# Similar matrices please check my proof.

1. Dec 11, 2008

### Diffy

1. The problem statement, all variables and given/known data

Prove that every periodic matrix is similar to a diagonal matrix.

2. Relevant equations

I use ~ to denote A similar to B as A~B

3. The attempt at a solution

Let A be periodic, and let A^m = I.

If A^m = I this implies A^m ~ I.

Claim: if X~Y implies X^n ~ Y^n for n a positive integer.
By induction.

Base case n = 2.
If X~Y then there exists a Z such that X = ZY(Z^-1).
X^2 = X*X = ZY(Z^-1)ZY(Z^-1) = Z*(Y^2)*(Z^-1).

Inductive step Assume true for some n.

X^(n+1) = (X^n) * X = Z(Y^n)(Z^-1)* ZY(Z^-1) = Z (Y^(n+1))(Z^-1) and we are done.

So since A^m ~ I implies A^(m+1) ~ I ^2 implies A^(m+1) ~ I, but since A^(m+1) = A we have A~I and since I is a diagonal matrix we are done.

2. Dec 11, 2008

### Dick

No, that doesn't work. You've shown X~Y implies X^n~Y^n. That's fine. But applying that to A^m~I only gives you e.g. (A^m)^2~I^2 or A^(2m)~I. That doesn't help. Think about what the blocks must look like in the Jordan normal form of A.

Last edited: Dec 11, 2008