Eigenvalues of special block matrix

Question

Suppose we have a $2n\times 2n$ matrix: $$M=\begin{bmatrix}A&B\\B&-A\end{bmatrix},$$ where $A$ and $B$ are two $n\times n$ self-adjoint matrices: $$A^* =A \;,\quad B^* =B$$ We know that the eigenvalues (and eigenvectors) of $M$ exists in pairs: $$\begin{bmatrix}A&B\\B&-A\end{bmatrix} \begin{bmatrix}c_1\\c_2 \end{bmatrix}=\lambda\begin{bmatrix}c_1\\c_2 \end{bmatrix}. $$ Multipying by $\begin{bmatrix} 0 & -1\\1&0 \end{bmatrix}$ from the left, and inserting $\begin{bmatrix} 0 & -1\\1&0 \end{bmatrix} ^{-1}\begin{bmatrix} 0 & -1\\1&0 \end{bmatrix}$ leads to: $$\begin{bmatrix}A&B\\B&-A\end{bmatrix} \begin{bmatrix}-c_2\\c_1 \end{bmatrix}=-\lambda\begin{bmatrix}-c_2\\c_1 \end{bmatrix}. $$ Is there a way to get the eigenvalues of $M$ ?

Answer 1

Let $J:=\begin{bmatrix}0 & -I \\ I & 0\end{bmatrix}$. Note that $J^{-1}=J^T=-J$ and $J^2=I$. It is easy to see that $$J^{-1}MJ=-M$$ This means $M$ and $-M$ are similar. So if $\lambda$ is an eigenvalue of $M$, it must also be an eigenvalue of $-M$. Also, $-\lambda$ must be an eigenvalue of $-M$ and vice versa. This means the eigenvalues of $M$ are symmetric across imaginary axis. Also, since $M$ is Hermitian, its eigenvalues are real.