Question on stochastic matrix with stirctly positive entries

Question

Let $M$ be a stochastic matrix and assume all its entries are striclty positive. Does this imply that the limit as $n \rightarrow \infty$ of $M^n \vec{x}$ is a vector with all entries equal?
I know that if we assume the entries are just positive (not strictly positive), this is certainly not true by considering for instance the matrix $$ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} , $$ for which the limit does not exist.

Answer 1

Yes and this is a consequence of the basic limit theorem of MC theory. This MC has a unique stationary distribution. $M^{n}$ converges to a matrix in which each row is the stationary distribution so $\lim M^{n} x$ has all its components equal (to $\sum \pi_i x_i$ where $(\pi_i)$ is the stationary distribution).

Answer 2

If you know a bit about Markov chains, your question is basically answered by Theorem 15.3 in https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch15.pdf. It asserts that something stronger is true, which I will restate in simpler terms here:

Theorem: Suppose $P$ is a $n\times n$ stochastic matrix with strictly positive entries $(P_{ij})_{1\leq i,j\leq n}$. Then, $$\lim_{n\to\infty} P^n=A,$$ where $A$ is a $n\times n$ matrix such that in each column of $A$, all entries in that column are equal.

In particular, for any $x\in\mathbb{R}^n$, the vector $Ax$ will have all of its entries equal to each other.