I would like to extend the question asked from the following past post, which proved that a finite irreducible Markov chain of period $d>1$ has exactly $d$ eigenvalues (counting multiplicity) which are roots of $z^d=1$.
However, is it possible to show that each of these roots appear exactly once as an eigenvalue? I don't think the proof offered on the previous question show this.
(In brief, the previous proof goes along the lines of: Let $P$ represent the Markov chain. $P^d$ is block diagonal with each of the $d$ block representing an irreducible aperiodic chain, so each block contributes exactly one count of eigenvalue 1.)
Edit: I'm actually mostly interested in showing that an irreducible Markov chain, even if $d>1$, has a unique stationary distribution (just the uniqueness part).