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I am interested in computing the sum of squares of determinants of principal minors. Let $A$ be an $n\times n$ positive semidefinite matrix and $A_S$ be a principal minor of $A$ indexed by the set $S \subseteq \{1,\ldots,n\}$. The classical result (without squares) is:

$\sum_{S \subseteq \{1,\ldots,n\}} \det(A_S) = \det(A+I)$

Are there any results on computing

$\sum_{S \subseteq \{1,\ldots,n\}} \det^2(A_S)$

or any other powers?

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Do you have a reference for the classical result? It does not ring a bell somehow... –  Thierry Zell Apr 11 '11 at 3:12
    
Horn & Johnson has some background: books.google.com/… –  Ben Apr 11 '11 at 6:05
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@Thierry. Imagine that you compute instead $\det(A+XI)$. Then expand in the indeterminate $X$. –  Denis Serre Apr 11 '11 at 6:07
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@Ben. I don't think that positive semidefiniteness plays a role here. –  Denis Serre Apr 11 '11 at 6:07
    
I agree, it's not necessary for the classical result. My particular problem has the additional psd constraint, which may or may not make the extension easier. –  Ben Apr 11 '11 at 6:16

1 Answer 1

up vote 5 down vote accepted

The identity you mention does generalize to sums of powers, but I don't know if it can give you anything computationally efficient. Given a set $X\subset \mathbb R$, let $D(X^n)$ denote all $n\times n$ diagonal matrices with diagonal elements from $X$. Then if you take $X_k=\{1,\omega,\cdots,\omega^{k-1}\}$ the $k$-th roots of unity, the following holds $$\sum_{S\subset \{1,2,\cdots,n\}}\det(A_S)^k=\frac{1}{k^n}\sum_{M\in D(X^n)}\det(A+M)^k$$ the proof is basically the same as the one for the case $k=1$ with a few more algebraic manipulations.

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Thanks, that's interesting! Not sure how to make it computationally efficient. –  Ben Apr 11 '11 at 6:06
    
BTW, do you have a reference for this result? –  Ben Apr 12 '11 at 23:47

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