Let H_n be the set of n× n positive semidefinite Hermitian matrices. For matrices A=[a_i,j] and B=[b_i,j] in H_n, define the Hadamard product by (A circ B)=[a_i,jb_i,j]. Bapat–Sunder's conjecture. If A,B∈H_n, then operatornameper(A circ B)≤ operatornameper(A)∏_i=1^n b_i,i. This is the permanental counterpart of Oppenheim's determinant inequality. The conjecture was disproved by Drury; the status is therefore refuted.

Bapat–Sunder's permanent Hadamard-product inequality

About 5 years old · traced to 2021

Let $\mathcal{H}_n$ be the set of $n\times n$ positive semidefinite Hermitian matrices. For matrices $A=[a_{i,j}]$ and $B=[b_{i,j}]$ in $\mathcal{H}_n$ , define the Hadamard product by $(A\circ B)=[a_{i,j}b_{i,j}]$ . Bapat–Sunder's conjecture. If $A,B\in\mathcal{H}_n$ , then

\operatorname{per}(A\circ B)\leq \operatorname{per}(A)\prod_{i=1}^n b_{i,i}.

This is the permanental counterpart of Oppenheim's determinant inequality. The conjecture was disproved by Drury; the status is therefore refuted.

References

Primary source

Léo Pioge, Kamil K. Pietrasz, Benoit Seron, Leonardo Novo and Nicolas J. Cerf, “A logical implication between two conjectures on matrix permanents”, arXiv:2508.00111 (2025).

Additional references

2 papers in this index state this conjecture (2021–2025). The statement above is taken from the most recent of them; the others are arXiv:2101.03428.

hadamard-products matrix-permanents positive-semidefinite-matrices

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The conjecture is false: papers report Drury’s counterexample, so the inequality fails in general.

Bapat and Sunder conjectured that the permanent of a Hadamard product is bounded by the permanent of one factor times the product of the other factor’s diagonal entries. The conjecture is now reported to be disproved by Stephen W. Drury.

Known results

The inequality holds for rank-one $A$ and for $A$ with nonnegative entries; characterizing all such $A$ remains open.

Drury counterexample: 2021 report, reaffirmed in 2025

A 2021 paper gives a positive-semidefinite Hermitian $7\times 7$ matrix with diagonal entries $1$ and ratio $\operatorname{per}(A\circ A^T)/\operatorname{per}(A)=1237/1152>1$ , violating the conjecture. A 2025 paper records Drury’s disproof and studies its relation to a second Bapat–Sunder conjecture. These reports are treated here as unverified.

Current status (as of September 2026): The inequality is refuted by Drury’s reported counterexample; the broader characterization of matrices satisfying it remains open.

Sources

No solutions have been posted yet.