# A square matrix $A$ is such that $\mathrm{rank}(A^{k}) = \mathrm{rank}(A^{k+1})$

A square matrix $$A$$ is such that $$\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+1})$$. Then $$\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+i})$$ for all $$i$$.

I think we prove this by induction on $$i$$.

Let $$i = 2$$. We show that rank $$(A^{k})$$ = rank $$(A^{k+2})$$.

Now, we know that rank $$(A^{k+2})$$ = rank $$(A^{k+1}A)$$ $$\le$$ min (rank$$(A^{k+1})$$, rank$$(A)$$) $$\le$$ rank $$(A^{k+1})$$ = rank $$(A^{k})$$.

However, I cannot proceed further.

Think geometrically. We have $A^{k+1}=A^kA$, so the image of $A^{k+1}$ is contained in the image of $A^k$. If the rank of the two are equal, that means the image is the same. Since $A^{k+1}=AA^k$, this means that $A$ is a bijection on said image.
This can be nicely shown by thinking of $A$ as a linear transformation $f \colon K^n \to K^n$ and looking at the decreasing chain $\operatorname{im} f^p \supseteq \operatorname{im} f^{p+1}$ as follows:
If $A \in \operatorname{Mat}_n(K)$ then let $f \colon K^n \to K^n$ with $f(x) = Ax$ be the associated endomorphism of $K^n$, and let $R_p := \operatorname{im} f^p$ for all $p \geq 0$. Then $$\operatorname{rank} A^p = \dim \operatorname{im} f^p = \dim R_p \quad \text{for all p \geq 0},$$ as well as $$f(R_p) = f(\operatorname{im} f^p) = \operatorname{im} f^{p+1} = R_{p+1} \quad \text{for all p \geq 0}.$$
Because we have a decreasing chain $$K^n = R_0 \supseteq R_1 \supseteq R_2 \supseteq R_3 \supseteq \dotsb$$ it follows from $\operatorname{rank} A^{k+1} = \operatorname{rank} A^k$ that $\dim R_{k+1} = \dim R_k$ and thus $R_{k+1} = R_k$.
It then further follows that $$R_{k+2} = f(R_{k+1}) = f(R_k) = R_{k+1},$$ and we find inductively that $R_{k+j} = R_{k+j+1}$ for all $j \geq 0$. So we have $$R_k = R_{k+1} = R_{k+2} = R_{k+3} = \dotsb$$ and thus $R_{k+i} = R_k$ for all $i \geq 0$.
so you know now that $rank (A^k)=rank (A^{k+1})=rank (A^{k+2})$. you just can take $k'=k+1$ and do the same work for $A^{k'}$. You obtain $$rank (A^k)=rank (A^{k+1})=rank (A^{k'})=rank (A^{k'+1})=rank (A^{k'+2})= rank (A^{(k+1)+2}) = rank (A^{k+3})$$ and so on