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Matrix Spencer: Eight Standard Deviations Suffice and an Almost-Linear Time Algorithm for Dense Input

Zhao Song, Lichen Zhang

Source record

Source: arXiv

Published: Sep 14, 2026

arXiv: 2609.15025

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Source abstract

The Matrix Spencer conjecture asserts that for all symmetric matrices A1,,AnRn×nA_1,\ldots,A_n\in\mathbb{R}^{n\times n} with Ai1\|A_i\|\le1 there are signs ε1,,εn{1,1}\varepsilon_1,\ldots,\varepsilon_n\in\{-1,1\} with i=1nεiAi=O(n)\|\sum_{i=1}^n\varepsilon_iA_i\|=O(\sqrt n). Random signs give only the matrix-concentration bound O(nlogn)O(\sqrt{n\log n}), and the conjecture was known only under rank, block-diagonal, or Frobenius-norm restrictions. We prove it: a signing of discrepancy below 8n8\sqrt n always exists. We also give a randomized algorithm that finds a signing of discrepancy below 12n12\sqrt n with failure probability at most pp using n3+o(1)polylog(1/p)n^{3+o(1)}\operatorname{polylog}(1/p) arithmetic operations in the real-arithmetic model, which matches the size n3n^3 of the dense input up to subpolynomial factors. The existence proof is a partial-coloring argument with one new estimate: a hereditary Gaussian small-ball bound for the spectral body {xRn:ixiAiR}\{x\in \mathbb{R}^n:\|\sum_ix_iA_i\|\le R\}, proved by interpolating a log-partition function from a diagonal model to the noncommutative one under a matrix-weighted Poincaré inequality. Proving the conjecture and bringing the constant below 88 are different problems. The partial-coloring argument loses a large factor twice, when it turns Gaussian measure into signs through a union bound and when it proves the small-ball estimate at a radius far larger than necessary. We remove the first loss by a lossless coding of Gaussian measure into signs and the second by smooth spectral barriers with certified coefficients. The algorithms project Gaussian points onto a smoothed version of the spectral body, whose derivatives are traces against one Gibbs matrix. Four successively cheaper ways of maintaining that matrix bring the running time down to n3+o(1)n^{3+o(1)}.

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