Pate's two-index permanent inequality challenge

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Let A∈HnA\in{\cal H}_n, and let A(i∣j)A(i|j) denote the matrix obtained by deleting row ii and column jj. Pate's inequality challenge.

a11perA(1∣1)−a12perA(1∣2)−a21perA(2∣1)+a22perA(2∣2)⩽2perA.a_{11}\mathop{\rm per} A(1|1)-a_{12}\mathop{\rm per} A(1|2)-a_{21}\mathop{\rm per} A(2|1)+a_{22}\mathop{\rm per} A(2|2)\leqslant2\mathop{\rm per} A.

The source presents this as a consequence of the now-refuted k=1k=1 eigenvalue conjecture, but does not state its independent resolution.

References

Primary source

Ian M. Wanless, “Lieb's permanental dominance conjecture”, arXiv:2202.01867 (2022).

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