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Let $A$ be a matrix of order $ n \times n$ and let $ \lim_{ k \rightarrow \infty} A^k = L > 0.$ I want to show that there exists some power $m$ such that $A^m>0$ and $ A^{m+i} > 0$ for any $ i \geq m.$ By a matrix $L>0$ and $A>0$ I mean all entries of the matrices are positive. Any hint or a proof please?

Thank you in advance.

2 Answers 2

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Since $\lim\limits_{k\to\infty} A^k=L$, then this holds for all entries of matrix $A$. In other words $ \lim\limits_{k\to\infty} A_{ij}^k=L_{ij}>0\tag{1} $ for all $i$ and $j$. Now take arbitrary $i,j$, then from $(1)$ it follows that there exist $m_{ij}\in\mathbb{N}$ such that $k\geq m_{ij}$ holds $A_{ij}^k>0$. Now consider $m=\max_{i,j} m_{ij}$, then for all $k>m$ and all $i,j$ we have $A_{ij}^k>0$. Which means that there exist $m\in\mathbb{N}$ such that for all $k\geq m$ holds $A^k>0$

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    You don't need to multiply. I've proved that A^k>0 for all k>m. Substitute k=m+i>m, and get the result.2012-11-13
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For $1 \leq i,j \leq n$, consider $\pi_{ij}$, the map which project a matrix on its $(i,j)$ component. The set of positive matrices in the sense you define is in fact $\bigcap_{1\leq i,j\leq n}\pi_{ij}^{-1}(\mathbb R_0^+).$ Since $\pi_{ij}$ is continuous, this is open (finite intersection of open subsets), hence the result.

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    I edited my answer to show that the set is indeed open.Equivalence of norms is of no particular importance here, but it is a good thing to know when speaking of topology in finite dimensional vector spaces.2012-11-15