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Suppose I have a special block, Hermitian matrix $H = \begin{bmatrix} A & B \\ B^* & A^* \end{bmatrix}$, where * denotes conjugate transpose. The blocks $A$ and $B$ are themself Hermitian in this case. Are there any theorems considering the eigenvalues and eigenvectors for this special matrix?

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    If they are Hermitian, then why do you write $A^*,B^*$, and not just $A,B$?2012-06-25
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    If $H$ is hermitian, it is clear that off-diagonal blocks must be conjugate transposed, but there is an additional property for the diagonal blocks2012-06-25
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    It is not true that the eigenvalues of $A$ are always eigenvalues of $H$. Consider $A=B=1$. Then $A$ has eigenvalue $1$, while $H=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ has eigenvalues $0$ and $2$.2012-06-25
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    @Martin Based on your above comment and what you typed, you should see now that $B$ need not be Hermitian. For example $H=\begin{bmatrix}1&i\\-i&1\end{bmatrix}$ with $B=[i]$. If you intended the blocks themselves to be Hermitian, please make that clearer in the statement, and possibly eliminate the asterisks.2012-06-25
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    Generally B need not to be Hermitian, but in case of my matrix it is. So does A. If I would write the matrix as $H = \begin{bmatrix} A & B \\ B & A \end{bmatrix}$ then H might not be Hermitian. Sorry for the confusion.2012-06-25
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    @Martin: Hi, and welcome to math.SE! Users with any number of "reputation points" [can comment on their own questions and answers](http://meta.stackexchange.com/questions/19756) (once you obtain 50 points, you gain the ability to comment anywhere), but you were not able to comment because you were not signed into the account that asked the question. I've now merged your duplicate account into the original.2012-06-25

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