Column stochastic matrix/Positive row/Convergence/Fact/Proof

Proof

Let $w\in {ℝ}^n$ be the stationary distribution, which is uniquely determined because of fact  (3). We set

$U=\{\left(\begin{array}{c}{u}_1\\ ⋮\\ u_n\end{array}\right)\mid \sum _{i=1}^n{u}_i=0\}\subseteq {ℝ}^n.$

This is a linear subspace of ${ℝ}^n$ of dimension $n-1$. Due to fact  (2), $w$ has only non-negative entries; therefore, it does not belong to $U$. Because of

$\begin{array}{cc}\sum _{i=1}^n(Mu{)}_i& =\sum _{i=1}^n\left(\sum _{j=1}^n{a}_{ij}{u}_j\right)\\ & =\sum _{j=1}^n{u}_j\left(\sum _{i=1}^n{a}_{ij}\right)\\ & =\sum _{j=1}^n{u}_j,\end{array}$

$U$ is invariant under the matrix $M$. Hence,

$V=ℝw\oplus U$

is a direct sum decomposition into invariant linear subspaces. For every $u\in U$ with $\mid \mid u\mid {\mid }_{\mathrm{sum}}=1$, we have

$\mid \mid Mu\mid {\mid }_{\mathrm{sum}}<1$

due to fact  (2). The sphere of radius $1$ is compact with respect to every norm; therefore, the induced maximum norm of $M{|}_U$ is smaller than $1$. Because of fact and fact, the sequence $M^{n}u$ converges for every $u\in U$ to the zero vector.

Let now $v\in V$ be a distribution vector; because of

$\sum _{i=1}^n{v}_i=1=\sum _{i=1}^n{w}_i,$

we can write

$v=w+u$

with $u\in U$. Because of

$M^{m}v=M^m(w+u)=M^{m}w+M^{m}u=w+M^{m}u,$

and the reasoning before, this sequence converges to $w$.

To fact