I am trying to understand the following proof that every finite-state Markov chain has a stationary distribution. The proof is from [here](http://homepages.math.uic.edu/~furman/preprints/whatis-published.pdf).

Let $P$ be the $k \times k$ (stochastic) transition probability matrix for our Markov chain. Now, 

> 
> ... $1$ is an eigenvalue for $P$ and therefore also for $P^t$ .
>  Writing a $P^t$ invariant $v$ as $v = v^+ − v^−$ with $v^+ , v^− \in (
> > \mathbb{R}\_+ )^k$ , we obtain $P^t v^± = v^±$ because $P^t$ preserves
>  the positive cone; if $v^+\neq 0$ take $ν = ( \sum v^+\_i )^{-1} · v^+,$
>  otherwise normalize $v^−$.
> 
> 
> 

The main thing I don't understand is 

> 
> we obtain $P^t v^± = v^±$ because $P^t$ preserves
>  the positive cone
> 
> 
> 

Why is this true?

I also don't understand why $( \sum v^+\_i )^{-1} · v^+$ works if $v^+ \neq 0$.

Is there any easier way to show that every finite state Markov chain has a stationary probability distribution?