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The well-known Oppenheim inequality says that for two positive definite matrices $A,B$ it holds that $\det(A \circ B) \geq (\prod{a_{ii}})\det(B)$.

There has been a lot of beautiful work done extending it to cases when $A$ or $B$ or both of them are $M$-matrices or their inverses, or totally nonnegative.

My question is: do you know of other extensions, in which $A$ is non-symmetric in an "interesting" way?

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    $\begingroup$ what is the notation $A \circ B$? $\endgroup$
    – JHM
    Commented Aug 30, 2012 at 3:13
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    $\begingroup$ Entrywise product of matrices. $\endgroup$
    – Felix Goldberg
    Commented Aug 31, 2012 at 15:01

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This is not a generalization to other matrix classes, but replacing determinant by permanent. Actually, it is a conjecture made by Bapat and Sunder: Under the same conditions $per(A \circ B) \leq (\prod{a_{ii}})per(B)$.

...but the following result due to Jiao [On a conjecture of H. Minc, Linear and Multilinear Algebra 32 (1992) 103–105.] couldn't surprise me more $$per(A \circ B)+per (A) per (B) \geq (\prod{a_{ii}})per(B)+(\prod{b_{ii}})per(A).$$

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    $\begingroup$ After this inequality, and that of Fallat & Johnson (see my answer below), could it be that for every immanent $I$ and every positive definite symmetric matrices $A,B$, we have $$I(A\circ B)+I(A)I(B)\ge(\prod a_{ii})I(B)+(\prod b_{ii})I(A)\qquad ?$$ $\endgroup$
    – Denis Serre
    Commented Feb 5, 2013 at 23:14
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    $\begingroup$ This would be a big conjecture. I don't know the answer. Would you post it as a new problem. $\endgroup$
    – Betrand
    Commented Feb 6, 2013 at 20:27
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This is not a generalization to other matrix classes, but a generalization of the inequality, within the same class of Hermitian positive definite (or semi-definite) matrices. The flaw of Oppenheim's inequality is that the right-hand side is not symmetric in $A$ and $B$, unlike the left-hand side. Instead, S. Fallat & C. Johnson proved a symmetric form of OI: $$\det(A \circ B)+\det A\det B \geq (\prod{a_{ii}})\det(B)+(\prod{b_{ii}})\det(A).$$ See Exercise 285 in my List of exercises on Matrices.

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  • $\begingroup$ Thanks, Dennis, that's a really beautiful one. (Looks a bit like a modular formula to me, but probably it's just my weird mathsight). Do you know perhaps of a useful way to reduce the non-Hermitian case the Hermitian that might help out here? $\endgroup$
    – Felix Goldberg
    Commented Aug 29, 2012 at 9:22
  • $\begingroup$ @Felix. I am not rich enough to afford two n's in my first name. $\endgroup$
    – Denis Serre
    Commented Aug 29, 2012 at 14:41
  • $\begingroup$ Sorry <blush>.... $\endgroup$
    – Felix Goldberg
    Commented Aug 31, 2012 at 15:01
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    $\begingroup$ Shall we pass around a hat to buy Denis another n? $\endgroup$
    – Gerry Myerson
    Commented Feb 5, 2013 at 23:54
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    $\begingroup$ That inequality is not due to S. Fallat & C. Johnson. It is called the Oppenheim-Schur inequality; see page 509 of [R. A. Horn, C. R. Johnson, Matrix Analysis, Cambridge University Press, 2nd ed., 2013.] $\endgroup$
    – Russel
    Commented Apr 26, 2014 at 19:07

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