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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.

Greg
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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 $\langle A w, w\rangle \ge \langle A v, v\rangle$.

orangeskid
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  • If I understood, $w$ is in the unit sphere and so, $\langle Aw, w \rangle$ must be the maximum of $\langle Av,v \rangle$ in the unit sphere, that is, $w$ must be an eigenvector associate to $\lambda_1$. – Greg Oct 05 '19 at 04:29
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    This ideia seems similar to Rayleigh quotient for self-adjoint maps. – Greg Oct 05 '19 at 04:31
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    @Greg: Yes, it is enough to get the largest value on the sphere, attained for a positive entry vector, and check that it is an eigenvector, like in the theory – orangeskid Oct 05 '19 at 05:23
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    Shouldn't it be $|\alpha_i|^2$ since $\alpha_i\in\mathbb C$? – Matija Aug 28 '23 at 17:44
  • @Matija: Yes, indeed! Will correct it right away, thank you! – orangeskid Aug 29 '23 at 03:05