Let $A$ a $n\times n$ quasi-positive matrix, i.e. $A$ has all non-negative entries $\ge 0$ and exists $n > 0$ such that $A^n$ is positive. We further assume that it has a $0$ eigenvalue, whose eigenvector $w$ is positive. We observe that $A\operatorname{diag}(w)$ is a matrix whose rows are of the form $(a_{j1}w_1, a_{j2}w_2, \dots, a_{jn}w_{n})$, because $w$ is positive and $A$ is non-negative, we have that each entry is non-negative. Moreover, $$\|e_j^t A\operatorname{diag}(w)\|_1 = 0$$ for any $j=1,.., n$, and $e_j$ be an element of the canonical orthonormal basis (because by definition of $0$ eigenvector $\sum_{i=1}^n a_{ji}w_{i} = 0$ for any choice of $j$). Then, because all elements in $A \text{diag}(w)$ are positive anyway, $$0 = \sum_{j=1}^n b_{j} \|e_j^t A\operatorname{diag}(w)\|_1 = \left\|\sum_{j=1}^n b_{j} e_j^t A\operatorname{diag}(w)\right\|_1$$ i.e. for any vector $v \in \mathbb{R}^n$ whose coordinates are non-negative, we have $v^t A \operatorname{diag}(w) = 0$. Finally, this proves that the rank is $0$, concluding that such an $A$ cannot exist.
Is this reasoning correct?
Edit: I initially removed the condition on quasi-positive matrix, because I didn't think was relevant, but comments indeed gave me a counter-example for $A$ non-negative with a positive eigenvalue i.e. $A = 0$ and any vector thanks to Anne Bauval. I also edited the equation on the norms requiring the $b_j$ to be positive.
Since $A\geqslant 0$, we have that only the spectral radius of $A$ can correspond to positive eigenvectors. Furthermore, since $A^n>0$ for some $n$, the spectral radius of $A$ is positive.