# Sufficient Condition for Spectral radius to be greater than or equal to 1

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 ?

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

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
$$\max_{v \in \mathcal{P}} \underset{v_j \ne 0}{\min_{1\le j \le n}} \frac{(Tv)_j}{v_j} = \text{the spectral radius of}\; T.$$
• I think you mean $j$ instead of $i$ in your equation. Commented Nov 16, 2022 at 20:50