The Maximum Diagonal Product of a Stochastic Matrix of Bounded Rank
2026-10-21
Research publication
The Maximum Diagonal Product of a Stochastic Matrix of Bounded Rank
Preprint
- Original date
- Authors
- Yair Lavi
- Matrix theory
- Stochastic matrices
- Nonnegative matrices
- Matrix inequalities
About this work
A proof of the second of my two conjectures from 2018, the one about diagonal products. The first one, about permanents, is proved in The Maximum Permanent of a Stochastic Matrix of Bounded Rank.
Let \(A=(a_{ij})\) be a row- or column-stochastic \(n\times n\) matrix with \(\operatorname{rank}A\leq k\), where \(1\leq k\leq n\). Write \(n=qk+s\), where \(0\leq s<k\). I conjectured in 2018 that, for every permutation \(\sigma\) of \(\{1,\ldots,n\}\),
\[ \prod_{i=1}^n a_{i,\sigma(i)}\leq \left(\frac{1}{q^q}\right)^{k-s} \left(\frac{1}{(q+1)^{q+1}}\right)^s, \]
with equality if and only if
\[ A=P^{\mathsf T}\left(J_q^{\oplus(k-s)}\oplus J_{q+1}^{\oplus s}\right)P\,\Pi_\sigma, \]
where \(P\) is a permutation matrix, \(\Pi_\sigma\) is the permutation matrix of \(\sigma\), and \(J_t\) is the \(t\times t\) matrix with every entry \(1/t\). We prove this conjecture in full.