TITLE: Sum of squares of determinants of principal minors
QUESTION [5 upvotes]: 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?

REPLY [7 votes]: 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.