Let $\rho : G \rightarrow GL_n(\mathbb{C})$ be a representation show that $|\operatorname{tr} X| \leq \dim \rho$

Question

Let $G$ be a finite group.

Let $\rho : G \rightarrow GL_n(\mathbb{C})$ be a representation, pick $g \in G$ and write $X=\rho(g)$. Prove that all eigenvalues of $X$ are roots of unity, and deduce that $\det X$ is a root of unity as well, and that $|\operatorname{tr} X| \leq \dim \rho$.

I believe that "all eigenvalues are roots of unity" will follow from the finiteness of the group. I dont believe it works otherwise.

$\det X$ will be root of unity as as determinant is the product of its eigenvalues.

I am stuck on to do $|\operatorname{tr} X| \leq \dim \rho$, I think there was some relationship between determinant and trace http://en.wikipedia.org/wiki/Determinant#Relation_to_eigenvalues_and_trace, but I dont think the general case helps here.

Answer 1

the trace of $X$ is the sum of the eigenvalues. $|\operatorname{tr} X|=|\sum_{I=1}^n\lambda_i|$ therefore $|\operatorname{tr}X|\leq\sum_{I=1}^n|\lambda_i|$ and $|\lambda_i|=1$ thus the result expected