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
2 A 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 2 A cubed plus A minus 3 times
the identity.
OK.
What do we get here?
Well, the point is what M times A cubed times M inverse,
we can just write that as three M A M inverses.
Similarly, we have an M*A M 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 2 B cubed-- three B's-- 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 T S inverse
A times T S inverse inverse, and you'll
get B. This follows 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,
J_1 and J_2.
The first thing you should see is
that these two are Jordan blocks-- sorry,
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 J_1 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 J_2 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 J_1 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 J_1 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.