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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.

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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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  • $\begingroup$ If we consider a matrix $ A = \begin{bmatrix} 0&0.1&0.5\\0.2&0&0\\0&0.3&0 \end{bmatrix}$ we see that $A^5>0$ but some $A_{ij}^5 =0$ for some $i,j.$ In fact $A_{ij}^k = 0$ for any power $k$ for some entries $A_{ij}$ of $A.$ So how did you get Equation(1)? $\endgroup$
    – Zizo
    Nov 13, 2012 at 10:58
  • $\begingroup$ $$A=\left( \begin{array}{ccc} 0.0012 & 0.00454 & 0.0002 \\ 0.00008 & 0.0012 & 0.003 \\ 0.0018 & 0.00012 & 0.0006 \end{array} \right)$$ so all the entries are positive. So something wrong with your computations $\endgroup$
    – Norbert
    Nov 13, 2012 at 11:02
  • $\begingroup$ Thanks! I now understood what you mean, there was miss reading because I read it $[A_{ij}]^k$ not $[A^k]_{ij}.$ I just cannot see when the entries $A_{ij}^m>0$, then $A^{k}>0$ for any $ k > m.$ $\endgroup$
    – Zizo
    Nov 13, 2012 at 11:24
  • $\begingroup$ Okay okay! I think I can see it! $\endgroup$
    – Zizo
    Nov 13, 2012 at 11:27
  • $\begingroup$ How $[A^k]_{ij}>0$ for any $k \geq m$ lead to $A^k>0$ while $A^k$ might does not have the entries $[A^k]_{ij}.$ I am just thinking that $A^k = A^{k-m}A^m$ so $A^{k-m}$ might have entries which are zero, in other words is it always true when you multiply a matrix $B \geq 0$ by $C>0$ then $BC>0.$ Is not always true I think. $\endgroup$
    – Zizo
    Nov 13, 2012 at 11:35
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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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  • $\begingroup$ Can you elaborate on this? Can you show that such matrices are an open set. I know the equivalence of norms in matrices space. How would this help me? $\endgroup$
    – Zizo
    Nov 13, 2012 at 11:45
  • $\begingroup$ 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. $\endgroup$
    – YoungFrog
    Nov 15, 2012 at 13:10

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