectors the same
00:07:07.550 --> 00:07:11.795
and adds the number c to
each of the eigenvalues.
00:07:15.260 --> 00:07:18.830
And I want each one of
these to be non-negative.
00:07:18.830 --> 00:07:21.730
For that to be
true, I have to have
00:07:21.730 --> 00:07:26.470
c greater than or equal to 1.
00:07:26.470 --> 00:07:29.580
c greater than or equal to 1.
00:07:29.580 --> 00:07:32.480
So the smallest value
that c can take that
00:07:32.480 --> 00:07:38.670
will make the matrix A
positive semidefinite is 1.
00:07:38.670 --> 00:07:43.110
That solves parts A
and B of this question.
00:07:43.110 --> 00:07:45.224
There is a part C
to this question.
00:07:45.224 --> 00:07:46.140
Let me show it to you.
00:07:50.700 --> 00:07:54.960
It's says: starting with one of
these three vectors, [3, 0, 0],
00:07:54.960 --> 00:08:01.490
[0, 3, 0], or [0, 0, 3], and
with u_(k+1) equals to a half
00:08:01.490 --> 00:08:06.490
of A times u_k, what is the
limit behavior of u_k as k goes
00:08:06.490 --> 00:08:09.300
to infinity?
00:08:09.300 --> 00:08:13.340
I've written the matrix one half
of A here for your convenience.
00:08:13.340 --> 00:08:15.900
And now, you can hit
pause and work on it.
00:08:15.900 --> 00:08:20.830
And when you're ready, we'll
get back and solve it together.
00:08:30.910 --> 00:08:34.070
I hope you managed
to solve this one.
00:08:34.070 --> 00:08:37.330
Now let's do it together.
00:08:37.330 --> 00:08:41.240
Well, if you've noticed,
the matrix one half of A
00:08:41.240 --> 00:08:44.120
is a Markov matrix.
00:08:44.120 --> 00:08:50.000
So there are all these
results about Markov matrices
00:08:50.000 --> 00:08:52.210
and steady states and so on.
00:08:52.210 --> 00:08:56.130
Usually, Markov matrices
have a unique steady state,
00:08:56.130 --> 00:09:00.300
but that is only true when
there are no non-zero entries.
00:09:00.300 --> 00:09:01.290
But here, there are.
00:09:01.290 --> 00:09:05.620
So we can't guarantee that
there's a unique steady state.
00:09:05.620 --> 00:09:08.190
What we can do is look
at the eigenvalues
00:09:08.190 --> 00:09:10.840
and see if this is
still true nonetheless.
00:09:10.840 --> 00:09:14.540
What are the eigenvalues
of A-- of one half of A?
00:09:14.540 --> 00:09:22.130
Well, if you remember from
part A, the eigenvalues of A
00:09:22.130 --> 00:09:27.350
were 1, 2, and -1.
00:09:27.350 --> 00:09:32.630
So the eigenvalues of one
half of A-- taking a multiple
00:09:32.630 --> 00:09:36.330
does not change the eigenvector,
but it changes the eigenvalue
00:09:36.330 --> 00:09:38.266
by the same multiple.
00:09:38.266 --> 00:09:45.690
It would be 1/2, 2 divided
by 2 is 1, and minus 1/2.
00:09:45.690 --> 00:09:48.200
So here are the
eigenvalues of A.
00:09:48.200 --> 00:09:50.380
And there's only
one eigenvalue that
00:09:50.380 --> 00:09:52.950
has absolute value equal to 1.
00:09:52.950 --> 00:09:57.050
So you actually still get a
unique steady state vector.
00:09:57.050 --> 00:09:59.050
So everything is fine.
00:09:59.050 --> 00:10:01.620
We can proceed as usual.
00:10:01.620 --> 00:10:06.590
And the usual procedure is
you find the eigenvector
00:10:06.590 --> 00:10:09.930
corresponding to
that eigenvalue, 1.
00:10:09.930 --> 00:10:13.380
And that will be
the limit behavior
00:10:13.380 --> 00:10:16.590
as k goes to infinity of u_k.
00:10:16.590 --> 00:10:18.980
So what is the eigenvector
corresponding to 1?
00:10:21.540 --> 00:10:22.040
Eigenvector.
00:10:27.020 --> 00:10:30.130
Well, you already
know how to do this,
00:10:30.130 --> 00:10:32.420
so I will just
write the solution.
00:10:32.420 --> 00:10:38.300
It is [1, 1, 1].
00:10:38.300 --> 00:10:44.620
That means that u_k,
as k goes to infinity,
00:10:44.620 --> 00:10:52.470
will converge to some
appropriate multiple
00:10:52.470 --> 00:10:56.180
of this eigenvector [1, 1, 1].
00:10:56.180 --> 00:10:58.700
How do you know which
multiple to use?
00:10:58.700 --> 00:11:01.690
Well, as usual in
Markov matrices,
00:11:01.690 --> 00:11:04.290
when you do an iteration
of the process,
00:11:04.290 --> 00:11:08.370
when you do u_(k+1) is equal
to one half of A times u_k,
00:11:08.370 --> 00:11:13.170
that does not change the sum of
the entries of the vector u_k.
00:11:13.170 --> 00:11:17.110
So whatever the sum was here,
it will still be the same here.
00:11:17.110 --> 00:11:19.220
If you go all the
way back and you
00:11:19.220 --> 00:11:23.850
start with u_0, whatever the
sum of the entries was here,
00:11:23.850 --> 00:11:28.970
that's what it will be all
the way through u_1, u_2, u_3,
00:11:28.970 --> 00:11:32.930
and so on, all the way to
the steady state, u_infinity.
00:11:32.930 --> 00:11:38.240
So whatever the
multiple of [1, 1, 1],
00:11:38.240 --> 00:11:45.850
it has to have the sum of
these entries add up to 3.
00:11:45.850 --> 00:11:48.490
Well, that's already there.
00:11:48.490 --> 00:11:51.282
We already happened to pick
the correct eigenvector,
00:11:51.282 --> 00:11:52.365
so that's very convenient.
00:11:56.510 --> 00:12:01.930
The correct multiple is
simply the vector [1, 1, 1].
00:12:01.930 --> 00:12:06.460
So the limit behavior of
u_k as k goes to infinity
00:12:06.460 --> 00:12:13.180
is u_infinity equal
to [1, 1, 1].
00:12:13.180 --> 00:12:14.540
We're done.
00:12:14.540 --> 00:12:16.350
Thank you.