# Square root of the determinant of AB+I where A, B are skew-symmetric

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 $x$?
• 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! – darij grinberg Aug 16 '17 at 14:28
• 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. – darij grinberg Aug 16 '17 at 14:36
• @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, $\det M=\det[(A)(B)-(-I)(I)]=\det(AB+I)$. – user1551 Aug 16 '17 at 15:45
• 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. – Louk Rademaker Aug 17 '17 at 0:42

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 $$\operatorname{Pf}(A,B)^{2} = \det(I - AB).$$ (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 $$\operatorname{Pf}(A,B) = \sum_{S} \operatorname{Pf}(A_{S}) \operatorname{Pf}(B_{S}),$$ 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 $$\frac{\operatorname{Pf}(A^{-1} -B)}{\operatorname{Pf}(A^{-1})} = \operatorname{Pf}(A,B).$$ holds, (Proposition II.6, page 7).