How to find the eigenvalue of this block matrix. Here is a $4\times 4$ matrix:
\begin{pmatrix}
\xi\delta_{ij} &\Delta_{ij}\\
\Delta_{ij}^\dagger& -\xi\delta_{ij}
\end{pmatrix}
In the above, $\delta_{ij}$ represent a $2\times2$ identity matrix, $\Delta_{ij}^\dagger$ is the hermitian conjugate of the $2\times2$ matrix $\Delta_{ij}$.
Question is: how to get the eigenvalue of the above matrix and the corresponding matrix that diagonalize it.
 A: Let $A=\begin{pmatrix}\xi I_2&\Delta\\\Delta^*&-\xi I_2\end{pmatrix}$. Then $\det(A-\lambda I_4)=\det((\lambda^2-\xi ^2)I_2-\Delta\Delta^*)$. Let $(\sigma_i)_{i\leq 2}$ be the singular values of $\Delta$ and $(\lambda_j)_{j\leq 4}=spectrum(A)$; since $\det(\sigma_i^2 I_2-\Delta\Delta^*)=0$, one obtains $\lambda^2-\xi^2=\sigma^2$, that is, $\lambda_1=\sqrt{\sigma_1^2+\xi^2},\lambda_2=-\lambda_1,\lambda_3=\sqrt{\sigma_2^2+\xi^2},\lambda_4=-\lambda_3$.
Let $\Delta=U\Sigma V^*$ be a SVD of $\Delta$. We assume that $\sigma_1>\sigma_2>0$.
Let $[x,y]^T$ be an eigenvector of $A$ associated to $\lambda=\pm\sqrt{\sigma^2+\xi^2}$. One has $\xi x+\Delta y=\lambda x,\Delta^*x-\xi y=\lambda y$. Then $\Delta y=(\lambda-\xi)x$ and $\Delta^*\Delta y=\sigma^2 y$; thus $y$ is an eigenvector of $\Delta^*\Delta$, that is a column of $V$; in the same way, $x$ is the column of $U$ with the same index as the previous column, that defines $x$ up to signum; the previous signum is given by the relation $x=1/(\lambda-\xi)\Delta y$.
