A question on rank of powers of matrices

Question

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?

Answer 1

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 $\text{rk}(A^{k+1}) = \text{rk}(A^{k+2})$.

Answer 2

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?