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Let $A$ be a non-negative primitive matrix. Then $$\lim_{n\to\infty}\left[\frac{A}{\rho(A)}\right]^n=xy^T,$$ where $x, y$ are the Perron roots of $A$ and $A^T$ respectively, they satisfy $x^Ty=1$.

Here are the notations.

  1. $A$ is non-negative if any entry of $A$ is greater than or equal to $0$.

  2. $A$ is primitive if $A$ is non-negative irreducible, and the number of eigenvalues of $A$ with modulus equal to $\rho(A)$ (the spectral radius of $A$) is $1$.

  3. $A$ is irreducible if $A$ is not reducible; and $A$ is reducible if there exists a permutation matrix $P$ such that $$P^T AP=\begin{pmatrix} B&0\\ C&D\end{pmatrix},$$ or equivalently, there exists a permutation $\sigma$ of $\{1,2,\cdots,n\}$ and a $1\leq k\leq n-1$ such that the sub-matrix of $A$ in rows $\sigma(1),\cdots,\sigma(k)$ and columns $\sigma(k+1),\cdots,\sigma(n)$ being $0$.

  4. The Perron root $x$ of $A$ is an eigenvector $x$ corresponding to the eigenvalue $\rho(A)$, the entries of $x$ are positive.

  5. $A^T$ is the transpose of $A$.

It is easy to show that the limit exists. In fact, we could just use Jordan carnonical form to find there exists a invertible matrix $T$ such that $$T^{-1}AT=\begin{pmatrix} \rho(A)&0\\ 0&*\end{pmatrix},$$ and thus $$ \lim_{n\to\infty}T^{-1}\left[\frac{A}{\rho(A)}\right]^nT =\begin{pmatrix} 1&0\\ 0&0\end{pmatrix}.$$ However, I could not prove that the limit if $xy^T$.

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Let $T=\begin{pmatrix}a&l\\c&U\end{pmatrix},T^{-1}=\begin{pmatrix}a'&l'\\c'&U'\end{pmatrix},B=\lim (\dfrac{A}{\rho(A)})^n$. Clearly $a'a+l'c=1$ and $x=[a,c^T]^T$ is an eigenvector of $A$ associated to $\rho(A)$. Moreover $B=[a,c^T]^T[a',l']$. Since $T^TA^T{T^{-1}}^T=\begin{pmatrix}\rho(A)&0\\0&V\end{pmatrix}$, with $\rho(V)<1$, $y=[a',l']^T$ is an eigenvector of $A^T$ associated to $\rho(A^T)=\rho(A)$. As required, $B=xy^T$ with $x^Ty=y^Tx=1$.

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