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How does one show that any finite-state time homogenous Markov Chain has at least one stationary distribution in the sense of $\pi = \pi Q$ where $Q$ is the transition matrix and $\pi$ is the stationary distribution? My instinct is it involved eigenvalues and the Perron Frobenius theorem but I'm having trouble completing the argument.

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There are a number of ways to prove this. In a recent article "What is a stationary measure?", Alex Furman outlines a straightforward proof using only linear algebra. He notes that the existence of an invariant probability measure also follows from the Brouwer fixed point theorem.

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    $\begingroup$ The article cited does not seem to provide an elementary proof as there is no(?) justification for the statement: if $P^Tv = v$ then $P^Tv^+ = v^+$. $\endgroup$ – Ben Dec 7 '18 at 9:12
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Isn't it simple? Suppose you take the matrix to a power k where k-> infinity. Then all eigenvalues approach 0 while eigenvalue =1 is stationary. Note that the image is always the span of colloms so for eigenvalue 1 the picture must be one dimensional.

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Any stochastic matrix $P$ has at least one stationary distribution $\pi$ in the sense that $\pi P=\pi$. This can be shown by the following statement from Wikipedia.

Brouwer Fixed Point Theorem (applied to the compact convex set of all probability distributions of the finite state space $\Omega$) implies that there is some left eigenvector which is also a stationary probability vector.

Of course, we could also prove this by Perron Frobenius theorem.

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