Singular values of $A$ and eigenvalues of $\left[\begin{smallmatrix} 0 & A \\ A^T & 0 \end{smallmatrix}\right]$. 
(Roger p.418) Let $A$ be $m \times n$ matrix, $q=\min(m,n)$, and
   $B=\begin{bmatrix}0&A\\A^*&0\end{bmatrix}$. Let $\sigma_1
 ,\ldots,\sigma_q \ge 0$. The singular values of $A$ are $\sigma_1
 ,\ldots,\sigma_q$ if and only if $m+n$ eigenvalues of $B$ are
   $\sigma_1 ,\ldots,\sigma_q, -\sigma_1 ,\ldots,-\sigma_q,$ and $|m-n|$
  additional 0's.

I found the above theorem in the book. But instead if $B=\begin{bmatrix}0&A\\A^T&0\end{bmatrix}$, is there some relations between singular values of $A$ and eigenvalues of $B$? I don't know what to do, or maybe the problem is not correct. 
 A: Following a very good comment by Branimir, we can construct an example:
$$A = \begin{bmatrix} 8+10{\rm i} & 5 \\ 10-8{\rm i} & -5{\rm i} \end{bmatrix}.$$
Constructing $B$ as in the OP's question, we get


*

*Singular values of $A$: $0$, $3\sqrt{42}$,

*Eigenvalues of $A$: $0$, $8+5{\rm i}$, but

*Jordan normal form of $B$:
$$\begin{bmatrix} 0 \\ & 0 & 1 \\ && 0 & 1 \\ &&& 0 \end{bmatrix}$$
So, $B$ is nondiagonalizable nilpotent matrix, even though $A$ is diagonalizable with some singular values and some eigenvalues being non-zero.
I think that this example suggests that no simple, yet useful relation exists (I don't count "all singular values of $A$ are zero, hence the eigenvalues of $B$ are zero" as "useful").
Mathematica code to verify the above:
BlockMatrix[arr_List] := Flatten[arr, {{1, 3}, {2, 4}}];
Print["A = ", (A = 5 {{8/5 + 2 I, 1}, {2 - (8 I)/5, -I}}) // MatrixForm]
Z1 = ConstantArray[0, {Dimensions[A][[1]], Dimensions[A][[1]]}];
Z2 = ConstantArray[0, {Dimensions[A][[2]], Dimensions[A][[2]]}];
Print["B = ", (B = BlockMatrix[{{Z1, A}, {Transpose[A], Z2}}]) // MatrixForm]
Print["Nonzero singular values of A: ", SingularValueList[A]]
Print["Jordan decomposition of A:\n", 
 Map[MatrixForm, JordanDecomposition[A]]]
Print["Jordan decomposition of B:\n", 
 Map[MatrixForm, JordanDecomposition[B]]]

with the output:

