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 $0$.