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 $\lim_{n\rightarrow\infty}\frac{A^nu}{||A^nu||_1}=\frac{v}{||v||_1}$?
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2$\begingroup$ $A = \left(\begin{smallmatrix}0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$, $u = (1,0)$ $\endgroup$– Nate EldredgeCommented Jan 28, 2020 at 17:22
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$\begingroup$ @NateEldredge thank you for your comment I added some more details. $\endgroup$– A. BatsisCommented Jan 28, 2020 at 18:13
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$\begingroup$ Search Perron's theorem on Google. $\endgroup$– Alexandre EremenkoCommented Jan 29, 2020 at 0:58
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$\begingroup$ @AlexandreEremenko (and at those who voted to close): I think the problem is more subtle than just a standard applications of Perron-Frobenius theory. If $A$ were irreducible, it would indeed be quite simple, but as $A$ is not assumed to be irreducible, various problems occur: for instance, $A$ can have more than one eigenvalue on the spectral circle, and these eigenvalues might not be semi-simple. Still, the assumption that $A(1,1) > 0$ and that $u(1) = 1$ gives some additional structure to the problem - so the answer might still be "yes", but I don't think it's in the standard references. $\endgroup$– Jochen GlueckCommented Jan 29, 2020 at 8:47
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1 Answer
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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}