Suppose A $ \in \mathbb C^{m\times m}$ has an $SVD: A = U\sum V^*$. Find an eigenvalue
decomposition form of the $2m \times 2m$ hermitian matrix
$$
B=\begin{bmatrix}
0&A^*
\\
A&0
\end{bmatrix}
$$
I cannot get the eigenvalue decomposition form of $ B=X\sum X^*$. How to do that?
Best Answer
We have $$ B=\begin{pmatrix}0&V\\U&0\end{pmatrix}\begin{pmatrix}0&\Sigma\\\Sigma^*&0\end{pmatrix}\begin{pmatrix}0&U^*\\V^*&0\end{pmatrix}. $$ and $$ W\begin{pmatrix}0&\Sigma\\\Sigma^*&0\end{pmatrix}W^*=2\begin{pmatrix}(\Sigma\Sigma^*)^{1/2}&0\\0&-(\Sigma^*\Sigma)^{1/2}\end{pmatrix}, $$ where $$ W=\begin{pmatrix}(\Sigma^*)^{1/2}&(\Sigma^*)^{-1/2}\\ \Sigma^{1/2}&-\Sigma^{-1/2}\end{pmatrix}. $$ It remains to compute $W^{-1}$.