Monotonicity of the number of positive entries in nonnegative matrix powers

Let A be a nonnegative matrix of order n and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$f(A)$\end{document}f(A) denote the number of positive entries in A. We prove that if \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$f(A)\leq3$\end{document}f(A)≤3 or \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$f(A)\geq n^{2}-2n+2$\end{document}f(A)≥n2−2n+2, then the sequence \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\{f(A^{k})\}_{k=1}^{\infty}$\end{document}{f(Ak)}k=1∞ is monotonic for positive integers k.


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 ≥ B or B ≤ 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 Problem ([4], p. 233) Characterize those sign patterns of square nonnegative matrices A such that the sequence {f (A k )} ∞ k=1 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 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 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.

Main results
Let A be a nonnegative square matrix. We will use the fact that if A 2 ≥ A (A 2 ≤ A), then A k+1 ≥ A k (A k+1 ≤ A k ) for all positive integers k and thus {f (A k )} ∞ k=1 is increasing (decreasing).
Thus, for k = 2, 3, . . . , is invariant under permutation similarity or transpose of A, it suffices to consider the following cases. ( It can be seen that in each case {f (A k )} ∞ k=1 is decreasing. This completes the proof.
Proof Under permutation similarity and transpose, it suffices to consider the following cases. ( k=1 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×n is defined as where S n is the set of permutations of the integers 1, 2, . . . , n. First we have the following important fact. Proof Since A is a 0-1 matrix with per A > 0, there exists a permutation matrix P such that

Lemma 4 Let
Theorem 5 Let A be a 0-1 matrix of order n. If f (A) ≥ n 2 -2n + 2, then the sequence is increasing. Next suppose per A = 0. Then by the Frobenius-König theorem [4, p. 46], A has an r × s zero submatrix with r + s = n + 1. Since f (A) ≥ n 2 -2n + 2, A has at most 2n -2 zero entries.
Thus rs ≤ 2n -2. It can be seen that r and s must be one of the following solutions.
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 B C 0 0 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 ≥ Q. Note that for all positive integers k, which implies that {f (A k )} ∞ k=1 is increasing.
If r = 2, s = n -1 or r = n -1, s = 2, then A is permutation similar to one of the matrices is increasing. This completes the proof.

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 is monotonic. We prove that if f (A) ≤ 3 or f (A) ≥ n 2 -2n + 2, then the sequence must be monotonic. Some examples show that if 4 ≤ f (A) ≤ n 2 -2n + 1 when n ≥ 3, then the sequence may not be monotonic.