This is a branched question of my previous question What's the condition for a matrix $A$ ($2N\times 2N$ dimension) to have eigenvalues in pairs $\pm\lambda$?.
Let's look at this $2N\times2N$ matrix $A$:
$A=\left(\begin{array}{cc}
B & C\\
-C^{*} & -B^{*}
\end{array}\right)$,
where $B$ is a hermitian matrix, $C$ is a complex symmetric matrix, and $M^\ast$ denotes the complex conjugate (but not the conjugate transpose) of a matrix $M$.
If $A$ is a $2\times2$ matrix, the eigenvalues are obviously in
pairs, and I also proved the $4\times4$ case by showing the determinant
of $A-\lambda I$ only depends on $\lambda^{2}$.
Then I use Mathematica to calculate the eigenvalues of this
matrix with randomly generated matrices $B$ and $C$ for higher dimension
cases, the eigenvalues also come in pairs. So I guess this matrix
always has eigenvalues in pairs, but I don't know how to prove it
analytically.
Best Answer
Suppose you are given an eigenvector relative to $\lambda$ and write it as $\left[\begin{smallmatrix}v\\ w\end{smallmatrix}\right]$: $$ \lambda\begin{bmatrix} v \\ w \end{bmatrix} = \begin{bmatrix} B & C \\ -C^* & -B^* \end{bmatrix} \begin{bmatrix} v \\ w \end{bmatrix} = \begin{bmatrix} Bv+Cw \\ -C^*v-B^*w \end{bmatrix} $$ This means $$ \begin{cases} Bv+Cw=\lambda v\\[4px] -C^*v-B^*w=\lambda w \end{cases} $$ Since $C$ is symmetric and $B$ is hermitian, we have $$ A^T= \begin{bmatrix} B^T & -(C^*)^T \\ C^T & -(B^*)^T \end{bmatrix} = \begin{bmatrix} B^* & -C^* \\ C & -B \end{bmatrix} $$ and we have $$ A^T\begin{bmatrix} w \\ -v \end{bmatrix} = \begin{bmatrix} B^* & -C^* \\ C & -B \end{bmatrix} \begin{bmatrix} w \\ -v \end{bmatrix} = \begin{bmatrix} B^*w+C^*v \\ Cw+Bv \end{bmatrix} = \begin{bmatrix} -\lambda w\\ \lambda v \end{bmatrix}= -\lambda\begin{bmatrix}w\\-v\end{bmatrix} $$ Therefore $-\lambda$ is an eigenvalue of $A^T$. It is well known that $A$ and $A^T$ have the same eigenvalues.
Important note. In the above, as explained in the comments to the question, $B^*$ denotes the matrix obtained by conjugating each entry in $B$, so the condition for $B$ being hermitian is $B^T=B^*$. Similarly, $(C^*)^T=(C^T)^*=C^*$ since $C$ is symmetric.