I'd like to know if there's a nice way to obtain this result from the Banach Fixed Point Theorem:
Let $A$ be a (stochastic*) matrix and $v$ a vector such that
$$\lim_{n\to\infty}A^nv = w.$$
Then $w$ is an eigenvector of $A$ with eigenvalue 1.
I've produced a proof of this another way but I think the Banach theorem could make it cleaner. How does the following sketch look?
Let $|\cdot|$ be a norm on the vector space which induces a matrix norm on $A$. Let $d$ be the standard metric w.r.t. this norm; then since A is stochastic
$$d(Av,Aw) = |Av - Aw| \leq |A||v-w| \leq |v-w| = d(v,w).$$
Unfortunately since $|A|$ = 1, it's not quite a contraction mapping. Is there still somewhere we can go with this line of thinking?
*Feel free to weaken this condition if possible.
**If someone has a better title go ahead and change it.
I think any application of the Banach theorem here is overkill. Here's a nice proof of the desired result:
We are given the existence of the limit $w = \lim_{n \to \infty}A^n v$. The mapping $v \mapsto Av$ is continuous (since $A$ is a matrix, i.e. a linear map between finite dimensional normed vector spaces). It follows that $$ Aw = A \left(\lim_{n \to \infty}A^n v \right) = \lim_{n \to \infty}A(A^n v) = \lim_{n \to \infty}A^{n+1} v = \lim_{n \to \infty}A^{n} v = w $$