# Monotonicity of the number of positive entries in nonnegative matrix powers

## Abstract

Let A be a nonnegative matrix of order n and $$f(A)$$ denote the number of positive entries in A. We prove that if $$f(A)\leq3$$ or $$f(A)\geq n^{2}-2n+2$$, then the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is monotonic for positive integers k.

## 1 Introduction

A matrix is nonnegative (positive) if all of its entries are nonnegative (positive) real numbers. Nonnegative matrices have many attractive properties and are important in a variety of applications [1, 2]. For two nonnegative matrices A and B of the same size, the notation $$A\geq B$$ or $$B\leq A$$ means that $$A-B$$ is nonnegative.

A sign pattern is a matrix whose entries are from the set $$\{ +, -, 0\}$$. In a talk at the 12th ILAS conference (Regina, Canada, June 26â€“29, 2005), Professor Xingzhi Zhan posed the following problem.

### Problem

([4], p. 233)

Characterize those sign patterns of square nonnegative matrices A such that the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is nondecreasing.

A nonnegative square matrix A is said to be primitive if there exists a positive integer k such that $$A^{k}$$ is positive. If we denote by $$f(A)$$ the number of positive entries in A, it seems that the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is increasing for any primitive matrix A. However, Å idÃ¡k [3] observed that there is a primitive matrix A of order 9 satisfying $$f(A)=18>f(A^{2})=16$$. This is the motivation for us to investigate the nonnegative matrix A such that $$\{f(A^{k})\}_{k=1}^{\infty}$$ is monotonic. It is reasonable to expect that the sequence will be monotonic when $$f(A)$$ is too small or too large.

Since the value of each positive entry in A does not affect $$f(A^{k})$$ for all positive integers k, it suffices to consider the 0â€“1 matrix, i.e., the matrix whose entries are either 0 or 1. Denote by $$E_{ij}$$ the matrix with its entry in the ith row and jth column being 1 and with all other entries being 0. For simplicity we use 0 to denote the zero matrix whose size will be clear from the context.

## 2 Main results

Let A be a nonnegative square matrix. We will use the fact that if $$A^{2}\geq A\ (A^{2}\leq A)$$, then $$A^{k+1}\geq A^{k}\ (A^{k+1}\leq A^{k})$$ for all positive integers k and thus $$\{f(A^{k})\}_{k=1}^{\infty}$$ is increasing (decreasing).

### Theorem 1

Let A be a 0â€“1 matrix of order n. If $$f(A)\leq2$$, then the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is decreasing.

### Proof

The case $$f(A)=0$$ is trivial.

If $$f(A)=1$$, then $$A=E_{ij}$$, $$1\leq i,j\leq n$$. Thus, for $$k=2,3,\ldots$$â€‰,

$$A^{k}=E_{ij}^{k}= \textstyle\begin{cases} E_{ii},& i=j;\\ 0,& i\neq j, \end{cases}$$

which implies that $$\{f(A^{k})\}_{k=1}^{\infty}$$ is decreasing. Next suppose $$f(A)=2$$.

Since $$\{f(A^{k})\}_{k=1}^{\infty}$$ is invariant under permutation similarity or transpose of A, it suffices to consider the following cases.

(1) $$A=E_{11}+E_{22}$$. Then $$A^{2}=A$$.

(2) $$A=E_{11}+E_{12}$$. Then $$A^{2}=A$$.

(3) $$A=E_{11}+E_{23}$$. Then $$A^{2}=E_{11}\leq A$$.

(4) $$A=E_{12}+E_{13}$$. Then $$A^{2}=0$$.

(5) $$A=E_{12}+E_{21}$$. Then $$A^{k}=E_{11}+E_{22}$$ for all even k, $$A^{k}=A$$ for all odd k.

(6) $$A=E_{12}+E_{23}$$. Then $$A^{2}=E_{13}$$, $$A^{3}=0$$.

(7) $$A=E_{12}+E_{34}$$. Then $$A^{2}=0$$.

It can be seen that in each case $$\{f(A^{k})\}_{k=1}^{\infty}$$ is decreasing. This completes the proof.â€ƒâ–¡

### Theorem 2

Let A be a 0â€“1 matrix of order n. If $$f(A)=3$$, then the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is monotonic.

### Proof

Under permutation similarity and transpose, it suffices to consider the following cases.

(1) $$A=E_{11}+E_{22}+E_{33}$$. Then $$A^{2}=A$$.

(2) $$A=E_{11}+E_{22}+E_{12}$$. Then $$A^{2}=A+E_{12}\geq A$$.

(3) $$A=E_{11}+E_{22}+E_{13}$$. Then $$A^{2}=A$$.

(4) $$A=E_{11}+E_{22}+E_{34}$$. Then $$A^{2}=E_{11}+E_{22}\leq A$$.

(5) $$A=E_{11}+E_{12}+E_{13}$$. Then $$A^{2}=A$$.

(6) $$A=E_{11}+E_{12}+E_{21}$$. Then $$A^{2}=A+E_{11}+E_{22}\geq A$$.

(7) $$A=E_{11}+E_{12}+E_{31}$$. Then $$A^{2}=A+E_{32}\geq A$$.

(8) $$A=E_{11}+E_{12}+E_{23}$$. Then $$A^{k}=E_{11}+E_{12}+E_{13}$$ for all $$k\geq2$$.

(9) $$A=E_{11}+E_{12}+E_{32}$$. Then $$A^{2}=E_{11}+E_{12}\leq A$$.

(10) $$A=E_{11}+E_{12}+E_{34}$$. Then $$A^{2}=E_{11}+E_{12}\leq A$$.

(11) $$A=E_{11}+E_{23}+E_{24}$$. Then $$A^{2}=E_{11}\leq A$$.

(12) $$A=E_{11}+E_{23}+E_{32}$$. Then $$A^{k}=E_{11}+E_{22}+E_{33}$$ for all even k, $$A^{k}=A$$ for all odd k.

(13) $$A=E_{11}+E_{23}+E_{34}$$. Then $$A^{2}=E_{11}+E_{24}$$, $$A^{k}=E_{11}$$ for all $$k\geq3$$.

(14) $$A=E_{11}+E_{23}+E_{45}$$. Then $$A^{2}=E_{11}\leq A$$.

(15) $$A=E_{12}+E_{13}+E_{14}$$. Then $$A^{2}=0$$.

(16) $$A=E_{12}+E_{13}+E_{21}$$. Then $$A^{k}=E_{11}+E_{22}+E_{23}$$ for all even k, $$A^{k}=A$$ for all odd k.

(17) $$A=E_{12}+E_{13}+E_{41}$$. Then $$A^{2}=E_{42}+E_{43}$$, $$A^{3}=0$$.

(18) $$A=E_{12}+E_{13}+E_{23}$$. Then $$A^{2}=E_{13}\leq A$$.

(19) $$A=E_{12}+E_{13}+E_{24}$$. Then $$A^{2}=E_{14}$$, $$A^{3}=0$$.

(20) $$A=E_{12}+E_{13}+E_{42}$$. Then $$A^{2}=0$$.

(21) $$A=E_{12}+E_{13}+E_{45}$$. Then $$A^{2}=0$$.

(22) $$A=E_{12}+E_{21}+E_{34}$$. Then $$A^{k}=E_{11}+E_{22}$$ for all even k, $$A^{k}=E_{12}+E_{21}$$ for all odd $$k\geq3$$.

(23) $$A=E_{12}+E_{23}+E_{31}$$. Then

$$A^{k}= \textstyle\begin{cases} E_{11}+E_{22}+E_{33},& k\equiv0\ (\mathrm{mod} 3);\\ A, & k\equiv1\ (\mathrm{mod} 3);\\ E_{13}+E_{21}+E_{32},& k\equiv2\ (\mathrm{mod} 3). \end{cases}$$

(24) $$A=E_{12}+E_{23}+E_{34}$$. Then $$A^{2}=E_{13}+E_{24}$$, $$A^{3}=E_{14}$$, $$A^{4}=0$$.

(25) $$A=E_{12}+E_{23}+E_{45}$$. Then $$A^{2}=E_{13}$$, $$A^{3}=0$$.

(26) $$A=E_{12}+E_{34}+E_{56}$$. Then $$A^{2}=0$$.

Since in each case $$\{f(A^{k})\}_{k=1}^{\infty}$$ is either increasing or decreasing, this completes the proof.â€ƒâ–¡

### Corollary 3

Let A be a 0â€“1 matrix of order 2. Then the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is monotonic.

### Remark

When A is of order $$n\geq3$$ with $$f(A)=4$$, the following example shows that $$\{f(A^{k})\}_{k=1}^{\infty}$$ may not be monotonic. Consider

$$A=E_{12}+E_{13}+E_{21}+E_{31}.$$

Direct computation shows that

$$A^{2}=2E_{11}+E_{22}+E_{23}+E_{32}+E_{33},\qquad A^{3}=2A.$$

Thus $$f(A)=4< f(A^{2})=5>f(A^{3})=4$$.

On the one hand, Theorems 1 and 2 show that $$\{f(A^{k})\}_{k=1}^{\infty}$$ is monotonic when $$f(A)\leq3$$. On the other hand, $$\{f(A^{k})\}_{k=1}^{\infty}$$ is expected to be also monotonic when $$f(A)$$ is large enough. Next we discuss the number of positive entries that A has to guarantee the sequence increasing.

The permanent of a matrix $$A=(a_{ij})_{n\times n}$$ is defined as

$$\operatorname {per}A=\sum_{\sigma\in S_{n}}\prod_{i=1}^{n}a_{i,\sigma(i)},$$

where $$S_{n}$$ is the set of permutations of the integers $$1,2,\ldots,n$$. First we have the following important fact.

### Lemma 4

Let A be a 0â€“1 matrix of order n. If $$\operatorname {per}A>0$$, then the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is increasing.

### Proof

Since A is a 0â€“1 matrix with $$\operatorname {per}A>0$$, there exists a permutation matrix P such that $$A\geq P$$. Now let $$A=P+B$$, where B is also a 0â€“1 matrix. Then $$A^{k+1}=A\cdot A^{k}=(P+B)A^{k}=P\cdot A^{k}+B\cdot A^{k}\geq P\cdot A^{k}$$ for all positive integers k. Thus $$f(A^{k+1})\geq f(P\cdot A^{k})=f(A^{k})$$, which implies that $$\{ f(A^{k})\}_{k=1}^{\infty}$$ is increasing.â€ƒâ–¡

### Theorem 5

Let A be a 0â€“1 matrix of order n. If $$f(A)\geq n^{2}-2n+2$$, then the sequence $$\{f(A^{k})\}_{k=1}^{\infty}$$ is increasing.

### Proof

First if $$\operatorname {per}A>0$$, by Lemma 4, $$\{f(A^{k})\} _{k=1}^{\infty}$$ is increasing.

Next suppose $$\operatorname {per}A=0$$. Then by the Frobeniusâ€“KÃ¶nig theorem [4, p. 46], A has an $$r\times s$$ zero submatrix with $$r+s=n+1$$. Since $$f(A)\geq n^{2}-2n+2$$, A has at most $$2n-2$$ zero entries. Thus $$rs\leq2n-2$$. It can be seen that r and s must be one of the following solutions.

(1) $$r=1$$, $$s=n$$;

(2) $$r=n$$, $$s=1$$;

(3) $$r=2$$, $$s=n-1$$;

(4) $$r=n-1$$, $$s=2$$.

If $$r=1$$, $$s=n$$ or $$r=n$$, $$s=1$$, i.e., A has a zero row or a zero column, then A is permutation similar to a matrix of the form

$\left[\begin{array}{cc}B& C\\ 0& 0\end{array}\right]$

or its transpose, where B is of order $$n-1$$ and C is a column vector. Since A has at most $$2n-2$$ zero entries, B has at most $$n-2$$ zero entries. Then there exists a permutation matrix Q of order $$n-1$$ such that $$B\geq Q$$. Note that

$\begin{array}{rl}{\left[\begin{array}{cc}B& C\\ 0& 0\end{array}\right]}^{k+1}& =\left[\begin{array}{cc}B& C\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}B& C\\ 0& 0\end{array}\right]}^{k}=\left[\begin{array}{cc}B& C\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{B}^{k}& {B}^{kâˆ’1}C\\ 0& 0\end{array}\right]\\ & â‰¥\left[\begin{array}{cc}Q& C\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{B}^{k}& {B}^{kâˆ’1}C\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}Q{B}^{k}& Q{B}^{kâˆ’1}C\\ 0& 0\end{array}\right].\end{array}$

Thus

$\begin{array}{rl}f\left({A}^{k+1}\right)& =f\left({\left[\begin{array}{cc}B& C\\ 0& 0\end{array}\right]}^{k+1}\right)â‰¥f\left(\left[\begin{array}{cc}Q{B}^{k}& Q{B}^{kâˆ’1}C\\ 0& 0\end{array}\right]\right)\\ & =f\left(Q{B}^{k}\right)+f\left(Q{B}^{kâˆ’1}C\right)=f\left({B}^{k}\right)+f\left({B}^{kâˆ’1}C\right)\\ & =f\left(\left[\begin{array}{cc}{B}^{k}& {B}^{kâˆ’1}C\\ 0& 0\end{array}\right]\right)=f\left({\left[\begin{array}{cc}B& C\\ 0& 0\end{array}\right]}^{k}\right)=f\left({A}^{k}\right)\end{array}$

for all positive integers k, which implies that $$\{f(A^{k})\} _{k=1}^{\infty}$$ is increasing.

If $$r=2$$, $$s=n-1$$ or $$r=n-1$$, $$s=2$$, then A is permutation similar to one of the matrices $$A_{1}, A_{2}, A_{1}^{T}, A_{2}^{T}$$, where

${A}_{1}=\left[\begin{array}{cccc}0& â‹¯& 0& 1\\ 0& â‹¯& 0& 1\\ 1& â‹¯& 1& 1\\ â‹®& â‹±& â‹®& â‹®\\ 1& â‹¯& 1& 1\end{array}\right],\phantom{\rule{2em}{0ex}}{A}_{2}=\left[\begin{array}{cccc}1& 0& â‹¯& 0\\ 1& 0& â‹¯& 0\\ 1& 1& â‹¯& 1\\ â‹®& â‹®& â‹±& â‹®\\ 1& 1& â‹¯& 1\end{array}\right].$

Direct computation shows that $$A_{1}^{2}\geq A_{1}, A_{2}^{2}\geq A_{2}$$. Thus $$\{ f(A^{k})\}_{k=1}^{\infty}$$ is increasing. This completes the proof.â€ƒâ–¡

### Remark

When $$f(A)=n^{2}-2n+1$$, the following example shows that $$\{f(A^{k})\}_{k=1}^{\infty}$$ may not be increasing. Consider

$A=\left[\begin{array}{cccc}0& 0& â‹¯& 0\\ 1& 0& â‹¯& 0\\ 1& 1& â‹¯& 1\\ â‹®& â‹®& â‹±& â‹®\\ 1& 1& â‹¯& 1\end{array}\right].$

Direct computation shows that $$f(A)=n^{2}-2n+1>f(A^{2})=n^{2}-2n$$.

## 3 Conclusion

This paper considers the number of positive entries $$f(A)$$ in a nonnegative matrix A and deals with the question of whether the sequence $$\{f(A^{k})\} _{k=1}^{\infty}$$ is monotonic. We prove that if $$f(A)\leq3$$ or $$f(A)\geq n^{2}-2n+2$$, then the sequence must be monotonic. Some examples show that if $$4\leq f(A)\leq n^{2}-2n+1$$ when $$n\geq3$$, then the sequence may not be monotonic.

## References

1. Bapat, R.B., Raghavan, T.E.S.: Nonnegative Matrices and Applications. Cambridge University Press, Cambridge (1997)

2. Berman, A., Plemmons, R.J.: Nonnegative Matrices in the Mathematical Sciences. SIAM, Philadelphia (1994)

3. Å idÃ¡k, Z.: On the number of positive elements in powers of a non-negative matrix. ÄŒas. PÄ›st. Mat. 89, 28â€“30 (1964)

4. Zhan, X.: Matrix Theory. Grad. Stud. Math., vol.Â 147. Amer. Math. Soc., Providence (2013)

### Acknowledgements

The author would like to express her sincere thanks to referees and the editor for their enthusiastic guidance and help.

Not applicable.

## Funding

This research was supported by the National Natural Science Foundation of China (Grant No. 71503166).

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Correspondence to Qimiao Xie.

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Xie, Q. Monotonicity of the number of positive entries in nonnegative matrix powers. J Inequal Appl 2018, 255 (2018). https://doi.org/10.1186/s13660-018-1833-5