How prove this matrix limit is $\lim_{m\to\infty}A^mx=\left(\dfrac{e}{n}\right)$ Question:

let $A$ is Doubly stochastic matrix,and the eigenvalue such
  $$\lambda_{1}=1,|\lambda_{j}|<1,(j=2,3,\cdots,n)$$
  and the $$e=(1,1,1,\cdots,1)^T$$
  show that : for any 
  vector $$x=(a_{1},a_{2},\cdots,a_{n})^T,a_{1}+a_{2}+\cdots +a_{n}=1,a_{i}\ge 0$$, have
  $$\lim_{m\to\infty}A^mx=\left(\dfrac{e}{n}\right)$$

where Doubly stochastic matrix some  properties and some result 
 can see this
My idea: since
$$|A|=\lambda_{2}\lambda_{3}\cdots\lambda_{n}<1$$
then I can't find this limit.
and can't solve this problem,
 A: As hinted in the comments, this is false.
Set $A:=\begin{bmatrix} \frac 1 2 & \frac 1 2\\ \frac 1 2 & \frac 1 2\end{bmatrix}$. Clearly $A^m=A$, for all $m\in \mathbb N$. Therefore with $x=\begin{bmatrix} 1\\ -1\end{bmatrix}$ the equality doesn't hold.
You need to require $x\ge 0$.

Layout of answer to the question post edit: The proof goes something like this. 


*

*Firstly prove that the sequence $\left(A^m\right)_{n\in \mathbb N}$ converges. To do this you need to use the fact $1$ is the only eigenvalue of absolute value $1$, that $1$ is the largest eigenvalue and that its algebraic multiplicity is $1$.

*The above implies that $\lim \limits_{m\to +\infty}\left(A^my\right)$ exists for all $y\in \mathbb R^{n\times 1}$. Next, letting $B$ be $\lim \limits_{m\to +\infty}\left(A^m\right)$, prove that $AB=B$. This shows that $B$'s columns are all eigenvectors of $A$ associated to the eigenvalue $1$.

*Prove that $B$ is stochastic by columns. To do this first prove that for all $m\in \mathbb N$, $A^m$ is stochastic by columns.

*From the fact that the columns of $B$ are eigenvectors associated to the eigenvalue $1$ and from the fact that $B$ is column-stochastic, using the hypothesis that $x$'s entries when summed equal $1$, conclude.



You could have asked me for more details instead of wasting reputation points in an unawarded bounty.

Answer: There exists $n\in \mathbb N$ such that $A$ is $n\times n$.


*

*The sequence $\left(A^m\right)_{n\in \mathbb N}$ converges if, and only if, $\left(J^m\right)_{n\in \mathbb N}$, where $J$ is a canonical jordan normal form of $A$. Since $1$ is the only eigenvalue of absolute value $1$, since $1$ is the largest eigenvalue and its algebraic multiplicity is $1$, it's clear from looking at the powers of $J$ that $\left(J^m\right)_{n\in \mathbb N}$ converges and consequently so does $\left(A^m\right)_{n\in \mathbb N}$.

*Therefore $\lim \limits_{m\to +\infty}\left(A^my\right)$ exists for all $y\in \mathbb R^{n\times 1}$. Define $B:=\lim \limits_{m\to +\infty}\left(A^m\right)$. Multiplying this definition by $A$ on the left yields $AB=A\lim \limits_{m\to +\infty}\left(A^m\right)=\lim \limits_{m\to +\infty}\left(A^{m+1}\right)=B$. Since $AB=A\left[\text{Col}_1(B)\mid \ldots \mid \text{Col}_n(B) \right]=\left[A\text{Col}_1(B)\mid \ldots \mid A\text{Col}_n(B) \right]$, it follows that the columns of $B$ are eigenvectors of $A$ associated to $1$.

*Note that any $n\times n$ matrix $M$ is column-stochastic if, and only if, $(1,e)$ is an eigenpair of $M^T$. It's easy to prove by induction that $\forall m\in \mathbb N\left((A^m)^Te=e\right)$. This fact justifies the third equality below: $$B^Te=\lim \limits_{m\to +\infty}\left((A^m)^T\right)e=\lim \limits_{m\to +\infty}\left((A^m)^Te\right)=\lim \limits_{m\to +\infty}\left(e\right)=e.$$ Therefore $B$ is column-stochastic.

*From 2. and 3. follows that $B=\left[\frac en\mid \ldots \mid \frac en\right]$. Given $x$ as in the question, it holds that $$\lim \limits_{m\to +\infty}\left(A^mx\right)=Bx=\begin{bmatrix}\frac 1 n(a_1+\ldots +a_n)\\ \vdots \\ \frac 1 n(a_1+\ldots +a_n)\end{bmatrix}=\dfrac e n.$$

