Does from $P^n$ irreducible for all $n\in\mathbb{N}$ follow that $P$ is aperiodic?

Question

let $P$ be a transition matrix of a Markov chain with state space E, that is finite.

Does from $P^n$ irreducible for all $n\in\mathbb{N}$ follow that $P$ is irreducible and aperiodic?

The first thing is clear. If $P^n$ is irreducible for all $n\in\mathbb{N}$, then especially for $n=1$, so $P$ is irreducible.

But the aperiodicity is not clear to me. Do not see that and have no idea how to show it.

Answer 1

A matrix $P$ is called primitive, if there is a natural number $m$, such that $P^m$ has only positive entries.

If a matrix is primitive, it is also irreducible and aperiodic.

So you only have to show, that a number $m$ with a positive matrix $P^m$ exists to complete the proof.

Answer 2

Please look my answer to this question. In a nutshell, the following theorem should help you.

Theorem. (Gantmacher F.R. The Theory of Matrices, 1960, Vol. 2, P. 81, Theorem 9.) If $A \geqslant 0$ is an irreducible matrix and some power $A^q$ of $A$ is reducible, then $A^q$ is completely reducible, i.e., $A^q$ can be represented by means of a permutation in the form $A^q = \operatorname{diag} \{ A_1, \dots, A_d \}$, where $A_1, \dots, A_d$ are irreducible matrices having one and the same maximal characteristic value. Here $d$ is the greatest common divisor of $q$ and $h$, where $h$ is the index of imprimitivity of $A$.