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BEN HARRIS: Hi.

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Today we're going to do a
problem about similar matrices.

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The problem is right here.

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So the question asks, which
of the following statements

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are true, and it asks
you to explain why.

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The first statement is: if A
and B are similar matrices, then

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2 A cubed plus A minus 3
times the identity is similar

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to 2 times B cubed plus B
minus 3 times the identity.

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The second question
asks: if A and B

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are 3 by 3 matrices with
eigenvalues 1, 0, and -1,

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then they're similar.

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And the third part asks you
whether these two J matrices

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are similar.

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OK.

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I'll give you a moment to hit
pause and try it on your own.

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I'll be back in just a moment,
and we can do it together.

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And we're back.

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Let's start with part
A. Part A is true.

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Why?

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Well, what does it mean
for A and B to be similar?

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Right, we know there's
some matrix M, such

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that-- that's what
"st" means-- when

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I multiply A on the left by
M and the right by M inverse,

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I get B.

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OK.

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So let's take that same
matrix M and multiply it

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on the left and the right of
2 A cubed plus A minus 3 times

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the identity.

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OK.

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What do we get here?

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Well, the point is what M
times A cubed times M inverse,

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we can just write that
as three M A M inverses.

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Similarly, we have an M*A
M inverse and an M times

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the identity times M inverse.

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Good.

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Remember that M times
A times M inverse is B.

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So we just get 2 B cubed--
three B's-- plus B minus-- well,

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M times the identity is just
M. So we just get the identity

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back, and we have 3
times the identity.

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Good.

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And this is a general remark,
that if you have matrices

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A and B that are similar,
then any polynomials

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in these matrices A
and B will be similar.

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It's the exact
same justification.

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OK.

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Now let's go on to
part B. So now A and B

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are 3 by 3 similar matrices
with the same eigenvalues.

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And their eigenvalues
are distinct.

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So it turns out that
b is true as well.

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And why is that?

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A matrix with distinct
eigenvalues is diagonalizable.

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So we can write A as S lambda
S inverse, where lambda is just

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this eigenvalue matrix.

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We can also write B as T
lambda T inverse, where lambda

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is the same in both
cases because they

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have the same eigenvalues.

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Good.

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Now I'll let you-- so
before we check it,

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let's just say the point.

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The point is if two matrices
are similar to the same matrix,

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then they're similar
to each other.

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Similarity is a
transitive relation.

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And I'll just let you check
that you can take T S inverse

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A times T S inverse
inverse, and you'll

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get B. This follows directly
from these two relations.

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Good.

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Now let's take on part
C. Part C is false.

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Let's come back over here and
look at these two matrices,

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J_1 and J_2.

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The first thing
you should see is

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that these two are
Jordan blocks-- sorry,

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not Jordan blocks, they're
matrices in Jordan normal form.

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They're different matrices
in Jordan normal form,

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so they will not be similar.

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But let's actually see why.

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Let's look at-- remember,
one of the things

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that similarity preserves are
eigenvectors and eigenvalues.

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So let's look at the eigenspace
with eigenvalue minus one

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with these two matrices.

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So J_1 plus the identity--
let's look at the nullspace

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of this matrix.

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So this is just
0's on the diagonal

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and 1's right
above the diagonal.

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And J_2 plus the identity.

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this is just 0, 1, 0, 0.

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So the point is that the
nullspace of this matrix

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is just one-dimensional.

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So there's only one
independent eigenvector

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of J_1 with eigenvalue minus 1.

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Whereas the nullspace of this
matrix is two-dimensional.

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There are two
independent eigenvectors

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with eigenvalue minus 1.

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So the dimension-- the nullspace
of J_1 plus the identity,

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this is 1, and this is 2.

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So they cannot
possibly be similar.

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Good.

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So that completes the problem.

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It's a nice exercise to
do this more generally.

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And you can use these techniques
not just looking at the number

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of independent eigenvectors and
the nullspace of your J minus

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lambda*I matrix, but also
powers of J minus lambda I

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and their nullspaces.

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You can use this to
show that any two

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matrices in Jordan normal
form that are different

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are not similar.

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This same method works.

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And that's a nice
exercise if you

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want to go a little further
with similar matrices.

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Let's just recap the
properties we saw here.

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We saw that if we had
two similar matrices,

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then any polynomials in
those matrices were similar.

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And we saw that if we
have two matrices that

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have the same distinct
eigenvalues, then

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they're similar.

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And we saw that,
in a special case,

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we saw that two matrices
in Jordan normal form

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that are different
are not similar.

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Thanks.