Eigenvalues of a general block Hermitian matrix I have a question regarding the eigenvalues of a block Hermitian matrix as a function of the eigenvalues of the diagonal block matrices and the off-diagonal matrices. In particular, I am interested in the $2 \times 2$ block case.
I have checked some previous posts [1]: Eigenvalues of certain block Hermitian matrix and in Wikipedia, and it is clear to me that the solution for the case $$M_1 = \begin{pmatrix} A & B\\ B &A \end{pmatrix}$$ where $M_1$, $A$ and $B$ are Hermitian, can be derived. Nevertheless, I would like to know if it is possible, in the following case $$M_2 = \begin{pmatrix} A & B\\ B^{H} &C \end{pmatrix}$$ where $M_2$, $A$ and $C$ are Hermitian and $B$ corresponds to the off-diagonal block, to say something about the eigenvalues of $M_2$ as a function of the eigenvalues of $A$ and $C$ and the matrix $B$.
 A: If $x$ is not an eigenvalue of $C$, then $\det(M_2-xI)=\det(C-xI)\det(A-xI-B(C-xI)^{-1}B^*)$.
EDIT: moreover $M_2\geq 0$ iff $A\geq 0,C\geq 0,AA^+B=B,\sigma_{max}(\sqrt{A^+}B\sqrt{C^+})\leq 1$.
A: Let us consider a general $2\times2$ block Matrix
\begin{equation}
M = \begin{pmatrix}
    A  & B\\
    C & D 
    \end{pmatrix}
\end{equation}
Assuming $C$ and $D-xI$ to be invertable , we can express the determinant of the Eigenvalue Problem
\begin{align}
\det&(M-xI)\\
     &=\det{\begin{pmatrix}
    A-xI & B\\
    C & D-xI
    \end{pmatrix}}\\
    &= \det(D-xI)\det(A-xI-B(D-xI)^{-1}C)\\
    &= \det(C)\det((A-xI)C^{-1}(D-xI)-B)\\
    &= \det(C)\det(AC^{-1} D-B-x(AC^{-1}+C^{-1} D)+x^2C^{-1})\\
    &= \det(CAC^{-1} D-CB -x(CAC^{-1} +D)+x^2I)\\
    &:=\det(E+xF+x^2I) = 0 
\end{align}
with $E= CAC^{-1} D-CB  $ and $F = -CAC^{-1} -D $
This is known as a Quadratic Eigenvalue Problem.
In this Paper (The Quadratic Eigenvalue Problem, Tisseur and Meerbergen) on page 251 and 252 they describe how to solve it.
There is a closed form solution if $F$ and $E$ commute. Then the solution is given by the eigenvalues of the matrices
\begin{equation}
 S= \frac{-F\pm\sqrt{F^2-4E}}{2}
\end{equation}
Note that in this case $E$ and $F$ are simultaniously diagonizable. And that the Eigenvalues of M are therefore a function of Eigenvalues of $E$ and $F$. Also Note that this is more general than if we assume A and C to commute.
I don't see how we can use C = B^H to simplify the result. Maybe we can use that all eigenvalues are real and therefore $\det(E+xF+x^2I) = 0 $ and $\det(E^H+xF^H+x^2I) = 0 $ have the same solutions.
I hope this helps
