Show that there exists $A$ $(n×n)$ matrix with complex entries such that...

Question

We have $M\left(X\right)=\left\{X^{k}|k\in \mathbb{N}^{*}\right\}$ Show that there exists $A$ $(n×n)$ matrix with complex entries such that $M\left(A\right)$ is finite and it does not contain $0$ and $A^{m}≠A$ for every $m\geq2$, if and only if $n\geq3$.
We can say that A is not a periodic matrix and also it is not nilpotent. I dont really know how to counter this type of problems. Can someone give me a hint or help me?

Answer 1

Since $M(A)$ is finite, there must be some $r<s \in \Bbb{N}^*$ such that $A^r=A^s$. Let's pick out the smallest such pair and denote $k=s-r$. Thus $A^r=A^s \implies A^s - A^r=A^r(A^k-I)=0$.

Suppose the minimal polynomial of $A$ is $p(x) \in \Bbb{C}[x]$, then $p|x^r(x^k-1)$. Assume $p(x)=x^dq(x)$ where $q(x)|x^k-1$, then $p(x)|x^d(x^k-1) \implies A^d = A^{d+k}$. Due to the minimality of $r,s$, we must have $d=r,p(x)=x^rq(x)$.

As your proposition stated, $A$ is not nilpotent, so $q(x) \ne 1$; $A^m \ne A \; \forall m>1$, so $r \ne 1$. Therefore, $n \ge \deg p(x)=r + \deg q(x) \ge 2+1=3$. And for every $n \ge 3$, any complex matrix whose minimal polynomial is $x^2(x-1)$ will satisfy your requirement.