Eigenvalues of stochastic matrix proof Let $M$ be a (square) right-stochastic matrix. I am familiar with the following facts:


*

*Letting $e$ denote the vector of all ones, $Me=e$ and hence $1$ is an eigenvalue of $M$.

*Since the product of two (compatible) right-stochastic matrices is itself a right-stochastic matrix, if $(\lambda,v)$ is an eigenvalue-eigenvector pair for $M$ with $|\lambda|>1$, we have that
$$\Vert M^nv\Vert_1=|\lambda|^n\Vert v\Vert_1 \rightarrow \infty \text{ as } n \rightarrow \infty,$$
a contradiction. Therefore, $\rho(M)\leq 1$.



I've heard from various sources that $1$ is a simple eigenvalue of
  $M$. Certainly this is true when $M$ is irreducible (apply
  Perron-Frobenius), but I am not sure what the corresponding argument
  is otherwise. Can someone point me in the right direction?

 A: If you have a Markov chain on a finite state space whose transition graph is not weakly connected, then this Markov chain is literally the "disjoint union" of two completely parallel Markov chains. Consequently it will have multiple independent invariant distributions, which may be taken to be concentrated on each weakly connected component. 
In linear algebraic language, you can see this by permuting variables (i.e. both rows and columns simultaneously) in such a way that the transition matrix is block diagonal. Then when a diagonal block has an eigenvector $v$ of eigenvalue $\lambda$, the overall matrix does too: it is just $v$ extended with zeros above and below.
Note that this is not quite the converse of the result that you stated, because a Markov chain is irreducible if and only if its transition graph is strongly connected. (I find this terminology confusing, because the chain is not really "reducible", in the sense of being able to be factored into completely parallel chains, when the graph is weakly connected but not strongly connected.)
