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Monotonicity of the number of positive entries in nonnegative matrix powers
- Qimiao Xie^{1}Email author
https://doi.org/10.1186/s13660-018-1833-5
© The Author(s) 2018
- Received: 20 July 2018
- Accepted: 1 September 2018
- Published: 21 September 2018
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.
Keywords
- Nonnegative matrix
- Power
- Monotonicity
MSC
- 15B36
- 15B48
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 26–29, 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, Šidák [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 0–1 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 0–1 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.
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. □
Theorem 2
Let A be a 0–1 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 \(k\geq3\).
(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. □
Corollary 3
Let A be a 0–1 matrix of order 2. Then the sequence \(\{f(A^{k})\}_{k=1}^{\infty}\) is monotonic.
Remark
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.
Lemma 4
Let A be a 0–1 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 0–1 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 0–1 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. □
Theorem 5
Let A be a 0–1 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 Frobenius–König 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) \(r=n-1\), \(s=2\).
Remark
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.
Declarations
Acknowledgements
The author would like to express her sincere thanks to referees and the editor for their enthusiastic guidance and help.
Availability of data and materials
Not applicable.
Funding
This research was supported by the National Natural Science Foundation of China (Grant No. 71503166).
Authors’ contributions
The author read and approved the final manuscript.
Competing interests
The author declares that she has no competing interests.
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Authors’ Affiliations
References
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