*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]](https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem)


*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:**



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> How to show that the matrix B is a non-primitive matrix. 
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The product of non-primitive matrices with zero positions in common
To decide whether a matrix with nonnegative entries is primitive or not, all you have to know is where the positive entries are – their exact values are irrelevant. 


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. 


