If $A \in \mathcal M_n(\mathbb C)$ is of finite order then $\vert \operatorname{tr}(A) \vert \le n$ I've been given this problem;
let $A$ be an $n \times n$ matrix over $\mathbb{C}$ of finite order. Show that $|\text{tr}(A)|\leq  n$, and that $|\text{tr}(A)| = n$ if an only if $A$ is a scalar matrix.
I don't even know where to begin, partly because I can't see how it would be true. Any help/tips would be greatly appreciated 
 A: As $A$ is a finite order matrix, it exists $m \in \mathbb N$ such that $A^m-1=0$. Hence, the minimal polynomial $\mu_A$ of $A$ divides $P(x) = x^m-1$. As $P$ has simple roots, this is also the case of $\mu_A$ and $A$ is diagonalizable. The eigenvalues of $A$ which are roots of $P$, have a modulus equal to $1$.
$\operatorname{tr}(A)$ is the sum of the diagonal values of a diagonal matrix similar to $A$. All those diagonal values have modulus equal to $1$, as they are roots of $\mu_A$. Therefore $\vert \operatorname{tr}(A) \vert \le n$.
Moreover, for $z_1, \dots, z_n \in \mathbb C$, you have $\vert z_1 + \dots +  z_n \vert = \vert z_1 \vert + \dots +\vert z_n \vert$ only if all the $z_i$ have the same argument.
Having $\vert \operatorname{tr}(A) \vert = n = \vert \sum_i \lambda_i \vert = \sum_i \vert \lambda_i \vert$, where $\lambda_i$ are the diagonal values of $A$ in a diagonal basis implies with above statement that all the $\lambda_i$ have the same modulus equal to $1$ and the same argument. Therefore all the $\lambda_i$ are equal to the same value and $A$ is diagonal.
