Paper 2026/2193

Square-Root Log-Rank Standard Deviations for the Higher-Rank Kadison–Singer Problem: Polynomial-Time Algorithms via Schatten-Norm Potentials

Abstract

We study the matrix signing problem for positive semidefinite matrices, assigning one sign to each original matrix regardless of its rank. This includes the higher-rank Kadison--Singer partition problem [KS59] when the matrices sum to the identity. For arbitrary positive semidefinite matrices $A_1,\ldots,A_m\in\mathbb C^{n\times n}$, we prove that there are signs $\sigma_i\in\{-1,1\}$ such that $\|\sum_i \sigma_i A_i\|\le 3.34963\|\sum_i \operatorname{tr}[A_i]A_i\|^{1/2}$. We give a deterministic algorithm achieving this bound in $\widetilde O(mn^2+n^{4.75})$ real arithmetic operations. Inspired by the Lee--Sidford barrier, we use a potential function built from Schatten norms to obtain a square-root logarithmic dependence on the rank. Under the additional assumptions $\sum_iA_i=I$, $\|A_i\|\le\alpha$, and $\operatorname{rank}(A_i)\le r$, we obtain a discrepancy of at most $\min\{1,3.48745\sqrt{\alpha\log(2r)}\}$. We also give a deterministic algorithm achieving a discrepancy of at most $\min\{1,4.171\sqrt{\alpha\log(2r)}\}$ in $\widetilde O(mn^2+n^{6.38})$ real arithmetic operations. This dependence on $\alpha$ and $r$ is optimal up to an absolute constant when $0<\alpha\le1/256$ and $r\ge256\alpha^{-6}$: even diagonal matrices can force every signing to have a discrepancy of at least $\min\{1,\sqrt{\alpha\log(2r)}\}$. For a zero-one incidence matrix with at most $d$ ones in each row and each column, our deterministic algorithm achieves a discrepancy of $O(\sqrt{d\log(d)})$, recovering Harvey's bound [Har15] by a different method. We give a deterministic construction of a single spanning tree that is simultaneously spectrally thin with respect to any given collection of $k\ge1$ positive edge weightings of the same graph, assuming small edge leverage in every weighting. The tree retains the original edge weights, and its thinness bound grows only logarithmically with $k$.