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The problem is about $rank (\mathbf{A}^k)$ when $k \rightarrow \infty$ 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 $rank(\mathbf{A}^k) \rightarrow 0$ when $k \rightarrow \infty$.

And for a nonsingular matrix, $\mathbf{A}^k$ is also nonsingular and $rank(\mathbf{A}^k)$ is the No. of the columns/rows.

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So my question is that is there any general theorem that tells us the result for a general square matrix.

Thanks!

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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 $$\text{rank}(A^k)=\begin{cases}h-k,&\ k\leq \ h-m\\ m,&\ k>r\end{cases}$$

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