Suppose $T^{m}=T^{n}$ for some positive integers $m \neq n .$ Find conditions on $T, m$ and $n$ such that $T^{2}=T$. Problem: Let $V$ be a vector space over $\mathbf{C}$ and let $T: V \mapsto V$ denote a linear map satisfying $T^{m}=T^{n}$ for some positive integers $m \neq n .$ Find conditions on $T, m$ and $n$ such that $T^{2}=T$.
Attempt:
Let $A$ be the matrix representing $T$ with respect to any base. Then $A^{m}=A^{n}$. If $A=P R P^{-1}$ and $R$ is the Jordan matrix corresponding to $A$, then $R^{m}=R^{n} .$
I have no idea how to continue, I can't see if what I've done will even lead me anywhere. Any help?
Surely $ A=0,I $ are among the conditions but how to find the rest of them?
Thanks in advance for any help!

If $T^2=T$, the minimal polynomial $\mu_T(x)$ of $T$ divides $x(x-1)$. Hence $\mu_T$ splits in linear factors and $T$ is diagonalizable with only $0$ or $1$ on the diagonal.
Conversely, if $T$ is diagonalizable with only $0$ or $1$ on the diagonal, then $T^m = T^n$ for any $m,n \in \mathbb N$.
Hence $T$ diagonalizable with only $0$ or $1$ on the diagonal is the condition we're looking for.
