Pate's two-index permanent inequality challenge

From papers

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

a11perA(11)a12perA(12)a21perA(21)+a22perA(22)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.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

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

Solutions 0

No solutions have been posted yet.