PROFESSOR: Hi guys.
Today, we are going to play around
with the basics of eigenvalues and eigenvectors.
We're going to do the following problem,
we're given this invertible matrix A,
and we'll find the eigenvalues and eigenvectors
not of A, but of A squared and A inverse minus the identity.
So, this problem might seem daunting at first, squaring a 3
by 3 matrix, or taking an inverse of a 3 by 3 matrix
is a fairly computationally intensive task,
but if you've seen Professor Strang's lecture
on eigenvalues and eigenvectors you
shouldn't be all too worried.
So I'll give you a few moments to think
of your own line of attack and then you'll see mine.
Hi again, OK, so the observation that makes our life really easy
is the following one.
So say v is an eigenvector with associated eigenvalue lambda
to the matrix A. Then, if we hit v with A squared, well,
this we can write it as A times A*v, but A*v is lambda*v,
right?
So we have A*lambda*v.
Lambda is a scalar, so we can move it in front and get
lambda*A*v, and lambda*A*v is, when we plug in A*v, lambda*v,
is just lambda squared v. So, what we've find out is that
if v is an eigenvector for A then it's also an eigenvector
for A squared.
Just that the eigenvalue is the eigenvalue squared.
Similarly, if we hit A inverse-- if you hit v with A inverse.
So in this case we can write v as A*v over lambda,
given that of course, lambda is non-zero.
But the eigenvalues of an invertible matrix
are always non-zero, which is an exercise
you should do yourselves.
So if we just, then, take out the A
and combine it with A inverse, this is the identity,
and so we get 1 over lambda v. So v
is also an eigenvector for a inverse,
with eigenvalue the reciprocal of lambda.
OK, and from here, of course, A inverse minus the identity
is lambda inverse minus 1 v, so the eigenvalue
of A inverse minus the identity is 1 over lambda minus 1.
OK, so, what we've figured out is:
we just need to find the eigenvalues and eigenvectors
of A and then we have a way of finding what the eigenvalues
and eigenvectors of A squared and A inverse
minus the identity will be.
OK so, how do we find the eigenvalues?
Well, what does it mean for lambda
to be an eigenvalue of A?
It means that the matrix A minus lambda the identity
is singular, which is precisely the case when
its determinant is 0, OK?
So we need to solve the following equation:
1 minus lambda, 2, 3; 0, 1 minus lambda, -2; and 0, 1,
4 minus lambda.
OK, it's fairly obvious which column we should
use to expand this determinant.
We should use the first column, because we
have only one non-zero entry, and so this
is equal to 1 minus lambda times the determinant of the two
by two matrix 1 minus lambda, -2; 1, 4
minus lambda, which is, I'm going to do the computation up
here.
1 minus lambda, lambda squared minus 5 lambda plus 6.
Which is a fairly familiar quadratic,
and we can write it as the product of linear factors,
as lambda minus 2, lambda minus three.
So the three eigenvalues of A are 1, 2, and 3.
OK so, first half of our problem is done,
now we just need to find what the eigenvectors associated
with each of these eigenvalues are.
How we do that?
Well, let's see.
Let's figure out what the eigenvector associated
with lambda equals 1 is.
So, we know that the eigenvector needs
to be in the null space of A minus lambda
the identity, so A minus the identity, v,
so-- write this out-- it's, 0, 0, 3, 2, 3, 0, -2, 0, 1.
And we see that the first column is 0,
so the first variable will be our free variable
if we want to solve this linear system of equations.
And you can just set it to 1 and it's not
hard to see that the other two entries should be 0.
So we can do the same procedure with the other two eigenvalues
and we'll get an eigenvector for each eigenvalue.
And in the end-- let me go back here.
So I'm going to put our results in a little table.
So A squared, inverse minus the identity, so the first row
will be eigenvalues.
So it's going to be: if lambda is an eigenvalue for A,
then we saw that lambda squared will
be the eigenvalue for A squared and lambda inverse minus 1
will be the eigenvalue for A inverse minus the identity.
And the eigenvectors will be the same.
OK, we're done.