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PROFESSOR: Today, we're
going to be solving

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a problem from a final exam.

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And here it is.

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It's about a matrix A,
[1, 0, 1;  0, 1, 1; 1, 1, 0].

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And we know that this matrix
has two eigenvalues, 1 and 2.

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And we also know that
if we do elimination,

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the first two pivots
will be 1 and 1.

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And here are two questions
about this matrix.

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The first one is find lambda_3
and d_3, the third eigenvalue

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and the third pivot.

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And the second one asks you,
what is the smallest a_(3, 3)--

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so if you can change this entry,
what is the smallest number

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that you can put there that
will make the matrix A positive

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semidefinite?

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And also, if instead of changing
that entry, you do A plus c*I,

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what is the smallest number
c that will make that matrix,

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A plus c*I, positive
semidefinite?

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There's also a third
part to the question,

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but we'll get to that later.

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Why don't you hit pause and
work on these two parts?

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And when you're ready, come back
and I'll show you how I did it.

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Hi.

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I hope you managed to do parts
A and B. Let's work on it

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together.

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Part A. Well, we want to know
what the third eigenvalue is.

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And you know what
the first two are.

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What else do you know about
eigenvalues and the matrix?

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You know that the sum of all
the eigenvalues of the matrix

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is equal to the
trace of the matrix.

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So lambda_1 plus
lambda_2 plus lambda_3

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is equal to the
trace of the matrix.

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In this case, you have
1 plus 2 plus lambda_3

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equals to the trace.

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The trace is the sum of
the diagonal entries.

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So, come over here.

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The trace is 1 plus 1 plus 0.

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The trace is equal to 2.

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So we have 3 plus
lambda_3 is equal to 2.

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So lambda_3 is equal to minus 1.

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On to the third pivot.

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We don't really want
to do elimination.

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That would take too long.

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So there must be some
trick that we can use.

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Well, we have the
first two pivots,

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and we want to know the third.

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Remember, when you do
elimination steps, that

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does not change the
determinant of the matrix.

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And you're left with
an upper triangular.

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So the determinant
of that matrix

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will be d_1 times d_2 times d_3.

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And it will still be
equal to the determinant

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of A. I guess there's a small
caveat that I should point out.

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The pivots are not always
the diagonal entries.

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It might be that one of the
diagonal entries will be 0.

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That happens if the
matrix is singular.

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But here, all my three
eigenvalues are non-zero.

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They are 1, 2, and -1.

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So that won't happen.

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So this is actually
possible to do.

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The product of the
three pivots will

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be equal to the determinant
of A. And the determinant of A

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is the product of the
eigenvalues, 1 times 2 times

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-1.

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So it's equal to -2.

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1 times 1 times
d_33 is equal to -2.

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Here is your third pivot, d_3.

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That finishes part A. Is
that the result that you got?

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Let's do part B. What is the
smallest a_(3,3) entry that

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would make the matrix
positive semidefinite?

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Well, first of all, note that
A is not positive semidefinite

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yet.

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The eigenvalues
are 1, 2, and -1.

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-1 is negative, so the matrix
is not positive definite and not

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even positive semidefinite.

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Positive semidefinite means
that all the eigenvalues

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will be either positive or 0.

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That is, non-negative.

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So our matrix will be
1, 0, 1; 0, 1, 1; 1, 1,

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and we're allowed to
change this third entry.

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How do we figure out if this
matrix is positive semidefinite

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or not?

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Well, I was talking
about the eigenvalues.

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But maybe the easiest way is
to do the determinant test.

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The determinant of
the small one by one

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left uppermost matrix is 1.

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The determinant of the two
by two upper leftmost matrix

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is 1 times 1 minus 0
times 0, 1, also positive.

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So we need to check that the
determinant of the whole matrix

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will also be non-negative.

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So what is the determinant
of this matrix?

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It is equal to the
three by three matrix.

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So do you know how
to do it quickly?

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There's this way that only
works for three by three and not

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for bigger.

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Which is, the determinant will
be 1 times 1 times a_(3,3) plus

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0 times 1 times 1.

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That's 0.

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Plus 1 times 0 times
1 That's 0 again.

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Minus 1 times 1 times 1 minus
1 times 1 times 1 minus a_(3,3)

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times 0 times 0.

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That's 0.

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So this is the determinant.

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It's a_(3,3) minus 2.

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And I want it to be
greater than or equal to 0.

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This will guarantee
that the matrix

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is positive semidefinite.

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So a_(3,3) must be bigger
than or equal to 2.

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The smallest value for a_(3,3)
that will make the matrix

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positive semidefinite is 2.

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There's another part
to the question still,

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which is what is the smallest
c that will make the matrix A

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plus c*I positive semidefinite?

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How should we do this?

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The quickest way is to
do the eigenvalue test.

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A has eigenvalues 1, 2, and -1.

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So A plus c*I has
eigenvalues-- well,

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you're just adding
c*I to the matrix.

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And in this particular
case, you should know by now

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that that keeps the
eigenvectors the same

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and adds the number c to
each of the eigenvalues.

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And I want each one of
these to be non-negative.

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For that to be
true, I have to have

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c greater than or equal to 1.

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c greater than or equal to 1.

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So the smallest value
that c can take that

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will make the matrix A
positive semidefinite is 1.

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That solves parts A
and B of this question.

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There is a part C
to this question.

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Let me show it to you.

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It's says: starting with one of
these three vectors, [3, 0, 0],

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[0, 3, 0], or [0, 0, 3], and
with u_(k+1) equals to a half

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of A times u_k, what is the
limit behavior of u_k as k goes

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to infinity?

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I've written the matrix one half
of A here for your convenience.

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And now, you can hit
pause and work on it.

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And when you're ready, we'll
get back and solve it together.

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I hope you managed
to solve this one.

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Now let's do it together.

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Well, if you've noticed,
the matrix one half of A

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is a Markov matrix.

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So there are all these
results about Markov matrices

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and steady states and so on.

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Usually, Markov matrices
have a unique steady state,

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but that is only true when
there are no non-zero entries.

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But here, there are.

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So we can't guarantee that
there's a unique steady state.

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What we can do is look
at the eigenvalues

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and see if this is
still true nonetheless.

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What are the eigenvalues
of A-- of one half of A?

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Well, if you remember from
part A, the eigenvalues of A

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were 1, 2, and -1.

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So the eigenvalues of one
half of A-- taking a multiple

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does not change the eigenvector,
but it changes the eigenvalue

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by the same multiple.

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It would be 1/2, 2 divided
by 2 is 1, and minus 1/2.

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So here are the
eigenvalues of A.

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And there's only
one eigenvalue that

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has absolute value equal to 1.

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So you actually still get a
unique steady state vector.

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So everything is fine.

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We can proceed as usual.

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And the usual procedure is
you find the eigenvector

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corresponding to
that eigenvalue, 1.

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And that will be
the limit behavior

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as k goes to infinity of u_k.

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So what is the eigenvector
corresponding to 1?

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Eigenvector.

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Well, you already
know how to do this,

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so I will just
write the solution.

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It is [1, 1, 1].

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That means that u_k,
as k goes to infinity,

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will converge to some
appropriate multiple

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of this eigenvector [1, 1, 1].

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How do you know which
multiple to use?

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Well, as usual in
Markov matrices,

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when you do an iteration
of the process,

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when you do u_(k+1) is equal
to one half of A times u_k,

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that does not change the sum of
the entries of the vector u_k.

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So whatever the sum was here,
it will still be the same here.

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If you go all the
way back and you

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start with u_0, whatever the
sum of the entries was here,

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that's what it will be all
the way through u_1, u_2, u_3,

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and so on, all the way to
the steady state, u_infinity.

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So whatever the
multiple of [1,  1, 1],

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it has to have the sum of
these entries add up to 3.

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Well, that's already there.

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We already happened to pick
the correct eigenvector,

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so that's very convenient.

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The correct multiple is
simply the vector [1, 1, 1].

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So the limit behavior of
u_k as k goes to infinity

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is u_infinity equal
to [1, 1,  1].

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We're done.

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Thank you.