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PROFESSOR: Hi, and welcome.
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Today, we're going
to do a problem
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about powers of a matrix.
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Our problem is first to find
a formula for the k-th power
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of this matrix C. This is
a two by two matrix that
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depends on variables a and b.
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And the second
part of our problem
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is to calculate C to the
100th in the special case
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where a and b are -1.
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You can hit pause now,
and I'll give you a minute
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to do the problem yourself.
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And then, I'll come back
and we can do it together.
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OK.
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We're back.
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Now, what's the first step in
finding powers of a matrix?
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Well, we need to find the
eigenvalues and eigenvectors
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of this matrix.
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So how do we do that?
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We compute the determinant of
C minus lambda I, which is just
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the determinant of this matrix:
2b minus a minus lambda,
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2b minus 2a; a minus b, and
2a minus b minus lambda.
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OK.
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If you compute this, well, we
have a lambda squared term.
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OK.
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Our lambda term,
if you look at it,
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you'll see we get 2b
minus a plus 2a minus b,
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which is just a plus b.
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And we have a negative sign.
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And it's negative a
plus b times lambda.
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And our last term is a
little tougher to compute.
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So I'll let you do it yourself.
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But you're just going
to get plus a*b.
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And this will factor as lambda
minus a times lambda minus b.
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So our eigenvalues
are just a and b.
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Now we need to find
our eigenvectors.
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So how do we do that?
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Well, what we need
to do is we need
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to look at C minus a
times the identity.
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And we need to find the
null space of this matrix.
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So what do we get here?
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We get 2b minus 2a.
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And then our next entry here,
we get 2a minus b minus a.
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So this is a minus b.
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Good.
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So you can see that this
matrix has the same columns
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and the same rows.
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And so you can see that a
vector in the null space,
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since this column is
-2 times this column,
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we can see that our first
eigenvector is just-- or 1, 2,
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I should say.
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It's just [1, 2].
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Good.
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Well, I guess we have space
to do the second one too.
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Why not?
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So let's write out
the second one also.
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Here, we're subtracting
b instead of a.
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You get b minus a.
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You get 2b minus 2a.
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We get a minus b.
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And what do we have here?
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We have 2a minus 2b.
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So now, what's in the
null space of this matrix?
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Well, what you can see
is that this column
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is -1 times that column.
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So our second eigenvector
is just going to be [1, 1].
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And I should remind you that
if you have a harder example,
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you can just find these
null spaces by elimination
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like we always do.
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Great.
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Now we have our eigenvalues
and our eigenvectors.
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So now we can write C
in a nice easy way that
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allows us to take powers of it.
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So what's that way?
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So that's C equals
S lambda S inverse.
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So this is just, what is S?
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Remember, S is our
matrix of eigenvectors.
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So S is the matrix 1, 2; 1, 1.
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Good.
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Now what is lambda?
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Lambda is the matrix
of eigenvalues.
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Right?
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So it's just a and b.
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Those are the diagonal
entries of my lambda matrix.
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And then, we just
find S inverse.
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So we just take
negative signs here
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and recall that we have to
divide by the determinant.
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And the determinant of
this matrix is just -1.
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So we just change
the signs there.
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Good.
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So this is our nice
decomposition of C.
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Now how do we take powers of C?
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Well, C to the k is just S
lambda to the k S inverse.
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[1, 1; 2, 1]; a to the k,
b to the k; [-1, 1; 2, -1].
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Good.
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And multiplying these
matrices together,
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just do a little
arithmetic here.
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Got a bunch of
powers of a and b.
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Because we take powers
of the eigenvalues.
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We have here, we have 2 b
to the k minus a to the k.
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Have a to the k
minus b to the k.
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2 b to the k minus 2 a to the k.
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And finally, we get 2 a
to the k minus b to the k.
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And this is our
k-th power matrix.
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Good.
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A quick check.
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It's always good to
check your work here.
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Let's plug in k equals 1.
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And what do we get?
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We get 2b minus a, a minus b,
2b minus 2a, and 2a minus b.
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And if we can go all the
way back to our matrix
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at the very beginning,
all the way back here,
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that agrees perfectly
with what we started with.
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So that's good.
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That means that we did
this decomposition right.
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Good.
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So now, we've computed the
k-th power of this matrix.
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Let's do a particular example.
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So let's plug in a and b are -1.
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So a equals b equals -1.
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And k equals 100.
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Then what do we get?
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Well, -1 to the 100th is just 1.
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So we're just plugging in 1
for b to the k and a to the k
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everywhere.
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And we just get, in this
case, C to the 100th is just
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[1, 0; 0, 1].
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It's just the identity matrix.
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Great.
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Great.
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OK.
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Now to summarize, how do
we take powers of a matrix?
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Well, first we diagonalize.
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We write our matrix
as S lambda S inverse.
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And then, we just take powers
of the diagonal matrix.