Suppose we have the following block matrix, $B$:
$B = \left(\begin{array}{cc} 0 & A^*\\ A & 0 \end{array}\right)$
where $A \in \mathbb{C}^{m \times m}$. It is clear that $B$ is Hermitian and that $A$ does not necessarily have to be Hermitian. I want to show explicitly that the eigenvalues of $B$ are the $\pm \sigma_i$ where $\sigma_i$ are the singular values of $A$. To do so, I consider the SVD of $A$: $A=U\Sigma V^*$. Then, I took the "block" eigenvector:
$v = \left( \begin{array}{c} v_j \\ u_j \end{array} \right)$, where $v_j$ and $u_j$ are the respective columns of $U$ and $V$ from the SVD. I was able to show after substituting $A$ with $U\Sigma V^*$ and $A^*$ with $V \Sigma^* U^*$ and with some matrix algebra that:
$B v = \left( \begin{array}{c} \sigma_j v_j \\ \sigma_j u_j \end{array} \right)=\sigma_j v$.
It mostly came from the fact that $U$ and $V$ are both unitary. However, showing this for $-\sigma_j$ is giving me significant trouble. I cannot seem to define a set of eigenvectors as I had for $\sigma_j$ that shows that $-\sigma_j$ is an eigenvalue. Does anyone have any tips?
Edit: SOLVED. If you consider
$v = \left( \begin{array}{c} -v_j \\ u_j \end{array} \right)$. You can see why the other eigenvalues are $-\sigma_j$. In addition, this follows the property that Hermitian matrices have orthogonal eigenvectors.
Best Answer
Suppose $A=U \Sigma V^*$, then consider $\begin{bmatrix} 0 & V \\ U & 0 \end{bmatrix}^*\begin{bmatrix} 0 & U \Sigma V^* \\ V \Sigma U^* & 0 \end{bmatrix} \begin{bmatrix} 0 & V \\ U & 0 \end{bmatrix} = \begin{bmatrix} 0 & \Sigma \\ \Sigma & 0 \end{bmatrix}$.
Now note that if $s \in \{-1,+1\}$, we have $\begin{bmatrix} 0 & \Sigma \\ \Sigma & 0 \end{bmatrix} \begin{bmatrix} e_k \\ s e_k \end{bmatrix} = s \sigma_k \begin{bmatrix} e_k \\ s e_k \end{bmatrix}$, from which it follows that the eigenvalues are $\{ \pm \sigma_k \}_k$.