## Perron–Frobenius theorem

Today I attended a lecture given by Vaughn Climenhaga. He presented a proof of the following version of Perron–Frobenius theorem:

Let $\Delta\subset \mathbb{R}^d$ be the set of probability vectors, $P$ be a stochastic matrix with positive entries. Then
–there is a positive probability $\pi\in \Delta$ fixed by $P$
–the eigenspace $E_1=\mathbb{R}\pi$
–the spectra $\Sigma(P)\subset B(0,r)\cup\{1\}$ for some $r<1$
–for all $v\in\Delta$, $P^n v\to \pi$ exponentially as $n\to \infty$.

Proof. (1) Let $v\in\Delta$. Then $\sum_i v_i=1$, and
$\sum_i (Pv)_i=\sum_i \sum_j p_{ij}v_j=\sum_j v_j=1$. So $Pv\in \Delta$. Moreover, $Pv$ is positive and $P(\Delta)\subset \text{Int}(\Delta)$. Therefore there exists some point $\pi\in\text{Int}(\Delta)$ fixed by $P$.

(2). Suppose on the contrary that there exists $v\notin \mathbb{R}\pi$ that is also fixed by $P$. Then $P$ fixes every vector in the plane $\Pi:=\mathbb{R}v\oplus\mathbb{R}\pi$, in particular the points on $\Pi\cap \partial \Delta$. This contradicts (1).

(3). We use the norm $|v|=\sum|v_i|$. Note that $|Pv|=\sum_i |(Pv)_i|\le \sum_{ij}p_{ij}|v_j|=|v|$. So $\Sigma(P)\subset D(0,1)$. It suffices to show $\Sigma(P)\backslash\{1\}\cap S^1=\emptyset$. If not, pick one ,say $\lambda$, and $n\ge 1$ such that $\text{Re}(\lambda^n)1$ for any $\epsilon>0$.

Consider the matrix $A=P^n-\epsilon I$, which is positive if $\epsilon$ is small enough. Then we have $|A|\le |P_n|$ and hence $\Sigma(A)\subset D(0,1)$. This contradicts the fact $\lambda^n-\epsilon$ is an eigenvalue of $A$.

(4). Let $W\subset \mathbb{R}^d$ be the subset of vectors with zero mean: $\sum v_i=0$, and consider the decomposition $\mathbb{R}^d=\mathbb{R}\pi\oplus W$. Note that $PW\subset W$ and hence $\Sigma(P|_{W})\subset D(0,r)$. For any $v\in\Delta$, we have $v=\pi+w$ for some $w\in W$. Then $|P^nv-\pi|=|P^n(v-\pi)|=|P^nw|\le Cr^n|w|$.

Only light calculations are used in his lecture. As pointed by Vaughn, this approach does not give precise information of the $r$.