Find a matrix A such that $\operatorname{rank}{A} = \operatorname{rank}{A^2} \neq \operatorname{rank}{A^3}$ Let $A$ be a complex square matrix of order 2 ($A \in M_{2,2}$).
Then, does there exist $A$ such that $\operatorname{rank}{A} = \operatorname{rank}{A^2} \neq \operatorname{rank}{A^3}$?
If that doesn't exist, how can I prove it?

$\DeclareMathOperator{\rank}{rank}$
In general, for a linear transformation $T: V \to V$, one has $T^{k+1} V \subseteq T^k V$ for all $k \ge 0$.  If one ever has equality for a  particular $k$, i.e. $T^{k+1} V  =  T^k V$, then $T^{k + l} V = T^k V$ for all $l \ge 1$. In fact, this holds for $l =1$ by assumption, and if it holds for any particular $l$, then also $$T^{k + l + 1} V = T T^{k + l} V = T T^k V = T^{k + 1} V = T^k V.$$ 
So in particular, for any square matrix $A$ of any size, if $\rank(A^2) = \rank(A)$, then $\rank(A^l) = \rank(A)$ for all $l \ge 1$. 
