Let $\phi: G\to GL_n(\mathbb C)$ be an irreducible representation of a finite group $G$. Let $\chi: G\to \mathbb C$ be the character of $G$. Prove that:
$$\displaystyle \sum_{g\in G}{\chi(g)}=0 $$
I know that if $e$ is the identity then $\chi(e)=n$, if $g\in G$ then $\chi(g^{-1})=\overline{\chi(g)}$. Thus if $g=g^{-1}$ then $\chi(g)=\chi(g^{-1})=\overline{\chi(g)}$ then $\chi(g)\in \mathbb R$. If $g \ne g^{-1}$ then $\chi(g)+\chi(g^{-1})=2Re(\chi(g))$. Thus we can conclude at least that the sum: $$\displaystyle \sum_{g\in G}{\chi(g)}=n+\sum_{g\ne g^{-1}}{\chi(g)}+\sum_{g=g^{-1}}{\chi(g)} \in \mathbb R $$
But I don't know how to prove that it's in fact $0$.
Thanks!
Here is an insight "from above." The trace of a projection operator is equal to the dimension of its image. (Can you prove this? Hint: form a basis for its image and kernel, then compute the matrix of the operator wrt this basis.) The operator defined by $\frac{1}{|G|}\sum_{g\in G}g$ is a projection $V\to V^G$. And finally the trace of a sum of linear maps is the sum of their traces.