# asymptotic behaviour of coefficients in nonnegative matrix iteration

Let $A$ be a square matrix with nonnegative integer coefficients.

Is there a simple way to prove that there is a "period" $d$ such that for all $0\leq r<d$, the coefficient $a_{i,j,n}$ at position $(i,j)$ in the matrix $A^{nd+r}$ is asymptotically (with respect to $n$) equivalent to $x n^y z^n$, where $x,y,z$ are nonnegative constants depending on $i,j,d,r$.

The period is needed to avoid oscillating behaviours like $(-1)^n$, or cycles of bigger length.

I found this paper which should help, but I'm having trouble formalizing a proof: http://www.ams.org/journals/jams/2000-13-04/S0894-0347-00-00342-8/S0894-0347-00-00342-8.pdf).

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maybe you want the perron frobenius thm, en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem – mike Aug 7 '13 at 17:02
No I looked into it but it doesn't seem to solve the problem (unless I missed something). – Denis Aug 7 '13 at 17:03

For a primitive matrix $A$, $A = ||a_{i j}||$ (the one that has a single eigenvalue with maximum modulo) you can use a Perron's formula for elements of matrix $A^m$, $A^m = ||a_{ij}^{<m>}||$: \begin{equation*} a^{<m>}_{i j} = \sum_{h = 1}^\nu \frac{1}{(n_h - 1)!} \biggl[\frac{d^{n_h-1}}{d\lambda^{n_h-1}} \biggl(\frac{\lambda^m A_{i j}(\lambda)}{\Psi_h(\lambda)} \biggr) \biggr]_{\lambda = \lambda_h}, \end{equation*} where $A(\lambda) = |\lambda E - A|$; $\lambda_1, \dots, \lambda_\nu$ are roots of the matrix characteristic equation $A(\lambda) = 0$ of multiplicities $n_1, \dots, n_\nu$ respectively; $A_{i j}(\lambda)$ is an algebraic adjunct for the element $\lambda \delta_{i j} - a_{i j}$ in determinant $A(\lambda)$; $\Psi_h = A(\lambda) \cdot (\lambda - \lambda_h)^{-n_h}$.
If $A$ is an imprimitive matrix $A$ with index of imprimitivity $h$, then $A^h$ splits into $h$ primitive matrices with the same maximum eigenvalue: $A^h = \operatorname{diag} \{ A_1, \dots, A_h\}$. In this case you can use the same formula for matrices $A^{mh}$.