I am interested in
$$\lim_{k \to \infty} \operatorname{rank} \left( \mathbf{A}^k \right)$$
for a $n \times n$ matrix $\mathbf{A}$. I know that
for a nilpotent matrix, $\mathbf{A}^k=0$ when $k$ is big enough, which means $\operatorname{rank} (\mathbf{A}^k) \to 0$ when $k \to \infty$.
for a nonsingular matrix, $\mathbf{A}^k$ is also nonsingular and $\operatorname{rank} \left( \mathbf{A}^k \right)$ is the number of the columns/rows.
Is there any general theorem that tells us the result for a general square matrix?
Since the image of $A^{k+1}=A^k\,A$ is included in the image of $A^k$, we have that the function $k\longmapsto \text{rank}(A^k)$ is non-increasing. As the only possible values are $\{1,\ldots,n\}$ the numbers $\text{rank}(A^k)$ have to stabilize in a finite number of steps.
If the rank of $A^{k+1}$ is equal to the rank of $A^k$, this forces the image of $A^{k+1}$ to be equal to that of $A^k$, and thus the rank stabilizes.
So, the possible ranks will be a strictly decreasing sequence of positive integers until it stabilizes.
Here is an example that shows that one can start at any rank $h$, and decrease until rank $m$. Let $B$ be nonsingular $m\times m$ and $C$ an $(h-m+1)\times (h-m+1)$ Jordan block with zero diagonal. Consider $$ A=\begin{bmatrix}B&0\\0& C\end{bmatrix}. $$ Then $$\text{rank}(A^k)=\begin{cases}h-k,&\ k\leq \ h-m\\ m,&\ k>r\end{cases}$$