I need to probe the follow lemma
Lemma 2 if $\vec{v} = (v_i)^n_{i=1}$ is a eigenvector of the markov matrix $M$ asociate to the eigenvalue $\lambda$ with $|\lambda| < 1$, then $\sum^n_{i=1}v_i=0$.
Before this lemma I probe that
Lemma 1 If $M$ is a Markov matrix $n x n$ and $\vec{v} \in \mathbb{R}^n \setminus \{ \vec{0} \}$ , then $||M\vec{v}|| \leq ||\vec{v}||$.
How can I probe the first line (lema 2)?
Let $M\equiv(m_{ij})$ be a nonnegative matrix whose columns sum to one. Let $v$ be an arbitrary column vector. Then, denoting by $e$ the column vector of all ones, $$ e^{\intercal}Mv=\sum_{i}\sum_{j}m_{ij}v_{j}=\sum_{j}\sum_{i}m_{ij}v_{j}=\sum_{j}v_{j}=e^{\intercal}v. $$ By induction, $e^{\intercal}M^{k}v=e^{\intercal}v$ for any positive integer $k$. Next, let $w$ be a (right) eigenvector associated to an eigenvalue $\lambda<1$ of $M$. Then, $$ \lambda^{k}e^{\intercal}w=e^{\intercal}M^{k}w=e^{\intercal}w. $$ Taking limits with respect to $k$ on both sides, $$ 0=e^{\intercal}w=\sum_{j}w_{j} $$ as desired.