diagonalize block matrices with special type of unitary matrices

Question

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:

1. 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.

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

3. 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 $0$.

Answer 1

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 ${\rm SO}(2n)$.