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?
 A: You can use the Frobenius inequality:
$$
\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}\,B+\textrm{rank}(ABC)
$$
It implies that
$$
\textrm{rank}\,A^3\geq \textrm{rank}\,A^2+\textrm{rank}\,A^2-\textrm{rank}\,A
=\textrm{rank}\,A,
$$
but $\textrm{rank}\,A^3>\textrm{rank}\,A$ is impossible, thus, $\textrm{rank}\,A^3=\textrm{rank}\,A$.
A: $\DeclareMathOperator{\rank}{rank}$
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 $l \ge 1$. 
A: The only way for a matrix to lose rank as you take powers, is through its Jordan blocks corresponding to zero eigenvalues.
A: Suppose $A\in M_2(\mathbb{C})$ is such that
$$\text{rank}(A) = \text{rank}(A^2) \ne \text{rank}(A^3)$$

First, suppose $\text{rank}(A)=0$.

Then $A = 0$, hence $A^2 = A^3 = 0$, so $\text{rank}(A^2) = \text{rank}(A^3) = 0$, contradiction.

Next, suppose $\text{rank}(A)=2$.

Then $A$ is non-singular, hence $A^2$ and $A^3$ are also non-singular, so $\text{rank}(A^2) = \text{rank}(A^3) = 2$, contradiction.

Thus, we must have $\text{rank}(A) = 1$.

Since the rank of a product of two matrices is less than or equal to the rank of any of the factors, it follows that $\text{rank}(A) = \text{rank}(A^2) = 1$, and $\text{rank}(A^3)=0$.

From $\text{rank}(A^3)=0$, we get $A^3=0$, hence, letting $p(x)$ denote the minimal polynomial of $A$, it follows that $p(x)$ divides $x^3$. 

We can't have $p(x)=x$, since $A \ne 0$, hence, since $p$ is monic of degree at most $2$, we must have $p(x)=x^2$.

But then $A^2 = 0$, contrary to $\text{rank}(A^2)=1$.
 
It follows that there is no such matrix $A$.
