Let $n\ge2$. For any given $X \in M_{n,n}$ with complex values, let $M(X)$={$X^{k}$|$k \in N^{*}$}. If $M(A)$ is finite, show that rank($A$) = rank($A^{k}$) for any $k\ge1$ only and only if there exists $m\ge2$ so that $A^{m}=A$.
Ideas
Let $x_{n}=$ rank($A^{n+1}$)$-$rank($A^{n}$). No matter the case we are in, I have proven that $$ \lim\limits_{n\rightarrow \inf} x_{n} = 0.$$
As $x_{n}$ has only integer values, it means that from a finite variabile $r$, all values of $x_{n}=0$, which means that rank($A^{r}$) = rank($A^{r+1}$) = rank($A^{r+2}$) =...
So, for the case in which we know that there exists $m\ge2$ so that $A^{m}=A$, it becomes quite obvious how rank($A$) = rank($A^{k}$) for any $k \ge 1$. My question is, how can we prove it the other way around?
Let us first look at the case when $A$ is nonsingular. In this case given the condition that $M(A)$ is finite with size $p$, then there is a $1\leq k \leq p$ and a $j>p$ such that $A^j = A^k \implies A^{j-k+1} = A$. Clearly $m=j-k+1\geq 2$.
Now let us look at the singular case. If $\DeclareMathOperator{\rank}{rank}\rank(A)=\rank(A^k)$ for $k=1,2,\ldots$ then $$ \rank(A) = \rank(A^k)=\rank(A^{k-1})-\dim N(A)\cap R(A^{k-1}) \implies \dim N(A)\cap R(A^{k-1})=0\implies N(A^{k-1})\cap R(A)= \{0\}, k=2,3\ldots $$ In particular $R(A)\cap N(A) =\{0\}\implies \mathbb{C}^n=R(A)\oplus N(A)$ by the rank-nullity theorem. Now if $S = \begin{pmatrix}X & Y\end{pmatrix}$ is nonsingular such that $R(X) = R(A)$ and $R(Y) = N(A)$, then $$ AS = \begin{pmatrix}AX & 0\end{pmatrix} = \begin{pmatrix}XZ & 0\end{pmatrix} = S(Z\oplus 0_{n-r}), $$ where $r=\rank(Z)=\rank(A), Z\in M_r$. Now $M(A)$ finite implies that $M(Z)$ is finite, which allows us to come to the conclusion, by applying the nonsingular case to $Z$, that there is an $m$ such that $Z^m = Z$. Therefore, since $A$ is similar to $Z\oplus 0_{n-r}$, it follows that there exists an $m$, such that $A = A^m$.