linear algebra - Finding an eigenvalue decomposition of a $2mtimes 2m$ Hermitian matrix

Let $A$ be an $m\times m$ matrix with entries in $\mathbb{C}$ and with a singular value decomposition $A=U\Sigma V^*$. Find an eigenvalue decomposition of the $2m \times 2m$ Hermitian matrix:

$$\begin{bmatrix} O &A^* \\ A & O\end{bmatrix}.$$

What I did so far is to denote $M = \begin{bmatrix} O &A^* \\ A & O\end{bmatrix}.$ I know that I need to find a diagonal matrix $\Lambda$ with the eigenvalues of $M$ and a matrix $X$ with eigenvectors of $M$ such that $M = X \Lambda X^{-1}$.

So $Mx = \lambda x \implies \begin{bmatrix} O & A^* \\ A & O \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} =\lambda \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \implies A^*x_2 =\lambda x_1$ and $Ax_1 = \lambda x_2$.

From here, I don't really know where to go with how to find the eigenvectors and relating or using the SVD of $A$.

Is this a viable approach to proceed?

Answer

Note that if the SVD of $A$ is given by $A = U \Sigma V^*$, then we have the following system of equations,

$$\begin{align*} AV = U \Sigma \\ A^*U = V\Sigma. \end{align*}$$

We can write the above system in terms of block matrices,
$$\begin{align*} \begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix} \begin{bmatrix} U \\ V \end{bmatrix} = \begin{bmatrix} U\Sigma \\ V \Sigma \end{bmatrix}. \end{align*}$$
It is also easy to verify that

$$\begin{align*} \begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix} \begin{bmatrix} U \\ -V \end{bmatrix} = \begin{bmatrix} -U\Sigma \\ V \Sigma \end{bmatrix}. \end{align*}$$
Putting these together, we have
$$\begin{align*} \begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix} \begin{bmatrix} U & U \\ V & -V \end{bmatrix} = \begin{bmatrix} U \Sigma & -U \Sigma \\ V \Sigma & V \Sigma \end{bmatrix} = \begin{bmatrix} U & -U \\ V & V \end{bmatrix} \begin{bmatrix} \Sigma & 0 \\ 0 & \Sigma \end{bmatrix}. \end{align*}$$
Pushing the negative sign into the diagonal matrix of singular values, we conclude
$$\begin{align*} \begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix} \begin{bmatrix} U & U \\ V & -V \end{bmatrix} = \begin{bmatrix} U & U \\ V & -V \end{bmatrix} \begin{bmatrix} \Sigma & 0 \\ 0 & -\Sigma \end{bmatrix}. \end{align*}$$
Multiplying on the right by $\begin{bmatrix} U & U \\ V & -V \end{bmatrix}^{-1}$ yields the desired eigenvalue decomposition.