Let $M \in \mathcal{M}_n(\mathbb{C})$ symetric with $[M]_{ij} \geq 0$. Prove that $M$ has an eigenvector $(v_1,...,v_n)$ with $v_i \geq 0$.
This question is from my exercise list about Spectral Theory of Self-Adjoint and Normal Maps. My problem is: I really don't know how to approach that question using this theory. So, I don't want a solution, I think some hints are enough.
HINT:
$M$ has $n$ orthogonal unit eigenvectors $e_i$ for eigenvalues $\lambda_1 \ge \lambda_2\ge \cdots \ge\lambda_n$. For a unit vector $v=\sum \alpha_i e_i$ we have $$\langle A v, v\rangle = \sum |\alpha_i|^2 \lambda_i$$ Therefore, the maximum value of $\langle A v, v\rangle$ on the unit sphere is attained only for an eigenvector for value $\lambda_1$
Now note that for $v$ unit vector, the vector $w$ with components the absolute value of the components of $v$ is also unit and $\langle A w, w\rangle \ge \langle A v, v\rangle$.