If $Rank(A^2) = Rank(A)$, then $Rank(A^3) = Rank(A)$, where $A \in \mathbb R^{n\times n}$

Question

I want to determine if the above statement is true or false, and if it is false, give a counter-example. I know for a fact that it it true. But, I am not sure if this proof is entirely correct or not:
$Image(A^2) \subseteq Image(A)$, but $dim(Image(A^2))=dim(Image(A))$, so we must have that $Image(A^2)=Image(A)$, and thus, $A^2=A$. So, $A^3=A^2=A$, and hence, $Rank(A^3)=Rank(A)$.

I am not entirely sure if the reasoning $Image(A^2)=Image(A) \implies A^2=A$ is correct or not.

It would be greatly appreciated if anyone can help me with that!

Answer 1

You can also use Frobenius Rank Inequality as

$rank(A^3) \ge rank(A^2) + rank(A^2) - rank(A)$

$\implies rank(A^3) \ge rank(A)$

And also $rank(A^3) \le min\{rank(A^2),rank(A)\}$

$\implies rank(A^3) \le rank(A)$

And we get $rank(A^3)=rank(A)$.