A property of irreducible and aperiodic Markov chains Let $P$ denote the $s\times s$ Markov transition matrix. We know that irreducibility and aperiodicity implies the following:

There exists an integer $N\geq 1$, such that $[P^n]_{ij}&gt;0$ for all $i,j$ and all $n\geq N$.

Is the following property true for irreducible and aperiodic chains?

Fix $i,j$. Suppose $[P^n]_{ij}&gt;0$ for some integer $n$, then $[P^{m}]_{ij}&gt;0$ for all $m\geq n$.
Nevermind, the statement is incorrect. I found a counterexample:

Suppose $P$ is $3\times 3$. Let $P_{11}=P_{13}=P_{22}=0$ and $P_{12}=1$. All other entries are positive. This chain is aperiodic and irreducible. You can check that $[P^2]_{12}=0$ even though $P_{12}&gt;0$.
If the chain is irreducible, and we have $[P^n]_{ij}&gt;0$ for all $i,j$ and all $n\geq N$, it means we communicate states $i$ and $j$ with some probability.

Your assumption is false, since the value $P_{ij}^n$ doesn't make sense (in the probabilistic mean.). for $n=1$, this represents the probability of going from state $i$ to state $j$, but to power it, means what?.

On the other hand, what you may want to do, it to factorize you matrix, using it's eigenvalues and eigenvectors, so you can operate matrix power much faster. We have $$P^m =U^{-1} \Sigma^m U,$$ where the matrix $U$ is composed by the eigenvectors of $P$ and $\Sigma$ is the diagonal matrix with the eigenvalues of $P$. And the nice thing about $\Sigma$ is that $\Sigma^m= \left(\sigma_{i,j}^m \;;\; (i,j) \in |s|^2\right),$ where $\sigma_{i,j} = 0$ if $i \neq j$.