Perron-Frobenius Theorem. A particular case?

Question

Let $\{a_{i,j}\} =A \in \mathbb{R}^{N \times N}$ be a non-negative matrix, such that:

1. $a_{i,i} = 0 ~~ \forall i \in \{1, \ldots, N\}$
2. $a_{i,j} \geq 0 ~~ \forall i \neq j$

Given the previous hypotheses and using the Perron-Frobenius theorem (or also other results), can I claim one of the following statements?

1. There exists an eigenvalue $\lambda$ of $A$ such that $\lambda \in \mathbb{R}$ and $\lambda > 0$
2. There exists an eigenvalue $\lambda$ of $A$ such that $\lambda \in \mathbb{R}$ and $\lambda \geq 0$

Answer 1

1. It can not be claimed, as the zero matrix satisfies both 1. and 2.

2. This can be claimed using Perron-Frobenius (rather using the idea of its proof), since, it $0$ is not an eigenvalue of $A$, then $Ax\ne 0$, for every $x\in \mathbb R^n\{0\}$ with non-negative elements. Take $$ f(x)=\frac{Ax}{\ell(Ax)}, $$ where $$\ell(x_1,\ldots,x)=x_1+\cdots+x_n, $$ Then $f : \Delta^{n-1}\to\Delta^{n-1}$, is a continuous map, where $\Delta^{n-1}$ is the simplex $$\Delta^{n-1}=\{x=(x_1,\ldots,x_n): \ell x=1, x_i\ge 0\},$$ and thus it has a fixed point (according to Brouwer's Fixed Point Theorem, as the closed simplex is homeomorphic to the closed unit ball in $\mathbb R^{n-1}$): $$ f(x)=x, $$ or $$ Ax=\ell(Ax)x. $$ This means that $x$ is an eigenvector and $\ell(Ax)>0$, the corresponding eigenvalue.