# How to prove the eigenvalues of this matrix $A$ come in pairs $\pm\lambda$?

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.

• Is $B^*$ the conjugate transpose of $B$, i.e. $B^*=\overline{B}^T$? – marco trevi Apr 1 '16 at 7:41
• $B^{*}$ is the complex conjugate of $B$, and I use $B^{\dagger}$ as the conjugate transpose of $B$ – phchen Apr 1 '16 at 7:45

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.
• Oh, my bad. The OP defines $B^\ast$ as $\bar{B}$. Sorry for my previous comment. – user1551 Apr 1 '16 at 10:36