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