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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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what is the notation $A \circ B$? –  J. Martel Aug 30 '12 at 3:13
    
Entrywise product of matrices. –  Felix Goldberg Aug 31 '12 at 15:01
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2 Answers 2

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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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 $$I(A\circ B)+I(A)I(B)\ge(\prod a_{ii})I(B)+(\prod b_{ii})I(A)\qquad ?$$ –  Denis Serre Feb 5 '13 at 23:14
    
This would be a big conjecture. I don't know the answer. Would you post it as a new problem. –  Betrand Feb 6 '13 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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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? –  Felix Goldberg Aug 29 '12 at 9:22
    
@Felix. I am not rich enough to afford two n's in my first name. –  Denis Serre Aug 29 '12 at 14:41
    
Sorry <blush>.... –  Felix Goldberg Aug 31 '12 at 15:01
    
Shall we pass around a hat to buy Denis another n? –  Gerry Myerson Feb 5 '13 at 23:54
    
@Gerry. I have fought for decades to keep my first name in French style (one enn). Actually, it is ancient Greek style. Only wasps write it with two enns. Hope that there is a hole deep in the hat. –  Denis Serre Feb 6 '13 at 2:05
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