Eigenvalues of the transition matrix for a periodic, irreducible markov chain

Question

For an irreducible periodic Markov Chain with period $d >1$, the transition probability matrix will have $d$ eigenvalues with absolute value 1. Why ?

Answer 1

Theorem 1.8.4 here shows that the set of states in a chain that has period $d$ can be partitioned into $d$ parts, $C_0,$ $C_1,\ \ldots,$ $C_{d-1},$ such that states in $C_0$ only make transitions to states in $C_1,$ states in $C_1$ only make transitions to states in $C_2,\ \ldots,$ states in $C_{d-1}$ only make transitions to states in $C_0.$

This implies that, if the states are ordered so that states of $C_0$ are listed first, then states of $C_1,$ and so on, then $P^d$ will be block-diagonal with $d$ blocks, and each block will be stochastic and irreducible. Therefore $P^d$ will have $d$ eigenvalues $1$, and the remaining eigenvalues with modulus less than one. From this it follows that $P$ has exactly $d$ eigenvalues of magnitude $1.$