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.

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 $S$?
