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.
