diagonalize block matrices with special type of unitary matrices Problem
(this is a problem from Michael Nielsen's notes on fermions)
$\alpha$ and $\beta$ are $n\times n$ complex matrices, $\alpha$ is Hermitian and $\beta$ is anti-symmetric. $M$ is the following $2n \times 2n$ matrix (in the following, the dimensions of the sub-blocks is always $n\times n$ and will not be specified. )
\begin{equation}
  M = \begin{bmatrix}
    \alpha & - \beta^* \\
    \beta  & -\alpha^* \\
  \end{bmatrix}
\end{equation}
where $*$ is the complex conjugate(not Hermitian conjugate). It's easy to see $M$ is a Hermitian matrix and so has only real eigenvalues. 
The eigenvalues comes in pair
Define $S = \begin{bmatrix} 0 & I \\ I & 0  \end{bmatrix}$ where $I$ is the identity matrix.  $M$ has the following symmetry
\begin{equation}
S M^* S^{-1} = - M 
\end{equation}
then if $\lambda$ is the eigenvalue, then
\begin{equation}
0 = \det( \lambda I - M )  = \det( \lambda^* I - M^* ) =  \det( \lambda^* I - SM^*S^{-1} ) = \det( -\lambda I - M  )
\end{equation}
i.e. $-\lambda$ is also an eigenvalue. There are $2n$ eigenvalues, so if $\lambda = 0$, its multiplicity must be even. 
Therefore, $\exists $ unitary matrix $U$, s.t.
\begin{equation}
M = U
\begin{bmatrix}
d & 0\\
0 & -d \\
\end{bmatrix} U^{\dagger} = UD U^{\dagger}
\end{equation}
where $d$ is diagonal. 
Question: 


*

*Is it possible restrict $U$ to have the following form
\begin{equation}
 U = \begin{bmatrix}
   \gamma & \mu \\
   \mu^* & \gamma^*
     \end{bmatrix}
\end{equation}


In the notes I read, the author says with a beautiful application of cosine-sine decomposition, it takes a few steps to prove this statement. I'd like to see such a proof using cosine-sine decomposition. 


*Is there an (numerical) algorithm that efficiently construct $U \in B$? 

*What is this symmetry of $U$ and $M$? 
The required $U$ satisfies
\begin{equation}
U = S U^* S^{-1}
\end{equation}
Let $B = \{U | U = SU^* S^{-1}, U \in \text{U}(2n) \}$, then $B$ is a subgroup of ${\rm U}(2n)$, since multiplication, identity, inverse easily checks out. What is this group? Symplectic? 
Partial Solution:
Taking a arbitrary $U$ that diagonalize $M$, we have
\begin{equation}
\begin{aligned}
  M U = U D  &\implies S M^*S^{-1} S U^*S^{-1} = S U^*S^{-1} S D S^{-1}\\
 & \implies  M  S U^*S^{-1} =  S U^*S^{-1} D \\  
\end{aligned}
\end{equation}
in other words, the each column of $ S U^*S^{-1}$ and $U$ are eigenvectors of the same eigenvalue.
Update: Almost there!
Write $U$ as a block form
\begin{equation}
U = 
\begin{bmatrix}
U_1 & U_2 \\
U_3 & U_4 \\    
\end{bmatrix}\qquad
S^{-1} U^* S = 
\begin{bmatrix}
U_4^* & U_3^* \\
U_2^* & U_1^* \\    
\end{bmatrix}
\end{equation}
From the previous analysis of $S^{-1} U^* S$, construct
\begin{equation}
V = \begin{bmatrix}
  U_1 & U_3^*\\
  U_3 & U_1^*\\
  \end{bmatrix} 
\end{equation}
so that we have
\begin{equation}
MV = 
M \begin{bmatrix}
  U_1 & U_3^*\\
  U_3 & U_1^*\\
  \end{bmatrix} 
= \begin{bmatrix}
  U_1 & U_3^*\\
  U_3 & U_1^*\\
  \end{bmatrix} 
\begin{bmatrix}
  d & 0 \\
  0 & -d \\
  \end{bmatrix} = V\begin{bmatrix}
  d & 0 \\
  0 & -d \\
  \end{bmatrix}
\end{equation}
If $0$ is not the eigenvalue, then columns of $\begin{bmatrix} U_1 \\  U_3 \\  \end{bmatrix} $ and the corresponding columns of $\begin{bmatrix} U_3^* \\  U_1^* \\  \end{bmatrix} $ belongs to different eigenspace, and hence are automatically orthogonal to each other(its possible to make diagonal elements of $d$ to be non-negative), the $V \in B$.
I don't know how to deal with the case when the eigenvalues contain $0$.  
 A: This problem can be solved by making all the entries to be real. 
Define the $T$ matrix to be the following unitary matrix
\begin{equation}
T = \frac{1}{\sqrt{2}}
\begin{bmatrix}
I & i I\\
I & -i I\\
\end{bmatrix}
\end{equation}
For matrices transform like $S X^* S^{-1} = \pm X$, it is very convenient to use $T$ matrix to make them real. For example, using the fact that $ST^* =T $, and $T^T S = T^{\dagger} $, we have 
\begin{equation}
(T^{\dagger} iM T)^* = -iT^T M^* T^* = T^T S iM S T^* = T^{\dagger} iM T  \equiv \mathcal{M} 
\end{equation}
and for $U \in B$
\begin{equation}
(T^{\dagger} U T )^* = T^{T} U^* T  = T^T S U S T^* = T^{\dagger} U T  \equiv \mathcal{U}
\end{equation}
The diagonalization for these equivalent real matrices becomes
\begin{equation}
\mathcal{U}^{T} \mathcal{M} \mathcal{U} = 
\begin{bmatrix}
0 & -d\\
d & 0 \\
\end{bmatrix}
\end{equation}
where $\mathcal{M}$ is a skew-symmetric matrix and $\mathcal{U}$ is an element of ${\rm SO}(2n)$. 
We know it is always possible to use ${\rm SO}(2N)$ matrix to "diagonalize" skew-symmetric matrix such that only $2\times 2$ block $\begin{bmatrix} 0 & -\lambda \\ \lambda & 0\end{bmatrix}$ appears on the diagonal. In particular, this can be done by a real Schur decomposition. Then reordering these $\lambda$s gives us the desired form of $\begin{bmatrix}0 & -d\\d & 0 \\\end{bmatrix}. $
Then $U = T \mathcal{U} T^{\dagger}\in B $. Therefore we have successfully constructed a method to find $U \in B$ and identify $B$ to be a group isomorphic to ${\rm SO}(2n)$. 
