Something about irreducible matrix and spectral radius

Question

I have studied matrix analysis by myself recently. Now I have a question on my book.

Here are some definitions: $A\in M_n$ is reducible if $n=1$ and $A=0$, or $n\geq2$ and there is a $n*n$ permutation matrix $P$ such that $P^tAP=\left[\begin{array}{car} A_1 & A_2 \\ 0 & A_3 \end{array}\right]$ where $A_1,A_2$ are square matrix Otherwise, $A$ is irreducible.

$A\succeq0$ means that $A$ has all entries $a_{ij}\geq0$

Suppose that $A\succeq0$, $\rho(A)>0$, $x$ is a nonnegative vector with $x\neq0$, and $Ax = \rho(A)x$. Show that if $x$ is not positive, then $A$ is reducible. If $x$ is positive, must $A$ be irreducible?

I think it might have something to do with Perron-Frobenius theorem, but I cannot figure out how to continue. Please teach me how to solve this

Answer 1

1. If $A$ is $\geq 0$ and irreducible, then any eigenvector associated to the eigenvalue $\rho(A)$ has no zero entries. Here $x$ has a zero entry and consequently, $A$ is reducible.

2. The answer is no. Indeed, consider the matrix $A=I_2$. It is reducible and $x=[1,1]^T>0$ is an eigenvector associated to $\rho(A)=1$.