Unique right eigenvector for row stochastic matrix

Question

Suppose $P$ is a row-stochastic matrix. Then $1$ is a right-eigenvalue of $P$, and any constant vector $v$ will satisfy $Pv=v$. When is it the case that only constant vectors solve $Pv=v$? In particular, I am wondering if this is the case if if $P$ is irreducible and aperiodic.

Answer 1

The condition that $P$ is irreducible is enough to guarantee that property.

We need to show that $Pv=v$ implies $v_1=\ldots=v_n$. Let $v_p$ be the smallest component of $v$, so that $v_i \ge v_p$ for all $i$. Let $U \subseteq \{1,\ldots,n\}$ be the set of those indices for which $v_i=v_p$. Note that $U\neq \emptyset$ since $p \in U$. If $m \in U$, then $$ (P v)_m = v_m = v_p. $$ Now, because $P$ is row-stochastic, the left hand side of the above equation is a convex combination of the components of $v$, while the right hand side is a minimum-value component of $v$. The only way that a convex combination of elements can yield a value equal to the minimum element is when all elements with a positive coefficient (here, all $v_k$ such that $P_{km}>0$) are equal to the minimum. Thus, $m \in U$ implies $k \in U$ whenever $P_{km}>0$. Since $P$ is irreducible, this implies $U=\{1,\ldots,n\}$, hence $v_1=\ldots=v_n$.