The eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$ are the singular values of $A$ along with the negative signs. 
The eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$ are the singular values of $A$ along with the negative signs. 

Here $A$ is an $n \times n$ matrix, has $n$ singular values. Here we are consider both $\lambda $ and $- \lambda$ if $\lambda$ is an singular value of $A$. Thus we have a set of $2n$ elements and we have to show that these are the eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$.
Need some hint to proceed with the problem.
 A: A nice summary of the SVD:
$$Av_i=\sigma_i u_i \\
A^* u_i=\sigma_i v_i.$$ 
So build a single vector $x_i$ so that when you multiply out $Bx_i$, $A$ hits $v_i$ and $A^*$ hits $u_i$. Since the structure of $B$ already makes the two trade places, this will wind up giving you an eigenvector with eigenvalue $\sigma_i$. Now how do you get one with eigenvalue $-\sigma_i$?
A: Let $A$ be a $(n \times n)$-matrix and let $B$ be the matrix $$B = \begin{pmatrix} 0 & A \\ A^* & 0
\end{pmatrix}.$$
Let $A = U_1 \Sigma U_2^*$ be a singular value decomposition of $A$. Then we can decompose $B$ into a product $B = USU^*$ where $U$ is unitary and $S$ has a simpler structure than $B$. Let us show this explicitly:
We have
$$
B = \begin{pmatrix} 0 & U_1 \\ U_2 & 0 \end{pmatrix} \begin{pmatrix} 0 & \Sigma^* \\ \Sigma & 0 \end{pmatrix} \begin{pmatrix} 0 & U_2^* \\ U_1^* & 0 \end{pmatrix},
$$
so we can choose
$$ S = \begin{pmatrix} 0 & \Sigma^* \\ \Sigma & 0 \end{pmatrix}, \quad U = \begin{pmatrix} 0 & U_1 \\ U_2 & 0 \end{pmatrix}.$$
This means that the eigenvalues of $B$ are exactly the eigenvalues of $S$.
Let $\epsilon_i$ be the $i$-th unit vector of dimension $2n$ and $\sigma_1,\dots,\sigma_{n}$ be the singular values of $A$ resp. the entries of the diagonal matrix $\Sigma$. Now try to consider the vectors $\epsilon_k+\epsilon_{n+k}$ and $\epsilon_k - \epsilon_{n+k}$ for $k = 1,\dots,n$. What happens if you multiply them with our matrix $S$?
A: Suppose $A = U\Sigma V^*$ and the block matrix has eigen decomposition,
$$ 
\begin{bmatrix}0&A\\A^*&0\end{bmatrix} 
\begin{bmatrix}X_1 & X_3\\X_2 & X_4\end{bmatrix} 
= 
\begin{bmatrix}X_1 & X_3\\X_2 & X_4\end{bmatrix} 
\begin{bmatrix}\Lambda_1 & 0 \\ 0 & \Lambda_2 \end{bmatrix} 
$$
Now how can you relate these questions to the SVD of $A$
A: Suppose we have an eigenvector $(x, y)$ of the given matrix where $x, y \in \mathbb{C}^n$.  Then:
$$ \begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} A^* y \\ A x \end{bmatrix} = \lambda \begin{bmatrix} x \\ y \end{bmatrix}. $$
It follows that $Ax = \lambda y$ and $A^* y = \lambda x$, so $A^* A x = A^* (\lambda y) = \lambda^2 x$.  Therefore, $\lambda^2$ is an eigenvalue of $A^* A$: if $\lambda \ne 0$, then $x = 0$ would imply $y = 0$ also, giving a contradiction; whereas if $\lambda = 0$ and $x = 0$, then $y \ne 0$ and $A^* y = 0$, so $A^*$ is singular, implying $A$ is also singular and therefore has 0 as an eigenvalue.
Conversely, if $\lambda^2$ is an eigenvalue of $A^* A$, then choose $x \ne 0$ such that $A^* A x = \lambda^2 x$.  Then in the case $\lambda \ne 0$, $(x, \lambda^{-1} A x)$ is an eigenvector of $\begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix}$ with eigenvalue $\lambda$.  In the case $\lambda = 0$, $A^* A x = 0$ implies $A x = 0$, and so $(0, x)$ is in the null space of $\begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix}$.  (In fact, in the case $\lambda = 0$, we also have that $A^*$ has a nonzero vector $y$ in its null space, and then $(y, 0)$ is in the null space of the block matrix; so 0 is an eigenvalue of the block matrix with multiplicity at least two - reflecting the fact that 0 is a "double" square root of 0 in the problem statement.)
In conclusion, $\lambda$ is an eigenvalue of $\begin{bmatrix} 0 & A \\ A^* & 0 \end{bmatrix}$ if and only if $\lambda^2$ is an eigenvalue of $A^* A$, which is in turn equivalent to $\lambda$ being plus or minus a singular value of $A$.
