PROFESSOR: Today, we're
going to be solving
a problem from a final exam.
And here it is.
It's about a matrix A,
[1, 0, 1;  0, 1, 1; 1, 1, 0].
And we know that this matrix
has two eigenvalues, 1 and 2.
And we also know that
if we do elimination,
the first two pivots
will be 1 and 1.
And here are two questions
about this matrix.
The first one is find lambda_3
and d_3, the third eigenvalue
and the third pivot.
And the second one asks you,
what is the smallest a_(3, 3)--
so if you can change this entry,
what is the smallest number
that you can put there that
will make the matrix A positive
semidefinite?
And also, if instead of changing
that entry, you do A plus c*I,
what is the smallest number
c that will make that matrix,
A plus c*I, positive
semidefinite?
There's also a third
part to the question,
but we'll get to that later.
Why don't you hit pause and
work on these two parts?
And when you're ready, come back
and I'll show you how I did it.
Hi.
I hope you managed to do parts
A and B. Let's work on it
together.
Part A. Well, we want to know
what the third eigenvalue is.
And you know what
the first two are.
What else do you know about
eigenvalues and the matrix?
You know that the sum of all
the eigenvalues of the matrix
is equal to the
trace of the matrix.
So lambda_1 plus
lambda_2 plus lambda_3
is equal to the
trace of the matrix.
In this case, you have
1 plus 2 plus lambda_3
equals to the trace.
The trace is the sum of
the diagonal entries.
So, come over here.
The trace is 1 plus 1 plus 0.
The trace is equal to 2.
So we have 3 plus
lambda_3 is equal to 2.
So lambda_3 is equal to minus 1.
On to the third pivot.
We don't really want
to do elimination.
That would take too long.
So there must be some
trick that we can use.
Well, we have the
first two pivots,
and we want to know the third.
Remember, when you do
elimination steps, that
does not change the
determinant of the matrix.
And you're left with
an upper triangular.
So the determinant
of that matrix
will be d_1 times d_2 times d_3.
And it will still be
equal to the determinant
of A. I guess there's a small
caveat that I should point out.
The pivots are not always
the diagonal entries.
It might be that one of the
diagonal entries will be 0.
That happens if the
matrix is singular.
But here, all my three
eigenvalues are non-zero.
They are 1, 2, and -1.
So that won't happen.
So this is actually
possible to do.
The product of the
three pivots will
be equal to the determinant
of A. And the determinant of A
is the product of the
eigenvalues, 1 times 2 times
-1.
So it's equal to -2.
1 times 1 times
d_33 is equal to -2.
Here is your third pivot, d_3.
That finishes part A. Is
that the result that you got?
Let's do part B. What is the
smallest a_(3,3) entry that
would make the matrix
positive semidefinite?
Well, first of all, note that
A is not positive semidefinite
yet.
The eigenvalues
are 1, 2, and -1.
-1 is negative, so the matrix
is not positive definite and not
even positive semidefinite.
Positive semidefinite means
that all the eigenvalues
will be either positive or 0.
That is, non-negative.
So our matrix will be
1, 0, 1; 0, 1, 1; 1, 1,
and we're allowed to
change this third entry.
How do we figure out if this
matrix is positive semidefinite
or not?
Well, I was talking
about the eigenvalues.
But maybe the easiest way is
to do the determinant test.
The determinant of
the small one by one
left uppermost matrix is 1.
The determinant of the two
by two upper leftmost matrix
is 1 times 1 minus 0
times 0, 1, also positive.
So we need to check that the
determinant of the whole matrix
will also be non-negative.
So what is the determinant
of this matrix?
It is equal to the
three by three matrix.
So do you know how
to do it quickly?
There's this way that only
works for three by three and not
for bigger.
Which is, the determinant will
be 1 times 1 times a_(3,3) plus
0 times 1 times 1.
That's 0.
Plus 1 times 0 times
1 That's 0 again.
Minus 1 times 1 times 1 minus
1 times 1 times 1 minus a_(3,3)
times 0 times 0.
That's 0.
So this is the determinant.
It's a_(3,3) minus 2.
And I want it to be
greater than or equal to 0.
This will guarantee
that the matrix
is positive semidefinite.
So a_(3,3) must be bigger
than or equal to 2.
The smallest value for a_(3,3)
that will make the matrix
positive semidefinite is 2.
There's another part
to the question still,
which is what is the smallest
c that will make the matrix A
plus c*I positive semidefinite?
How should we do this?
The quickest way is to
do the eigenvalue test.
A has eigenvalues 1, 2, and -1.
So A plus c*I has
eigenvalues-- well,
you're just adding
c*I to the matrix.
And in this particular
case, you should know by now
that that keeps the
eigenvectors the same
and adds the number c to
each of the eigenvalues.
And I want each one of
these to be non-negative.
For that to be
true, I have to have
c greater than or equal to 1.
c greater than or equal to 1.
So the smallest value
that c can take that
will make the matrix A
positive semidefinite is 1.
That solves parts A
and B of this question.
There is a part C
to this question.
Let me show it to you.
It's says: starting with one of
these three vectors, [3, 0, 0],
[0, 3, 0], or [0, 0, 3], and
with u_(k+1) equals to a half
of A times u_k, what is the
limit behavior of u_k as k goes
to infinity?
I've written the matrix one half
of A here for your convenience.
And now, you can hit
pause and work on it.
And when you're ready, we'll
get back and solve it together.
I hope you managed
to solve this one.
Now let's do it together.
Well, if you've noticed,
the matrix one half of A
is a Markov matrix.
So there are all these
results about Markov matrices
and steady states and so on.
Usually, Markov matrices
have a unique steady state,
but that is only true when
there are no non-zero entries.
But here, there are.
So we can't guarantee that
there's a unique steady state.
What we can do is look
at the eigenvalues
and see if this is
still true nonetheless.
What are the eigenvalues
of A-- of one half of A?
Well, if you remember from
part A, the eigenvalues of A
were 1, 2, and -1.
So the eigenvalues of one
half of A-- taking a multiple
does not change the eigenvector,
but it changes the eigenvalue
by the same multiple.
It would be 1/2, 2 divided
by 2 is 1, and minus 1/2.
So here are the
eigenvalues of A.
And there's only
one eigenvalue that
has absolute value equal to 1.
So you actually still get a
unique steady state vector.
So everything is fine.
We can proceed as usual.
And the usual procedure is
you find the eigenvector
corresponding to
that eigenvalue, 1.
And that will be
the limit behavior
as k goes to infinity of u_k.
So what is the eigenvector
corresponding to 1?
Eigenvector.
Well, you already
know how to do this,
so I will just
write the solution.
It is [1, 1, 1].
That means that u_k,
as k goes to infinity,
will converge to some
appropriate multiple
of this eigenvector [1, 1, 1].
How do you know which
multiple to use?
Well, as usual in
Markov matrices,
when you do an iteration
of the process,
when you do u_(k+1) is equal
to one half of A times u_k,
that does not change the sum of
the entries of the vector u_k.
So whatever the sum was here,
it will still be the same here.
If you go all the
way back and you
start with u_0, whatever the
sum of the entries was here,
that's what it will be all
the way through u_1, u_2, u_3,
and so on, all the way to
the steady state, u_infinity.
So whatever the
multiple of [1,  1, 1],
it has to have the sum of
these entries add up to 3.
Well, that's already there.
We already happened to pick
the correct eigenvector,
so that's very convenient.
The correct multiple is
simply the vector [1, 1, 1].
So the limit behavior of
u_k as k goes to infinity
is u_infinity equal
to [1, 1,  1].
We're done.
Thank you.
