Given that $\sum_{j=1}^n v_j = 0$ and $\bf{Pv} = \bf{v}$, where $\bf{P}$ is an irreducible stochastic matrix, show that $\bf{v} = \bf{0}$.
Here's what I have:
Let $\displaystyle v_i = \max_{1\leq j \leq n} v_j$. Then $v_i \geq 0$. From hypothesis, $$ \begin{align} \displaystyle v_i &=- \sum_{j=1, j\neq i}^n v_j \quad \text{ and}\\ \displaystyle v_i &= \sum_{j=1}^n p_{ij} v_j \end{align} $$ Thus $$ \begin{align} \displaystyle -\sum_{j=1, j\neq i}^n v_j &= \sum_{j=1}^n p_{ij} v_j\\ \sum_{j=1, j\neq i}^n v_j (p_{ij} + 1) + p_{ii} v_i &= 0\\ \end{align} $$
I'm not sure what to do next to show that $\bf{v} = \bf{0}$.
Let us assume $P$ is aperiodic for otherwise we can replace $P$ by $1/2(P+I)$. Also assume $p_{ij} >0$ for all $i,j$ for otherwise, since $P$ is irreducible, we can choose $n$ large enough such that $P^n$ has this property.
Let $|v_i| = max_j |v_j|$. If $v_i = 0$, we are done. Without loss of generality, we can assume $v_i > 0$ since otherwise we can work with $- \boldsymbol{v}$. Now $$ v_i = |\sum_j v_j p_{ij}| \le \sum_{j} |v_j| p_{ij} \le \sum_j |v_i|p_{ij}=|v_i| $$ Since $p_{ij} >0$ for all $i,j$, this means $|v_i| = |v_j| =x$ for all $j$. This implies we can break up the state space into two parts, $A=\{i: v_i = x\}$ and $B =\{i: v_i = -x\}$. Now if $i \in A$, $$ x = \sum_{j \in A}xp_{ij} - \sum_{j \in B}xp_{ij} \quad ; \quad x= x \sum_{j}p_{ij} $$ which implies $\sum_{j \in A}p_{ij} = 1$ which implies $B = \varnothing$ since $p_{ij} >0$ for all $i,j$ by assumption. Hence $\sum_iv_i = 0$ implies $x =0$.