A question on rank of powers of matrices Let $A$ is a $n\times n$ matrix with complex entries and  $\operatorname{rank}(A^k)=\operatorname{rank}(A^{k+1})$ for some $k \in \mathbb{N}$.
Then prove that $\operatorname{rank}(A^{k+1})=\operatorname{rank}(A^{k+2})$.
Is there any easy way to show this?
 A: Let's look at the matrices as linear transformation $\mathbb C^n$.
We note that the rank equals the dimension of the image of the linear transformation. We also note that the dimension of the kernel equals $n-\text{dim}(\text{Im})$.
Now, for every $\text{ker}(A^k) \subset \text{ker}(A^{k+1})$.
Let's assume $\text{rk}(A^k) = \text{rk}(A^{k+1})$. Therefore, to prove $\text{rk}(A^{k+1}) = \text{rk}(A^{k+2})$, it is enough to show $\text{ker}(A^{k+2}) \subset \text{ker}(A^{k+1})$.
Let $x \in \text{ker}(A^{k+2})$. Then $A^{k+2}x = 0 = A^{k+1}(Ax)$, meaning $Ax \in \text{ker}(A^{k+1})$.
Since we assumed $\text{ker}(A^k) = \text{ker}(A^{k+1})$, we have $A^k(Ax) = 0 = A^{k + 1}x$, and therefore $x \in \text{ker}(A^{k+1})$, which proves that $\text{rk}(A^{k+1}) = \text{rk}(A^{k+2})$.
A: Hint. Note that for any nonnegative integer $m$,
$\mathbb{C}^n = \ker(A^m) \oplus \operatorname{im}(A^m)$.
Moreover, 
$
\ker(A^m) \subseteq \ker(A^{m+1})
$
, and 
$
\operatorname{im}(A^m) \supseteq \operatorname{im}(A^{m+1})
$.
Now remember that $\operatorname{rank}(A^m)=\dim(\operatorname{im}(A^m))$.
Can you take it from here?
