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There is a $2N\times2N$ matrix $A$, where $N$ is a positive integer, which is of the form:

$A=\left(\begin{array}{cc} B & C\\ -C^{*} & -B^{*} \end{array}\right),$

where $B$ is a hermitian matrix, and $C$ is a symmetric matrix.

What can we say about the eigenvalues of $A$ and it's diagonalization?

For example, if $N=1$:

$A=\left(\begin{array}{cc} b & c\\ -c^{*} & -b^{*} \end{array}\right),$

where $b$ must be a real number, and $c$ has no restriction .

The eigenvalues are the same if and only if $\mid c\mid=\mid b\mid$, and the eigenvalues are equal to $0$. But the matrix $A$ is not diagonalizable under this condition:

$A=b\left(\begin{array}{cc} 1 & e^{i\theta}\\ -e^{-i\theta} & -1 \end{array}\right)=b\left(\begin{array}{cc} -e^{i\theta} & -e^{i\theta}\\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 1\\ -e^{-i\theta} & -1 \end{array}\right).$

Is this an accident only for $N=1$ case? Is there some relation like this for general cases $N>1$?

wjmolina
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phchen
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  • I find the eigenvalues of this type of matrix are always in pairs $\pm\lambda$, but don't know how to prove it, then I ask this question – phchen Apr 01 '16 at 07:39
  • The previous property is proved in the question. And later on, I find the eigenvalues of this type of matrix are also in pairs of $\lambda$ and $\lambda^{*}$, which is able to be proved in the similar way. – phchen Apr 30 '16 at 03:35

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