Statement of the Perron–Frobenius theorem Let $A=(a_{{ij}})$ be an $n\times n$ positive matrix: $ a_{{ij}}>0$ for $1\leq i,j\leq n$.
On Wikipedia, the statement of Perron–Frobenius theorem indicates, among others, the following claims:


*

*There exists an eigenvector $v = (v_1,\dots,v_n)$ of $A$ with eigenvalue $r$ such that all components of $v$ are positive: $A v = r v$, $v_i > 0$ for $1 \leq i \leq n$. (Respectively, there exists a positive left eigenvector $w : w^T A = r w^T$, $w_i > 0$.)

*$\lim _{{k\rightarrow \infty }}A^{k}/r^{k}=vw^{T}$, where the left and right eigenvectors for A are normalized so that $w^Tv = 1$. 


I'm trying to prove the second claim, but so far I have not gone well. Any suggestion, please?
Thanks in advance.
 A: Perron-Frobenius theorem asserts more than what you said in your first bullet point. In fact, when $A$ is primitive (this includes the special case where $A$ is positive), $r=\rho(A)$, the spectral radius of $A$, is a simple, positive and dominant eigenvalue of $A$, and $A$ has a positive left-eigenvector and a positive right-eigenvector for this eigenvalue. (It follows that even when $A$ has more than one positive eigenvalues, $\rho(A)$ is the only eigenvalue that has a positive left/right-eigenvector. So, in your first bullet point, we are not talking about just any positive eigenvalue of $A$, but specifically the spectral radius $\rho(A)$.)
This means $A/r$ has a Jordan decomposition $A/r=P(1\oplus J)P^{-1}$, where $J$ is a Jordan normal form whose eigenvalues all have magnitudes strictly smaller than $1$. Let the first column of $P$ be $v$ and the first row of $P^{-1}$ be $w$. Then $v$ and $w$ are respectively right- and left-eigenvectors of $A$ for the eigenvalue $1$ and $w^Tv=1$.
It is well known that we can always make the super-diagonal of any non-trivial Jordan block arbitrarily small via a similarity transform using a diagonal matrix. That is, for any $\epsilon>0$, we can always pick a diagonal matrix $D$ (e.g. $D=\operatorname{diag}(1,\epsilon,\epsilon^2,\ldots,\epsilon^{n-1})$) such that $1\oplus J=D(1\oplus B)D^{-1}$ for some bidiagonal matrix $B$ whose diagonal entries (i.e. eigenvalues) are exactly those of $J$ and whose super-diagonal entries all lie inside $[0,\epsilon]$. Therefore, when $\epsilon$ is sufficiently small, the induced $\infty$-norm of $B$ (i.e. $\|B\|_\infty=\max_i\sum_j|b_{ij}|$) is strictly smaller than $1$. So, we see that $\lim_{k\to\infty} B^k=0$ and in turn
$$
\lim_{k\to\infty} (A/r)^k=PD(1\oplus 0)D^{-1}P^{-1}=P(1\oplus 0)P^{-1}=vw^T.
$$
