# Eigenvalues of special block matrix

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 $$M$$ ?

• Have you tried computing the characteristic polynomial? Commented Jan 17, 2020 at 11:54
• I think $M$ is a hermitian matrix so all the eigenvalues will be real Commented Jan 17, 2020 at 13:37

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.