Imagine I have two skew-symmetric square matrices $A$, $B$. (So $A^\intercal = -A$, etc.) Now I am interested in the square root of the determinant of $AB+I$, where $I$ is the identity matrix,

$$ x = \sqrt{ \det \left( AB + I \right) } $$

As quick inspection for small matrices suggests that this $x$ is a polynomial of the elements of $A$ and $B$, for example for $3 \times 3$ matrices we find

$$ x = 1 - a_{12} b_{12} - a_{13} b_{13} - a_{23} b_{23} $$

and I checked this analytically for matrices up to $6 \times 6$. It reminds me of the pfaffian of a skew-symmetric matrix, which is also a 'square root of a determinant' but nonetheless a polynomial in the matrix elements.

Now my questions are:

  • Does anyone know a proof that $x$ is a polynomial in the elements of $A$ and $B$, and if so, what is that polynomial?
  • Does anyone know an efficient (so not $O(n!)$) algorithm to compute $x$?
  • 1
    $\begingroup$ If you multiply $AB+I $ with the skew-symmetric matrix $A $, you get the skew-symmetric matrix $ABA+A$. This shows that $\det\left (AB+I\right) $ is a quotient of two squares when $n$ is even. When $n $ is odd, you can go up one dimension. But this is clearly not the "right" proof. Great question! $\endgroup$ – darij grinberg Aug 16 '17 at 14:28
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    $\begingroup$ By the proof of en.m.wikipedia.org/wiki/Sylvester%27s_determinant_identity , it suffices to prove that $\det M $ is a square, where $M = \begin{pmatrix}I_m & -A \\ B & I_n \end{pmatrix}$. Now, $M $ can be reduced easily to a skew-symmetric matrix (switch the first block column with the second one, then multiply said column by $-1$); this clears everything up. $\endgroup$ – darij grinberg Aug 16 '17 at 14:36
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    $\begingroup$ @darijgrinberg There is no need to do any row/column operations. As $A$ and $B$ have identical sizes, we may begin with $M=\pmatrix{A&-I\\ I&B}$. Since the two sub-blocks at the bottom commute, $\det M=\det[(A)(B)-(-I)(I)]=\det(AB+I)$. $\endgroup$ – user1551 Aug 16 '17 at 15:45
  • $\begingroup$ Thank you! As for a fast algorithm, now that $x$ is just the pfaffian of the matrix M that @user1551 defined, I can use the methods of Wimmer in this arXiv-post: link to compute it fast. $\endgroup$ – Louk Rademaker Aug 17 '17 at 0:42

The quantity you are computing is very close to the relative Pfaffian, as found in for example Section 2 of this paper.

Let $A,B$ be skew matrices acting on $\mathbb{C}^{2n}$, then there is something called the relative Pfaffian of $A$ with respect to $B$, denoted $ \operatorname{Pf}(A,B)$, which satisfies the relation \begin{equation} \operatorname{Pf}(A,B)^{2} = \det(I - AB). \end{equation} (So I guess you would be interested in $\operatorname{Pf}(-A,B)$.) An explicit formula for the relative Pfaffian is given in Definition II.7 on page 8 (or 355 if you prefer). It reads \begin{equation} \operatorname{Pf}(A,B) = \sum_{S} \operatorname{Pf}(A_{S}) \operatorname{Pf}(B_{S}), \end{equation} where the sum runs over non-empty even subsets $S$ of $\{1, ..., 2n\}$, and where $A_{S}$ denotes the restriction of $A$ to the subspace spanned by $\{e_{j}\}_{j \in S}$.

It is furthermore proven that if $A$ is invertible, then the relation \begin{equation} \frac{\operatorname{Pf}(A^{-1} -B)}{\operatorname{Pf}(A^{-1})} = \operatorname{Pf}(A,B). \end{equation} holds, (Proposition II.6, page 7).

Just in case that link rots, the article is called "Pfaffians on Hilbert space" and is written by Arthur Jaffe, Andrzej Lesniewski, and Jonathan Weitsman, published in the journal of functional analysis.

  • $\begingroup$ Are you sure you want to say "non-empty"? $\endgroup$ – darij grinberg Nov 21 '18 at 19:26

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