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math.CO — Combinatorics

A Spectral Blakley–Roy Inequality for Nonnegative Matrices

Contributed by Suvrit Sra

Let AA and BB be real symmetric matrices with nonnegative entries. We prove that ρ(AB)k≤ρ(AkBk)\rho(AB)^k\le\rho(A^kB^k) for the spectral radius ρ(⋅)\rho(\cdot) and every integer k≥1k\ge1; for B=uuTB=uu^T this is the Blakley–Roy inequality. We prove it via the more general inequality ∥(AB)m∥k≤∥(AkBk)m∥\|(AB)^m\|^{k}\le\|(A^kB^k)^m\| in the spectral norm, valid for all integers k,m≥1k,m\ge1. The even exponents case follows from standard tools; for odd exponents k=2r+1k=2r+1 the claim follows via the more general product inequality (uTC1⋯Ctv)k≤(uTu)r(vTv)r uTC1k⋯Ctkv(u^TC_1\cdots C_tv)^k\le(u^Tu)^r(v^Tv)^r\,u^TC_1^k\cdots C_t^kv for nonnegative symmetric CiC_i and nonnegative u,vu,v. Its proof is an entropy argument on paths that traverse each edge kk times, back and forth.

A mix of human-written and AI-generated textHuman understanding: all partsBlakley-Roy inequalitymatrix productsnonnegative matricesspectral radius

math.FA — Functional Analysis

On Lin's Arithmetic–Geometric Means Conjecture for Singular Values

Contributed by Suvrit Sra

Let AA and BB be positive semidefinite, and let 0≤t≤10\le t\le 1. Lin (2017) conjectured that λj((1−t)A+tB)≥σj1/2(A2(1−t)B2t)\lambda_j((1-t)A+tB)\ge\sigma_j^{1/2}(A^{2(1-t)}B^{2t}) for j=1,…,nj=1,\ldots,n. The case t=1/2t=1/2 is Drury's theorem, which answered a long-standing question of Bhatia and Kittaneh. Lin remarked that Drury's method does not seem to reach other values of tt. We prove his conjecture for every t∈[0,1]t\in[0,1]. The proof builds on Drury's argument, but replaces his block-matrix step (unique to the midpoint t=1/2t=1/2) by an eigenvalue count based on an inertia identity and a one-variable inequality.

A mix of human-written and AI-generated textHuman understanding: all partsLin's conjecturearithmetic-geometric mean inequalitymatrix inequalitiessingular valuesunitarily invariant norms

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