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