Perron-Frobenius theorem In the proof of the Perron-Frobenius theorem why can we take a strictly positive eigenvector corresponding to the eigenvalue $1$? Before that, why can we even take a non-negative eigenvector?
Books simply take such a vector, no explanation whatsoever. Pisses me off.
Wikipedia only proves it assuming the matrix is irreducible.
Any help? Thanks.
EDIT: Nevermind, got it. Delete this question, please.

It really depends on how your book does it.
The easiest way I know of, is to consider the simplex $S = \{ (x_i) | x_i \geq 0, \sum x_i = 1\} $, and consider the action of the matrix $M$ on this simplex given by $M(s) = Ms / ||Ms||$ (i.e. normalize it such that it lies back in the simplex).
This gives us a map of the simplex to the simplex, hence has a fixed point by Brouwer's fixed point theorem. It remains to show that the boundary of this simplex is not a fixed point, because the entries of the matrix are all positive. Hence, the fixed point lies in the interior, and corresponds to the eigenvector that you're looking for. 
Note that this doesn't prove the uniqueness, nor that the corresponding eigenvalue has strictly largest norm.
