 Solution

Write $Q=1r^T-RI$, with $r=R\pi$. Then ${\pi }^{T}1=1$ andThus $\pi$ is a stationary probability vector, a left eigenvector for eigenvalue zero. If any probability vector $\nu$ is stationary, the same calculation gives ${\nu }^{T}Q=r^T-R{\nu }^T=0$, forcing $\nu =\pi$. Hence 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.