Zero-Sum Squares in Bounded Discrepancy $\{-1,1\}$-Matrices

Abstract

For $n\geqslant 5$, we prove that every $n\times n$ matrix $\mathcal{M}=(a_{i,j})$ with entries in $\{-1,1\}$ and absolute discrepancy $\lvert\mathrm{disc}(\mathcal{M})\rvert=\lvert\sum a_{i,j}\rvert\leqslant n$ contains a zero-sum square except for the split matrices (up to symmetries). Here, a square is a $2\times 2$ sub-matrix of $\mathcal{M}$ with entries $a_{i,j}, a_{i+s,s}, a_{i,j+s}, a_{i+s,j+s}$ for some $s\geqslant 1$, and a split matrix is a matrix with all entries above the diagonal equal to $-1$ and all remaining entries equal to $1$. In particular, we show that for $n\geqslant 5$ every zero-sum $n\times n$ matrix with entries in $\{-1,1\}$ contains a zero-sum square.

Zero-Sum Squares in $\{-1, 1\}$-Matrices with Low Discrepancy

“…This was confirmed by Arévalo, Montejano and Roldán-Pensado in [1]. In fact, they proved that, except when n 4, every n×n non-split {−1, 1}-matrix M with | disc(M )| n has a zero-sum square.…”

“…The final lemma we will use to prove Theorem 2 is a variation on Claim 11 from [1]. The main difference between Lemma 5 and the result used by Arévalo, Montejano and Roldán-Pensado is that we will always find a square submatrix, which simplifies the proof of Theorem 2.…”

“…By repeating the argument when disc(A i ) is negative, it follows that A i and A i have the same sign for every i. In particular, two of the A i must have different signs, and we can apply an interpolation argument as in [1].…”

“…Recently, Arévalo, Montejano and Roldán-Pensado [1] initiated the study of a zerosum variant of Erickson's problem. Here we wish to avoid zero-sum squares, squares with entries that sum to 0.…”

“…In fact, they proved that, except when n 4, every n×n non-split {−1, 1}-matrix M with | disc(M )| n has a zero-sum square. They remark that it should be possible to extend their proof to give a bound of 2n, and they conjecture that the bound Cn should hold for any C > 0 when n is large enough relative to C. Conjecture 1 (Conjecture 5 in [1]). For every C > 0 there is an integer N such that whenever n N the following holds:…”