Does $A^2$ similar to $B^2$ imply that $A$ is similar to $B$?

Question

So we have square matrices $A$ and $B$. Now suppose $A^2$ and $B^2$ are similar, does it follow that $A$ and $B$ are similar?
I don't think so, but I'm having trouble showing is not.

My attempt:
If $A$ is similar to $B$, and $A$ is similar to $C$, then $B$ is also similar to C. More importantly, $A$ is similar to the diagonal matrix of its eigenvalues, then $B$ must also be. So $A$ and $B$ have the same eigen values. But then I realised that this question doesn't even make the assumption that $A$ and $B$ are invertible, so they may or may not have those eigen values.

Can someone give me some hints to this question?
If this is a proof by counterexample, is there at least some way to narrow the possible contradictions down by some theorem?

score 2 · Answer 1 · answered Sep 03 '15 at 21:43

2

Hint: You can assume that $A$ and $B$ are diagonal.

answered Sep 03 '15 at 21:43

Dominik

19,963

5

And $1 \times 1$. :) – Andrew D. Hwang Sep 03 '15 at 22:03
If you were proving this (were it true) certainly it would be insufficient to assume the matrices are diagonal. However, what you really mean is that it's not true and one can find a diagonal counterexample. – Matt Samuel Sep 03 '15 at 22:11
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@MattSamuel that's too much of a hint. – guest Sep 03 '15 at 23:06

Does $A^2$ similar to $B^2$ imply that $A$ is similar to $B$?

1 Answers1