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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).

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    $\begingroup$ If $\pi P=\pi$ and $P$ is irreducible, then $\frac 1 n\sum_{i=1}^n P^i\to (\pi^t|\dots|\pi^t)^t$ and since the limit on the left is unique, $\pi$ is unique. $\endgroup$ – A.S. Nov 27 '15 at 18:38
  • $\begingroup$ The linked solution assumes what you are trying to prove, so you can't rely on it in your proof. $\endgroup$ – user940 Nov 27 '15 at 19:52
  • $\begingroup$ @ByronSchmuland: How so? I can prove that if Q represents an irreducible aperiodic chain, then $Q^m$ is positive for some $m$. This allows me, with Perron-Frobenius, to prove everything I need about irreducible aperiodic chains without knowing anything about irreducible (possibly periodic) chains. $\endgroup$ – suncup224 Nov 28 '15 at 10:50
  • $\begingroup$ @suncup224 Oh, ignore my comment. I didn't realize that you were assuming Perron-Frobenius. $\endgroup$ – user940 Nov 28 '15 at 16:08
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The transition matrix has the "block cycle structure" $$P=\pmatrix{0&A_0&0&0&\cdots&0\cr 0&0&A_1&0&\cdots&0\cr 0&0&0&A_2&\cdots&0\cr \vdots&&&\ddots&&\vdots\cr A_{d-1}&0&0&0&\cdots&0}$$ For any $d$th root of unity $\omega$, define the block vector $x=(\omega^0|\omega^1|\omega^2|\cdots|\omega^{d-1})^T$ using powers of $\omega$. Then we have $Px=(\omega^1|\omega^2|\omega^3|\cdots|\omega^{0})^T=\omega x$ so $\omega$ is a eigenvalue of $P$.

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