Let $P$ be a stochastic matrix and $E$ the $n\times n$ identity matrix. Assume that $P^q$ = $E$ for some integer $q \geq 2$, but $P \neq E$. Find a steady-state vector for $P$. Give an example for such a $P$.
We know that $P$ is diagonizable since $P^q = E$. That means $P$ can be written as $AD^nA^{-1} = E$ which I believe implies $AD^n = A$. But this I believe implies $A$ has only the eigenvector 1 (implies identity), or only the eigenvector -1 which means no steady state vector.
The steady state vector is unique when $(*)$ the eigenvalue $1$ of $P$ has multiplicity $1$. Here $1$ may have a multiplicity $>1$ (cf. Ian 's comment in the case when there are several cycles).
Assume that $(*)$ is true; then the steady state $u$ is an eigenvector of $P^T$. Here $u$ is never (except if $P=I$) a distribution limit, that is, it does not satisfy: for every distribution vector $v$, $u=\lim_{\;k\rightarrow\infty} vP^k$ (cf. amd 's example).