1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\{ \pm \sigma_k \}_k$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.