Why is P irreducible and aperiodic?

Question

Let $P=(p_{ij})_{i,j\in E}$ denote the transition matrix of a Markov chain with finite state space $E$. Why does the following implication hold:

$$ \exists n\in\mathbb{N}: p_{ij}^{(n)}>0~\forall i,j\in E~~\implies P^n\text{ is irreducible for all }n\in\mathbb{N} $$

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As far as I see $P^n$ irreducible for a $n\in\mathbb{N}$ means that for each pair $(i,j)$ there exists a $m\in\mathbb{N}$ such that $$ p_{ij}^{(mn)}>0? $$

If yes, then my idea would be that because of the assumption it is $p_{ij}^{(n)}>0$ and $p_{jj}^{(n)}>0$ so that $$ p_{ij}^{(mn)}\geq \underbrace{p_{ij}^{(n)}\cdot p_{jj}^{(n)}\cdot\ldots\cdot p_{jj}^{(n)}}_{\text{m-times}}>0, $$ so that for each pair $(i,j)$ there is a $m\in\mathbb{N}$ such that $p_{ij}^{(mn)}>0$.

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Does this make sense or is it nonsense?

Answer 1

Consider the following transition matrix: $$P=\left(\begin{array}{cc}0 & 1 \\ 1 & 0 \end{array} \right) $$

Here I would say $P$ is irreducible but $P^2$ is not. More generally, $P^k$ is irreducible if $k$ is odd but not if $k$ is even. Do you agree, or am I missing something?

All of this is to say the RHS of your statement is false here but the LHS is true ($n=1$ works), so I would seem to have contradicted the statement. What am I missing?

I'm not sure what you're up to with the $m$'s. Could you add a little more text to motivate your operations?

Answer 2

I believe the answer is given by the theorem.

Theorem. (Gantmacher F.R. The Theory of Matrices, 1960, Vol. 2, P. 80, Theorem 8.) A matrix $A \geq 0$ is primitive if and only if some power of $A$ is positive: $A^p > 0$ ($p \geq 1$).

Proof of the theorem can be found in the book.

Actually the answer depends if the statement should be understood like this $$ (\forall i,j\in E)(\exists n\in\mathbb{N})[p_{ij}^{(n)}>0]~~\implies (P^n\text{ is irreducible for all }n\in\mathbb{N}), $$ or like this $$ (\exists n\in\mathbb{N})(\forall i,j\in E)[p_{ij}^{(n)}>0]~~\implies (P^n\text{ is irreducible for all }n\in\mathbb{N}). $$ The first interpretation is false and is hold for any irreducible matrix, while the second is true due to the theorem above.