Let $A\in M_n(\mathbb{C})$ be a matrix s.t $A^N=I_n$. Prove, using representation theory, that $A$ is diagonalizable.
My attempt: We look at $G=\langle A\rangle\subset GL_n(\mathbb{C})$. This is a finite, cyclic group. Let's define a representation $(\rho,\mathbb{C}^n)$ of $G$ by $\rho_{A^k}(v)=A^kv$. This is obviously a representation. Since $G$ is finite and $|G|\neq0\in\mathbb{C}$, by Maschke's theorem, $\rho\simeq\oplus_{i=1}^{m}\rho_i$, with $(\rho_i,W_i)$ irreducible representations ($W_i\subset\mathbb{C}^n$ $G$-invariant). But since $\mathbb{C}$ is algebraically close and $G$ is abelian, each $\rho_i$ is representation of dimension $1$, meaning $$\rho\simeq\oplus_{i=1}^{n}\rho_i$$, and hence: $$\mathbb{C}\simeq\oplus_{i=1}^{n}W_i$$ Since $W_i$ are one dimensional $G$-invariant subspaces, it's easy to show that they're eigenspaces. Therefore $A$ is diagonalizable.
Is this correct or am I missing something? (Just making sure I understood the meaning of a decomposition of a representation).