Finding an eigenvalue decomposition of a $2m\times 2m$ 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? 
 A: 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.
