A non negative matrix $A\in\mathbb{R}^n$ is said fully indecomposable if it doesn't exist permutation matrices $P,Q$ such that $$PAQ=\begin{bmatrix}X &0\\Y & Z\end{bmatrix}$$ where $X,Z$ are square matrices.
A non negative matrix is said primitive if there exists an integer $m>0$ such that $A^m$ all entries strictly positive.
I would like to prove that if $A$ is fully indecomposable then is primitive. I know that
- if a matrix is irreducible and has a positive element on the main diagonal then it is primitive;
- a matrix $A$ is fully indecomposable if and only if it exist a permutation matrix $P$ such that $PA$ is irreducible and its main diagonal is strictly positive;
From the first point I deduce that if $P$ is a permutation matrix and $A$ is irreducible, then A is primitive if and only if $PAP^T$ is primitive since simultaneous permutations of row and columns don't change the main diagonal.
Anyway I am not able to conclude, maybe I miss something. Could you help me?