BEN HARRIS: Hi.
Today we're going to do a problem
about similar matrices.
The problem is right here.
So the question asks, which of the following statements are
true, and it asks you to explain why.
The first statement is if A and B are similar matrices,
then 2A cubed plus A minus 3 times the identity is similar
to 2 times B cubed plus B minus 3 times the identity.
The second question asks if A and B are 3 by 3 matrices with
eigenvalues 1, 0, and -1, then they're similar.
And the third part asks you whether these two j matrices
are similar.
OK.
I'll give you a moment to hit pause and try it on your own.
I'll be back in just a moment, and we can do it together.
And we're back.
Let's start with part a.
Part a is true.
Why?
Well, what does it mean for A and B to be similar?
Right, we know there's some matrix M, such that-- that's
what st means--
when I multiply A on the left by M and the right by M
inverse,
I get B. OK.
So let's take that same matrix M and multiply it on the left
and the right of 2A cubed plus A minus 3 times the identity.
OK.
What do we get here?
Well, the point is when A times A cubed times m inverse,
we can just write that as 3 MAM inverses.
Similarly, we have an MAM inverse and an M times the
identity times m inverse.
Good.
Remember that M times A times M inverse is B. So we just get
2B cubed, 3B plus B minus--
well, M times the identity is just M. So we just get the
identity back, and we have 3 times the identity.
Good.
And this is a general remark, that if you have matrices A
and B that are similar, then any polynomials in these
matrices A and B will be similar.
It's the exact same justification.
OK.
Now let's go on to part B. So now A and B are 3 by 3 similar
matrices with the same eigenvalues.
And their eigenvalues are distinct.
So it turns out that b is true, as well.
And why is that?
A matrix with distinct eigenvalues is diagonalizable.
So we can write A as S, lambda, S inverse, where
lambda is just this eigenvalue matrix.
We can also write B as T, lambda, T inverse, where
lambda is the same in both cases because they have the
same eigenvalues.
Good.
Now I'll let you--
so before we check it, let's just say the point.
The point is if two matrices are similar to the same
matrix, then they're similar to each other.
Similarity is a transitive relation.
And I'll just let you check that you can take TS inverse A
times TS inverse inverse, and you'll get B. This falls
directly from these two relations.
Good.
Now let's take on part C. Part C is false.
Let's come back over here and look at these two
matrices, J1 and J2.
The first thing you should see is that these
two are Jordan blocks--
they're not Jordan blocks, they're matrices in Jordan
normal form.
They're different matrices in Jordan normal form, so they
will not be similar.
But let's actually see why.
Let's look at--
remember one of the things that similarity preserves are
eigenvectors and eigenvalues.
So let's look at the eigenspace with eigenvalue
minus one with these two matrices.
So J1 plus the identity--
let's look at the nullspace of this matrix.
So this is just 0's on the diagonal and 1's right above
the diagonal, and J2 plus the identity.
This is just 0, 1, 0, 0.
So the point is that the nullspace of this matrix is
just one dimensional.
So there's only one independent eigenvector of J1
with eigenvalue minus 1 whereas the nullspace of this
matrix is two dimensional.
There are two independent eigenvectors with
eigenvalue minus 1.
So the dimension--
the nullspace--
of J1 plus the identity, this is 1, and this is 2.
So they cannot possibly be similar.
Good.
So that completes the problem.
It's a nice exercise to do this more generally.
And you can use these techniques not just looking at
the number of independent eigenvectors and the nullspace
of your J minus lambda I matrix, but also powers of J
minus lambda I and their nullspaces.
You can use this to show that any two matrices in Jordan
normal form that are different are not similar.
This same method works.
And that's a nice exercise if you want to go a little
further with similar matrices.
Let's just recap the properties we saw here.
We saw that if we had two similar matrices, then any
polynomials in those matrices were similar.
And we saw that if we have two matrices that have the same
distinct eigenvalues, then they're similar.
And we saw that, in a special case, we saw that two matrices
in Jordan normal form that are different are not similar.
Thanks.