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Let $T$ be a square matrix with non-negative entries and $v$ be a non-negative vector such that $Tv \geq v$ (basically the inequality holds for each entry i.e., $(Tv)_j \geq v_j)$. Can we conclude that the spectral radius of $T$ is greater than or equal to 1 ?

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  • $\begingroup$ Are you saying the inequality holds for any nonnegative vector $v$? $\endgroup$
    – angryavian
    Commented Mar 18, 2020 at 17:32
  • $\begingroup$ It holds for some vector $v,$ not necessarily for all vectors. $\endgroup$
    – egt123
    Commented Mar 18, 2020 at 17:37
  • $\begingroup$ A dumb counterexample is $T$ being the zero matrix and $v = 0$. I feel there are also some other counterexamples where $v \ne 0$. $\endgroup$
    – angryavian
    Commented Mar 18, 2020 at 17:41
  • $\begingroup$ $v \neq 0.$ Is there a simple counterexample for non-zero $v$ ? $\endgroup$
    – egt123
    Commented Mar 18, 2020 at 17:44

1 Answer 1

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Your statement is true, it comes with the Perron-Frobenius Theorem and the Collatz–Wielandt Formula. The fact can be found here in Theorem III.2, part (iii), stated as follows:

Let $T \in \mathbb{R}^{n \times n}$ be a square matrix with non-negative entries and let $\mathcal{P} \subset \mathbb{R}^n$ be a set of all non-negative vectors (excluding the zero vector). Then

$$ \max_{v \in \mathcal{P}} \underset{v_j \ne 0}{\min_{1\le j \le n}} \frac{(Tv)_j}{v_j} = \text{the spectral radius of}\; T. $$

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  • $\begingroup$ I think you mean $j$ instead of $i$ in your equation. $\endgroup$ Commented Nov 16, 2022 at 20:50
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    $\begingroup$ @MarineGalantin Fixed, thank you. $\endgroup$
    – Zeekless
    Commented Nov 28, 2022 at 13:29

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