Suppose that $A$ is a symmetric 3 × 3 matrix with positive entries, such that the sum of every row is 1. Let $v=(1,\frac{1}{2},\frac{1}{3})$. What is $$ \lim\limits_{n\to\infty}A^nv $$
My solution is: $\sqrt{max\lambda}=||A||_2\leq||A||_{\infty}=1 \Rightarrow $ all $\lambda\leq1$, but I cannot prove that all $\lambda<1$ which can lead to $\lim\limits_{n\to\infty}A^nv=0$.
So I tried the second way and found that A could be regarded as a Markov transition matrix, but how to relate this to my problem?
Note that $A$ is an irreducible stochastic matrix with positive entries. There exists a unique stationary row vector $\pi$ such that $$ \pi A = \pi \tag{$\clubsuit$} $$ Additionally, $$ \lim_{n \rightarrow \infty} A^n = \mathbf1\pi $$ where $\mathbf 1$ is a column vector with all ones. See here.
Because $A$ is symmetric, the matrix$\ \mathbf 1 \pi\ $ is also symmetric. In other words, all entries in $\pi$ are equal, thus $$ \pi = [\frac{1}{3}, \frac{1}{3}, \frac{1}{3}] $$ We conclude $$ \lim_{n \rightarrow \infty}A^n v = \mathbf 1\pi v = \begin{bmatrix}11/18 \\ 11/18 \\ 11/18\end{bmatrix} $$