Given an orthogonal matrix $O$ with dimensions $4n \times 4n$ and $\det O = -1$, how to prove that
$\det[O\_{11} - O\_{22} + i (O\_{12} + O\_{21})] = 0$?


Here $O$ is a block matrix $[[O\_{11}, O\_{12}], [O\_{21}, O\_{22}]]$, and all blocks have equal size.


An equivalent statement is the following: if $\det O = -1$, then $\det(\Omega - O \Omega O^T)=0$, with $\Omega = [[0, I],[-I, 0]]$. The reason why I believe this is correct is due to the physical meaning of it (it's about presence of bound states in a certain system), and numerical checks. I am not sure why odd $n$ doesn't work, but it makes sense for physics reasons.


Example of a matrix for which it works is below:


$$\left(\begin{array}{cccc}
0.44090815& -0.71206204& -0.44576549& 0.31600755 \\\
0.12767731& 0.35584235& 0.19884606& 0.90417641 \\\
0.88038152& 0.229104 & 0.30519084& -0.28159952 \\\
0.11927654& 0.56022784& -0.81768693& -0.05749749
\end{array}\right)$$


Edit(Will Jagy):In particular, this fails for $(4n - 2) \times (4n - 2),$ as in the $ 2 \times 2$
$$ O \; \; = \; \; 
\left( \begin{array}{cc}
 \cos t & \sin t\\\
 \sin t & - \cos t 
\end{array} 
 \right) , $$
with the relevant $ 1 \times 1$ matrix of complex numbers
$$ \left( \cos t - ( - \cos t) + i ( \sin t + \sin t )\right) = \left( 2 e^{it} \right) . $$


Edit(Anton Akhmerov):
Yet another statement which could solve the problem is the following:
Prove that any $4n\times 4n$ orthogonal matrix $O$ can be brought to a block-diagonal form by a transformation $O\rightarrow S\_1 O S\_2$ with $S\_1$ and $S\_2$ symplectic matrices. If this was correct, the rest follows immediately.



Relationship between determinants.
I denote your matrix $\Omega$ by $W$ for the sake of brevity. Note that $W$ is both skew-symmetric and orthogonal, i. e. it satisfies $W=-W^T$ and $W^2=-I$. (And this is all I am going to use about $W$.)


The only thing I am going to use about the matrix $O$ is that $O^TO=I$. The assumption that $O$ is a real matrix will not be needed (it could be from any field of characteristic $\neq 2$).


We are going to use notion of the [Pfaffian](http://en.wikipedia.org/wiki/Pfaffian) of a skew-symmetric matrix.


**Lemma 1.** Let $R$ be a commutative ring with $1$. Let $A\in R^{2n\times 2n}$ be a skew-symmetric matrix, and $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, the matrix $B^TAB$ is skew-symmetric as well, and $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$.


*Proof of Lemma 1.* Clearly, the matrix $B^TAB$ is skew-symmetric. It remains to prove that $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$. Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Since the square of the Pfaffian of a skew-symmetric matrix is the determinant of this matrix, we have


$\left(\mathrm{Pf}\left(B^TAB\right)\right)^2=\det\left(B^TAB\right)=\underbrace{\det B^T}\_{=\det B}\cdot\underbrace{\det A}\_{=\left(\mathrm{Pf}A\right)^2}\cdot \det B = \left(\det B\cdot\mathrm{Pf} A\right)^2$.`


Thus, FOR EVERY $A$ and $B$, we have either $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$ OR $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$. By a Zariski-topological argument, we can interchange the words "for every" with the words "or" here, so we obtain: (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$) OR (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$). Since (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$) is wrong (take $A=W$ and $B=\mathrm{id}$), we thus must have (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$), and Lemma 1 is proven.


**Lemma 2.** Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ and $C\in R^{2n\times 2n}$ be arbitrary matrices. Then, $\det\left(W-BWC\right)=\det\left(W-CWB\right)$.


*Proof of Lemma 2.* We have $W-BWC=W\cdot\left(I+WBWC\right)$ (because $-I=W^2$), and thus
$\det\left(W-BWC\right)=\det\left(W\cdot\left(I+WBWC\right)\right)=\det W\cdot\det\left(I+WBWC\right)$
and similarly
$\det\left(W-CWB\right)=\det W\cdot\det\left(I+WCWB\right)$.
Thus, in order to prove Lemma 2, it remains to show that $\det\left(I+WBWC\right)=\det\left(I+WCWB\right)$. This follows from the general fact that if $U$ and $V$ are two square matrices of the same size, then $\det\left(I+UV\right)=\det\left(I+VU\right)$ (for a proof of this fact, apply Corollary 2 in [MathLinks post #1491761](http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1491761#p1491761) to $X=-1$). Thus, Lemma 2 is proven.


**Lemma 3.** Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$.


*Proof of Lemma 3.* Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Applying Lemma 2 to $C=B^T$, we obtain $\det\left(W-BWB^T\right)=\det\left(W-B^TWB\right)$. Since the determinant of a skew-symmetric matrix is the square of its Pfaffian, this rewrites as $\left(\mathrm{Pf}\left(W-BWB^T\right)\right)^2=\left(\mathrm{Pf}\left(W-B^TWB\right)\right)^2$. Again, a Zariski density argument like in the proof of Lemma 1 shows us that we must have $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$ (because we cannot have $\mathrm{Pf}\left(W-BWB^T\right)=-\mathrm{Pf}\left(W-B^TWB\right)$ for all $B$, since this would fail for $B=0$). This proves Lemma 3.


Now let us solve our problem:


We have $\mathrm{Pf}U=\mathrm{Pf}\left( -U\right) $ for every skew-symmetric $4n\times 4n$ matrix $U$ (because $\mathrm{Pf}$ is a homogeneous polynomial of degree $2n$). Now,


$\mathrm{Pf}\left( W-OWO^{T}\right) =\mathrm{Pf}\left( W-O^{T}WO\right) $ (after Lemma 3, applied to $B=O$)


$=\mathrm{Pf}\left( -\left( W-O^{T}WO\right) \right) $ (since $\mathrm{Pf}U=\mathrm{Pf}\left( -U\right) $ for every skew-symmetric $4n\times4n$ matrix $U$)


$=\mathrm{Pf}\left( O^{T}WO-W\right) $


$ =\mathrm{Pf}\left( O^{T} WO-O^{T}OWO^{T}O\right) $ (since $O^{T}O=I$ yields $W=O^{T}OWO^{T}O$)


$=\mathrm{Pf}\left( O^{T}\left( W-OWO^{T}\right) O\right) $


$=\underbrace{\det O}\_{=-1}\cdot\mathrm{Pf}\left( W-OWO^{T}\right) $ (by Lemma 1, applied to $B=O$ and $A=W-OW^{T}O^{T}$)


$=-\mathrm{Pf}\left( W-OWO^{T}\right) $,


so that $2\mathrm{Pf}\left( W-OWO^{T}\right) =0$ and thus $\mathrm{Pf}\left( W-OWO^{T}\right) = 0$, so that $\det\left( W-OWO^{T}\right) =\left(\mathrm{Pf}\left( W-OWO^{T}\right) \right)^2=0^2=0$, qed.


