= Solution
Write $Q=\mathbf1r^T-RI$, with $r=R\pi$. Then $\pi^T\mathbf1=1$ and
$$
\pi^TQ=r^T-R\pi^T=0.
$$
Thus $\pi$ is a stationary probability <vector>, a left <eigenvector> for <eigenvalue> zero. If any probability <vector> $\nu$ is stationary, the same calculation gives $\nu^TQ=r^T-R\nu^T=0$, forcing $\nu=\pi$. Hence \b[the <equilibrium> is unique], even if some $r_j$ vanish and the chain is not irreducible; those zero-rate states are transient or inaccessible rather than extra stationary classes.
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