0

Suppose $\{X_n\}$ is an irreducible Markov Chain on finite state space $S$. Then, the number of eigenvalues of the transition matrix with unit modulus is precisely equal to the period of the chain.

I don't really know where to start. I know that if $P$ is the transition matrix, then $P^d$ is block diagonal with diagonal blocks aperiodic and irreducible. Now, I was thinking of finding eigenvalues of unit modulus for $P^d$, which would in turn be eigenvalues with unit modulus of the block diagonal matrices.

But I don't think this yields anything.

Please give me a hint only, and NOT a proof.

Landon Carter
  • 13,462
  • 4
  • 37
  • 90

1 Answers1

1

You’re on the right track, I think. First, back up a step and label the states so that $P$ takes the form of a block cyclic permutation matrix $$P=\begin{bmatrix} 0 & P_1 & \ddots & \ddots & 0 \\ 0 & 0 & P_2 & \ddots & \ddots \\ \ddots & \ddots & \ddots & \ddots & \ddots \\ 0 & 0 & \ddots & \ddots & P_{d-1} \\ P_d & 0 & \ddots & \ddots & 0 \end{bmatrix}$$ and let $P_m'$ be the matrix consisting of block $P_m$ and zeros elsewhere (so that $P=\sum_{m=1}^d P_m'$). Letting $\hat{\mathbf\pi}^{(m)}$ be the (left) eigenvector of $1$ for the $m$th diagonal block of $P^d$, you can show that $\hat{\mathbf\pi}^{(m)} P_m'=\hat{\mathbf\pi}^{(m+1)}$ and then use these eigenvectors to construct eigenvectors $\mathbf\pi^{(k)}$ of $P$ that correspond to the $d$th roots of unity.

amd
  • 55,082