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.
