Why is the only eigenvector of the transition matrix for an irreducible Markov chain with eigenvalue $= 1$ the eigenvector with all ones?
2026-04-24 12:58:46.1777035526
Eigenvector of transition matrix for Markov chain
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Otherwise the transition matrix would have two linearly independent eigenvectors on the right for the eigenvalue $1$, hence it would also have two linearly independent eigenvectors $\pi$ and $\varrho$ on the left for the eigenvalue $1$, the eigenvector $\pi$, say, being a stationary distribution. Irreducibility would then ensure that every component of $\pi$ is positive hence some linear combination of $\pi$ and $\varrho$ would also be a stationary distribution, in contradiction with the uniqueness of the stationary distribution.