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 2x2 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 $M_1=\left(\begin{array}{} A & B\\ B &A \end{array}\right)$, 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: $M_2=\left(\begin{array}{} A & B\\ B^{H} &C \end{array}\right)$ 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 $B$.

Best regards.

  • $\begingroup$ Is there a simplification if we assume $A=C$? $\endgroup$ – Carlos Nov 5 '14 at 14:41

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 $A\geq 0,C\geq 0,AA^+B=B,\sigma_{max}(\sqrt{A^+}B\sqrt{C^+})\leq 1$.


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