Proof that there exists a non-negative eigenvector corresponding to eigenvalue 1 of stochastic matrix

Question

Let $P \in [0,1]^{n \times n}$ be a [irreducible or reducible] stochastic matrix where its rows sum to 1 i.e. $$ \forall i \in \{ 1 , \dots n \} \quad \sum_{j=1}^{n} P_{ij} = 1 $$ It is easy to show that $P$ always has an eigenvalue of 1 by using the above property.

However, I wanted to show that there exists a corresponding left eigenvector [to eigenvalue 1] with real entries, each of which has the same sign [zero entries would be allowed] i.e. non-negative.

Anybody have an idea on how to prove this? Any references would also be greatly appreciated.

Answer 1

Stochastic preserve the set of all stochastic vectors (vectors with non-negative entries that sum to unity), and this action is continuous. The set of stochastic vectors is closed and bounded, therefore compact. So we can use the Brouwer fixed point theorem from topology implies that there exists a fixed point. This fixed point has the properties you ask for.

Answer 2

Please look here: http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem. You should only to verify that your matrix P satisfies the conditions of this theorem but it shouldn't be so difficult.