I'm solving the exercises of chapter 14 in the book Representations and Characters of groups. (Gordon James, Martin Liebeck) Always working with $\mathbb{R}$ or $\mathbb{C}$.
One of them says:
Suppose that $\chi$ is a non-zero, non trivial character of $G$, and that $\chi(g)$ is a non-negative real number for all $g$ in $G$. Prove that $\chi$ is reducible
At this stage of the book, I think that the natural procedure is to calculate $<\chi,\chi>$ and see that it is $\neq 1$
I know that non-trivial means $\exists g \in G$ such that $\chi(g)\neq 1$, but since $\chi(1_G)\in \mathbb{N}$, I don't know what they want to say with the non-zero condition.
Anyway, I've doing the next:
By definition, $<\chi,\chi>=\displaystyle\dfrac{1}{|G|}\sum_{g\in G}\chi(g)\chi(g^{-1})$
I know that there exist a base $\mathcal{B}$ where $[g]_{\mathcal{B}}$ is diagonal, then $\chi(g)$ is sum of $m$th roots of the unity (considering the $order(g)=m$) so $\chi(g^{-1})=\overline{\chi(g)}$
But $\chi(g)$ is a non-negative real number for all $g$ in $G$.
So we obtain $<\chi,\chi>=\displaystyle\dfrac{1}{|G|}\sum_{g\in G}\chi(g)^2$
Since $\chi$ is non-trivial, $\exists g \in G$ such that $\chi(g)\neq 1$, but I'm stuck here. Maybe I'm missing something related to the non-zero condition.
Any advice would be welcome.