The product of non-primitive matrices with zero positions in common

Def.[1]:

A non-negative $$n \times n$$ matrix $$A$$ is called a non-primitive if there is no an integer $$k$$ such that all entries of $$A^k$$ are positive.[1]

Def.[2]:

Let $${\bf A}=(a_{i,j})$$ and $${\bf B}=(b_{i,j})$$ be two $$n \times n$$ non-negative matrix over $$\mathbb{R}$$. Let the positions of zero entries of $${\bf A}$$ and $${\bf B}$$ be the same ( if $$a_{i,j}=0$$ then $$b_{i,j}=0$$ and vice versa ). Then, we say all zero positions of $${\bf A}$$ and $${\bf B}$$ are in common.

Let $${\bf A}_1,{\bf A}_2, \cdots, {\bf A}_k$$ be $$n \times n$$ non-primitive matrices such that all zero positions of $${\bf A}_i$$'s are in common. Let $${\bf B}=\prod_{i=1}^k{\bf A}_i.$$

My question:

How to show that the matrix B is a non-primitive matrix.


If $$C$$ and $$D$$ have positive entries in the same places, then $$CD$$ and $$C^2$$ will have positive entries in the same places. More generally, if $$A_1,A_2,\dots,A_k$$ have positive entries in the same places, and $$B=A_1A_2\cdots A_k$$, then $$B$$ has positive entries in the same places as $$A_1^k$$. If $$A_1$$ is primitive, then so is $$A_1^k$$, so $$B$$ is primitive. And if $$A_1$$ is not primitive, then neither is $$A_1^k$$, so $$B$$ is not primitive.