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I have a square matrix $J \in \mathbb{C}^{2n \times 2n}$ where,

$J=\begin{pmatrix} A&B \\\bar{B} & A \end{pmatrix}$

$A \in \mathbb{R}^{n \times n}$ and is ${\bf diagonal}$.

$B \in \mathbb{C}^{n \times n}$ and is ${\bf symmetric}$ so its conjugate $\bar{B}= B^*$.

These properties make $J$ Hermitian and thus with real eigenvalues.

Now I want to separate the $n$ largest eigenvalues of $J$ in a way that can be expressed as eigenvalues of an $n \times n$ matrix constructed based on $A$ and $B$. The ultimate goal is to separate the positive and negative eigenvalues of $J$ when there are exactly $n$ positive and $n$ negative eigenvalues and identify the cases when this symmetry goes away by only inspecting the $n \times n$ matrices (functions of $A$ and $B$).

I already know that if $K=\begin{pmatrix} A&B \\B & A \end{pmatrix}$ (notice that no restriction is required on $A$ or $B$) the e-values of $K$ is the union of the e-values of $A+B$ and $A-B$. Can there be any similar formulation for $J=\begin{pmatrix} A&B \\\bar{B} & A \end{pmatrix}$ or $J=\begin{pmatrix} A&B \\\bar{B} & \bar{A} \end{pmatrix}$

I would appreciate any help with the above problem.

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Are you studying the signature of the matrix $J$ which is defined as the number of positive eigenvalues of J minus the negative eigenvalues of $J$? – Mustafa Said Mar 8 '14 at 21:42
Yes. If $B$ is real symmetric, there is symmetry in e-values with signature being (n,n,0). As B is perturbed in the complex domain, the perfect symmetry in the e-values of $J$ disappears and $J$ may become singular and at least one e-value changes sign. But to make sure that $J$ never becomes singular I want to enforce that the $n \times n$ matrix corresponding to positive e-values always remains positive definite and the other $n \times n$ matrix with the spectrum of the $n$ negative e-values of $J$ remains negative definite. But how can I construct these two matrices based on $A,B,\bar{B}$? – Sam1984 Mar 9 '14 at 19:36
Sam, I think that you did not work really about your question. In particular, $J$ is "almost" a generic hermitian matrix and then, without supplementary hypothesis about $A,B$ (positive, definite ?), why this matrix $J$ would have a signature equal to $(n,n)$ under the hypothesis: $B$ is a real matrix ? It is obviously false: when $n=1$, the eigenvalues are $a\pm|b|$. Moreover, assume that $J$ has signature $(n,n)$ ; if $B$ is SLIGHTLY perturbed in the complex domain, then its signature is invariant. Observe seriously the case when $B$ is real. – loup blanc Mar 10 '14 at 17:19
@Sam, can you show that $\det(J-xI)=\det((A-xI)(A-xI-\overline{B}(A-xI)^{-1}B))$ ? – loup blanc Mar 10 '14 at 17:20
loup blanc, you are right about real $B$. Only in a special case where $A$ is zero, the eigenvalues of $J$ are perfectly symmetrical. $A$ is in fact not only symmetric but also $\bf diagonal$ which I forgot to mention (I edited the problem). Does this more stringent condition help illuminate the problem. Also $det(J−xI)=det((A−xI)(A−xI−\bar{B}(A−xI)^{−1}B))$ holds for the $J$ in my problem but how is this helpful? May be I can provide more information about how A and B are constructed if it is really necessary. – Sam1984 Mar 10 '14 at 19:06

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