Does $\operatorname{rank} (A^2) = \operatorname{rank} (A)$ for any matrix $A\in \operatorname{Mat}_{n \times n}$?

Question

I wanted to prove/disprove whether for every matrix $A\in \operatorname{Mat}_{n \times n}$

$\operatorname{rank} (A^2) = \operatorname{rank}(A)$ for any matrix $A\in \operatorname{Mat}_{n \times n}$

I know that for two matrices $A, B$ (it actually doesn't matters what is the size of $B$ as long as $AB$ is defined)

$$\operatorname{rank}(AB) \leq \min \{ \operatorname{rank}(A), \operatorname{rank}(B)\}$$

Therefore, $$\operatorname{rank}(A^2) \leq \operatorname{rank} (A)$$

In order to prove that:

$$\operatorname{rank}(A^2) = \operatorname{rank}(A)$$

you'd have to prove that: $$\operatorname{rank}(A^2) \geq \operatorname{rank}(A)$$ which I don't know how to prove, or finding a different way to prove directly both directions (meaning; one direction is $\leq$ and the other is $\geq$)

Answer 1

It is not true that the square of a matrix has the same rank as the matrix itself. For instance, the $2\times 2$ matrix $A= \left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right]$ has rank $1$ and satisfies $A^2=0$.

Answer 2

First, it makes no sense to write $A^2$ if $A$ is not a square matrix. So you need $m=n$.

This being said, there is a class of matrices called nilpotent, from which it is trivial to build counterexamples:

$$\operatorname{rk}\left(\matrix{0&1\\0&0}\right)=1$$$$\operatorname{rk}\left(\matrix{0&1\\0&0}\right)^2=0$$