International Association for Cryptologic Research

International Association
for Cryptologic Research

IACR News item: 26 September 2026

Zhao Song, Song Yue
ePrint Report ePrint Report
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 $k$.
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