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]


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. 

1 Answer 1


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.

  • $\begingroup$ Thanks for your simple and nice answer. $\endgroup$
    – Amin235
    Apr 11, 2020 at 12:21

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