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Suppose $A,A_1,\ldots,A_{n-2}$ (resp. $B$) are (resp. is) real positive-definite (resp. arbitrary) symmetric $n\times n$ matrices and denote by $D(\cdot,\ldots,\cdot)$ the mixed discriminant. We have the following well-known Aleksandrov-Fenchel inequality

\begin{equation}\label{e} D(A,B,A_1,\ldots,A_{n-2})^2\geq D(A,A,A_1,\ldots,A_{n-2})D(B,B,A_1,\ldots,A_{n-2}), \end{equation} with equality iff $A$ and $B$ are proportional.

Now my question comes: does this inequality and equality case still hold if we assume these matrices are complex Hermitian matrices rather than real symmetric ones? I guess this is the case but I am not able to find a reference. The standard textbook of Schneider only treats the real symmetric case.

Many thanks in advance!

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  • $\begingroup$ I think this inequality even holds with quaternionic hermitian matrices. The above inequality can be most likely shown directly by mimicking the "usual" proof? $\endgroup$ – Suvrit Aug 23 '17 at 22:19
  • $\begingroup$ @Suvrit In section 5.5 of Shneider's book, the proof of this inequality is obtained as a special case of an inequality for hyperbolic polynomials. These polynomilas are defined on $\mathbb{R}^n$ and I don't think it can be treated similarly on $\mathbb{C}^n$, at least at the first glance. $\endgroup$ – Kevin Aug 24 '17 at 1:42
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The Van der Waerden Conjecture for Mixed Discriminants, by Leonid Gurvits (2004), proves a generalized Alexandrov-Fenchel inequality for semidefinite $n\times n$ Hermitian matrices:

Theorem 5.2, with equation (21), that for the special case $\alpha=(1,1,\ldots,1)$, $\alpha^{(1)}=(2,0,1,\ldots 1)$, $\alpha^{(2)}=(0,2,1,\ldots 1)$ reduces to the desired inequality with an extra factor $\exp(-2n^n/n!)$ on the right-hand side.

Not good enough, obviously, but I have not found a proof for the Hermitian case without this reduction factor.

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The answer is yes, as shown by Ping Li, theorem 3.1.

https://arxiv.org/pdf/1710.00520.pdf

or by Shenfeld and Van Handel

https://arxiv.org/pdf/1811.08710.pdf

Moreover the original proof by Alexandroff in the real case seems to work in the Hermitian case too.

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