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Low-rank matrices, tournaments, and symmetric designs

Published 25 Jan 2024 in math.CO | (2401.14015v4)

Abstract: Let a=(ai)<em>i1\mathbf{a} = (a_{i})<em>{i \geq 1} be a sequence in a field F\mathbb{F}, and f ⁣:F×FFf \colon \mathbb{F} \times \mathbb{F} \to \mathbb{F} be a function such that f(a</em>i,ai)0f(a</em>{i},a_{i}) \neq 0 for all i1i \geq 1. For any tournament TT over [n][n], consider the n×nn \times n symmetric matrix MT(f,a)M_{T}(f, \mathbf{a}) with zero diagonal whose (i,j)(i,j)th entry (for $i &lt; j$) is f(ai,aj)f(a_{i},a_{j}) if iji \to j in TT, and f(aj,ai)f(a_{j},a_{i}) if jij \to i in TT. It is known (cf. Balachandran et al., Linear Algebra Appl. 658 (2023), 310-318) that if TT is a uniformly random tournament over [n][n], then rank(MT(f,a))(12o(1))n\operatorname{rank}(M_{T}(f,\mathbf{a})) \geq (\frac{1}{2}-o(1))n with high probability when char(F)2\operatorname{char}(\mathbb{F}) \neq 2 and ff is a linear function. In this paper, we investigate the other extremal question: how low can the ranks of such matrices be? We work with sequences a\mathbf{a} that take only two distinct values, so the rank of any such n×nn \times n matrix is at least n/2n/2. First, we show that the rank of any such matrix depends on whether an associated bipartite graph has certain eigenvalues of high multiplicity. Using this, we show that if ff is linear, then there are n×nn \times n real matrices MT(f;a)M_{T}(f;\mathbf{a}) of rank at most n2+O(1)\frac{n}{2} + O(1). For rational matrices, we show that for each $\varepsilon &gt; 0$ we can find a sequence a(ε)\mathbf{a}(\varepsilon) for which there are n×nn \times n matrices MT(f;a(ε))M_{T}(f;\mathbf{a}(\varepsilon)) of rank at most (12+ε)n+O(1)(\frac{1}{2} + \varepsilon)n + O(1). These matrices are constructed from symmetric designs, and we also use them to produce bisection-closed families of size greater than 3n/22\lfloor 3n/2 \rfloor - 2 for n15n \leq 15, which improves the previously best known bound (cf. Balachandran et al., Electron J. Combin. 26 (2019), #P2.40).

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