$A$ is a Regular Transition Matrix $\Rightarrow$ $\lim\limits_{m \to \infty} A^m$ exists and rank 1

Question

A is a Regular Transition Matrix $\Rightarrow$ $\lim\limits_{m \to \infty} A^m$ exists and rank 1

At the above proposition, what does "regular" mean?

Answer 1

Def: $A$ is regular if some power of $A$ has all positive entries.

See also this question and here.

This comes up for regular Markov chains (i.e. Markov chains with a regular transition matrix).

Concerning the statement you reference: $$ A \;\,\text{regular} \;\;\implies\;\; \text{rank}\left(\lim_{m\rightarrow\infty} A^m\right)=1 $$ this is because any regular transition matrix $A$ satisfies $$ S:=\lim_{m\rightarrow\infty} A^m = \begin{bmatrix} s_1 & s_1 & \ldots & s_1 \\ s_2 & s_2 & \ldots & s_2 \\ \vdots & \vdots &\ddots & \vdots \\ s_n & s_n & \ldots & s_n \end{bmatrix} $$ where $s = (s_1, \ldots, s_n)$ is the steady state distribution of the Markov chain (with $||s||_1=1$).

Notice that the rank of $S$ is obviously 1, because there is only one linearly independent column. Intuitively, this means a regular Markov chain converges to a distribution $s$ after sufficient time passes, no matter what its initial distribution was and that this does not change as time passes further (i.e. $As=s$).