$rk(A^2)=rk(B^2) \implies rk(A)=rk(B)$ is it true?

Question

The original statement is this: given A,B matrices $n \times n$, if $A^2$ is "Left-Right equivalent" to $B^2$ then A is LR equivalent to B (is it true or false?)

I know that A is LR equivalent to B iff rk(A)=rk(B) so I decide to work with ranks.

I think it's true so I tried a proof by contradiction but something is wrong about it: if it were $rk(A)\neq rk(B)$ it's not in general $rk(A^2)=rk(B^2)$ (this should be the absurd) so it has to be $rk(A)=rk(B)$

But if the claim were the opposite ($rk(A^2)=rk(B^2) \implies rk(A)\neq rk(B)$) I can use again the contradiction: if it were $rk(A)=rk(B)$ it's not true in general that $rk(A^2)=rk(B^2)$ so it has to be $rk(A)\neq rk(B)$.

I see that I'm not able to apply properly the proof by contradiction. What's the error?

How can I solve the original question?

For completeness A is LR equivalent to B if there exist P,Q invertible matrices such that A=PBQ

Thank you.

Answer 1

Nope! The statement doesn't hold. Take $$ A = 0, \quad B = \pmatrix{0&1\\0&0} $$

Answer 2

The statement $\text{rank}(A^2) = \text{rank}(B^2) \implies \text{rank}(A) = \text{rank}(B)$ is not true.

Take $A = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)$ and $B = \left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right)$. Then $A^2 = B^2 = \left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right)$, so $\text{rank}(A^2) = \text{rank}(B^2) = 0$, but $\text{rank}(A) = 1$ and $\text{rank}(B) = 0$.