Let $A$ be a non-negative (entrywise) matrix such that $A(1,1)>0$. Set $u=(1,0,0,...,0)^T$. Is it always true that there exists a non-negative eigenvector $v$ of $A$ such that $\lim\_{n\rightarrow\infty}\frac{A^nu}{||A^nu||\_1}=\frac{v}{||v||\_1}$?



Matrix iteration for non-negative matrices. Does it converge to some eigenvector?
The statement is not true. Let $a>1$ and define 


\begin{equation}
A:=\begin{bmatrix}1&0&0\\1&0&a\\0&a&0 \end{bmatrix}.
\end{equation}


Suppose there is $v\in\mathbb{R}^n\backslash\{0\}$ such that $\lim\_{n\rightarrow\infty}\frac{A^nu}{||A^nu||\_1}=\frac{v}{||v||\_1}$. By a trivial induction argument we can prove that 


\begin{equation}
A^nu=\left(1,\frac{a^{n+1}-1}{a^2-1},\frac{a^{n}-a}{a^2-1}\right)
\end{equation}


when $n$ is odd bigger than $2$ and


\begin{equation}
A^nu=\left(1,\frac{a^{n}-1}{a^2-1},a\frac{a^{n}-1}{a^2-1}\right)
\end{equation}


when is $n$ even bigger than 3. Observe that $\frac{v(1)}{||v||\_1}=\lim\_{n\rightarrow\infty}\frac{A^nu(1)}{||A^nu||\_1}=0$ so $v(1)$=0. That means that $u(2)\neq 0 $ or $u(3)\neq 0$. 


From that we get 


\begin{equation}
 \lim\_{n\rightarrow\infty}\frac{A^nu(3)}{A^nu(2)}=\begin{cases}\frac{v(3)}{v(2)}\quad v(2)\neq 0\\\infty \quad v(2)=0\end{cases}.
\end{equation}


This gives a contradiction since 


\begin{equation}
 \lim\_{n\rightarrow\infty}\frac{A^{2n+1}u(3)}{A^{2n+1}u(2)}=a^{-1}
\end{equation}


and 


\begin{equation}
 \lim\_{n\rightarrow\infty}\frac{A^{2n}u(3)}{A^{2n}u(2)}=a.
\end{equation}


