1
$\begingroup$

Let $A$ be an $n \times n$ real orthogonal matrix, $A^T A = I$. For subsets $S, S'$ of $\{1,\dots, n\}$, let $A_{S,S'}$ denote the restriction of $A$ to rows indexed by $S$ and columns indexed by $S'$. I'm wondering if there is a general formula or bound for (the absolute value of) $$\sum_{S' \subseteq\{1,\dots, n\}:|S'| = |S|} \text{det}(A_{S,S'})$$ for a given $S \subseteq \{1,\dots, n\}$?

For instance, in the special case where $A$ is not only orthogonal, but also a permutation matrix, this sum evaluates to $+1$ or $-1$, since for fixed $S$, there is exactly one $S'$ for which $A_{S,S'}$ has a nonzero entry in each row (and for that $S'$, $A_{S,S'}$ is itself a permutation matrix). But more generally, for orthogonal $A$, I'm interested in whether (the absolute value of) this sum can be bounded (as a function of $n$ and/or $|S|$). Is there a known reference for this problem? Even just knowing whether it's polynomially or exponentially increasing in $n$ (and/or $|S|$) would suffice as a starting point.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

When $|S|=1$ we are asking for the upper bound of $x_1+x_2+\dots+x_n$ subject to $x_1^2+\dots+x_n^2=1$. This is achieved when the $x_i$ are all equal to $\frac{1}{\sqrt{n}}$: and so the upper bound is $\sqrt{n}$.

Now note that when $A$ is orthogonal so too is the matrix of $r\times r$ cofactors $A^{(r)}$: this follows from the general facts that $(XY)^{(r)}=X^{(r)}Y^{(r)}$, $(X^{T})^{(r)}=(X^{(r)})^T$ and $I^{(r)}=I$.

We can therefore apply the result for $|S|=1$ to the ${n\choose |S|}\times {n\choose |S|}$ orthogonal matrix $A^{({n\choose |S|})}$ and obtain the result that for arbitrary $|S|$ the upper bound is $\sqrt{n\choose |S|}$.

$\endgroup$
2
  • $\begingroup$ Do you have a reference for (the matrix of) r x r cofactors? How are they defined? $\endgroup$
    – user38762
    May 8, 2023 at 18:37
  • 1
    $\begingroup$ I usually consult Aitken's "Determinant and Matrices". But the Wikipedia articles on Cauchy-Binet, and on Minors may help. $\endgroup$ May 9, 2023 at 6:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .