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It is known that for a regular Markov matrix $M,$ $M^{n}$ has the steady-state vector as all of its columns as $n \to \infty.$ I learned this in class, but what if there is more than one steady-state vector? What form does $M^{n}$ take? Do they switch alternatively one after another?

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The regularity ensures that the process is irreducible and aperiodic, and in that case there's a single stationary distribution. – joriki Nov 14 '11 at 10:42
up vote 1 down vote accepted

From the Perron-Frobenius Theorem we have that there is maximal simple eigenvalue $\lambda$ for $M$, meaning that the eigenspace associated to $M$ is one-dimensional. That is, up to a non-zero constant multiple we have exactly one eigenvector $\bf{v}$ associated to $\lambda$. The long-term behavior of $M^n \bf{u}$ for a non-zero vector $\bf{u}$ is independent of $\bf{u}$ and we have that $M^n \bf{v}$ approaches $\bf{v}$(in direction) for large $n$.

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