Prove: every strongly connected graph has a single stationary distribution

Question

Let $G= \langle V,E ~\rangle$ be a finite graph. We define $A_g$ to be the matrix with $[A_g]_{i,j}=1$ if there is an edge $e=(i,j)$ and $0$ otherwise.

Let $P$ be the transition matrix for a random walk on the graph $G$. So $P_{i,j}=P(x_t=j|x_{t-1}=i)$

$P$ is a normalize version of $A_g$ such that every row sums to $1$.

We define $\pi$ to be a stationary vector if it holds that $\pi_i \geq 0$ and $\lVert \pi \rVert_1=1$ and satisfy $\pi^T=\pi^TP$

For ergodic graphs, I know there is a single stationary vector $\pi$ and for every vector $x$ we have $\lim_{t\to\infty}x^TP^t=\pi^T$

What I want to prove, is that for not ergodic graphs, but still strongly connected (for every vertex $i$ and vertex $j$ there is a way to get from $i$ to $j$, or in other word there is some $k$ that $[P^k]_{i,j} > 0$), there is only a single stationary vector $\pi$.

I know I won't be promised that for every $x$ we get $\lim_{t\to\infty}x^TP^t=\pi^T$, but still there is only 1 $\pi$

I can find a stationary vector with $\pi_i=\frac{deg(i)}{2\cdot |E|}$. One can multiply it by $P$ and see it works. But I am struggling to show only this $\pi$ can work with strongly connected graph.

Thanks

Answer 1

This is false for an infinity graph.

Take $G=(\mathbb{Z},E)$ with $(i,j) \in E $ if $|i-j|=1$

Then by an argument of symmetry, the probability measure should be the same on each integer but then can not sum to $1$.

Answer 2

For a finite graph there is only one stationnary vector $\pi$

You are looking for eigenvector for the matrix $P^T $ of eigenvalue $1$ (that is solving $P^T \pi= \pi$. But $P^T$ is a non negative matrix and primitive so the Perron-Frobeinus theorem state that there is only one eigenvector for the spectral radius eigenvalue (here $1$). https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem#Multiplicity_one for more details.