I posted the following question on MSE, feeling that it perhaps isn't *research level* mathematics, but didn't get any bites. So, I am crossposting here.

The following *ergodic theorem* is well known.

Ergodic Theorem.Let $X$ be an ergodic (ie, irreducible and aperiodic) Markov chain on a countable state space $I$. Suppose that $X$ has an invariant distribution—$\pi$, say. Let $\mu$ be any distribution on $I$. Then,$$ \mathbb P_i(X_t = j) \to \pi_j \quad\text{as}\quad t \to \infty \quad\text{for all}\quad i,j \in I. $$

In particular, since the limit exists, the invariant distribution is unique.

This is the way that I have seen uniqueness of the invariant distribution proved, eg via a coupling argument. It seems to me, though, that a direct, more algebraic, proof should exist.

Prove uniqueness of the invariant distirbution by direct, algebraic methods, not appealing to the probabilistic interpretation.

After all, it's just a system of linear equations! I haven't been able to find one, but below are some of my thoughts on the matter. They're not super insightful, though...

Irreducibility is necessary, but aperiodicity isn't

- $\pi P = P \iff \pi (I + P)/2 = \pi$, so can just make the chain lazy
- this removes aperiodicity issues, and implies that the real part of any eigenvalue is non-negative

$\pi P = P \iff \pi(I - P) = 0 \iff (I - P^T) \pi^T = 0$

- so, we want to show that $I - P^T$ has a one-dimensional kernel
- this is equivalent to showing that the multiplicity of eigenvalue $1$ of $P^T$ is $1$

An invariant distribution can be constructed via the expected return times

- this is just a
*sufficient*condition for $\pi P = \pi$ to hold - I'm after a
*necessary*condition

- this is just a
Naturally, I've also searched a lot online, including SE, but have not been successful

If anyone can point me to a good reference online, or give a proof—or even an outline—that would be appreciated!

"since irreducibility implies that $−1$": No, irreducibility is independent of the question whether $-1$ is an eigenvalue. (ii)"this lazification implies that we may assume that all eigenvalues of $P$ are non-negative": That's not quite correct, since there can be complex eigenvalues. $\endgroup$aperiodicitywith $-1$ eigenvalue more (eg, non-lazy random walk on a cycle). I almost always work with reversible chains in my research, so I often forget about complex eigenvalues! Still, at least all have non-negative real part... $\endgroup$