Compute the singular value of matrix I would like to prove that for a matrix $A$ with dimension $p \times q$, and dim$(A)=q$, define a p+q by p+q symmetric indefinite matrix B with zero diagonal blocks and with A and $A^T$ in the off-diagonal block positions,  show that B has 2q nonzero eigenvalues which are plus and minus the singular values of A, and p-q eigenvalues which are zero. 
 A: From what I can gather $B$ is symmetric,  so note it can be orthogonally diagonalized. 
First, $B^2 \in \mathbb{R}^{p+q}$ is a matrix that has zero off diagonal blocks and diagonal blocks given by $AA^T$ and $A^TA$. If we orthogonally diagonalize the blocks $AA^T$ and $A^TA$ then we get
$$B^2 = PDP^T$$ 
where $D$ is a matrix whose first diagonal block is $p \times p$ and has rank $q$ and second diagonal block is $q \times q$ and has full rank $q$ $(**)$. 
both blocks have $q$ diagonal entries which are the squares of the singular values of $A$, $\sigma_i$. Now by noting that $B$ itself can be orthogonally diagonalized 
$$B = E\tilde{D}E^T \qquad (*)$$
where $\tilde{D}$ are the eigenvalues ($\lambda_i$) of $B$. Now squaring 
$$B^2 = E\tilde{D}^2E^T$$
so therefore we have
$$\lambda_i^2=\sigma_i^2 \quad, i\in[1,2q]$$
BUT $A$ has only $q$ distinct eigenvalues and $B$ has rank $2q$ hence we can conclude that 
$$\lambda_i= \sigma_i, \quad i\in[1,q]$$
and
$$\lambda_i= -\sigma_i, \quad i\in[q+1,2q]$$
the $p-q$ eigenvalues being zero follows from $(**)$
"Alternative Argument"
using $$B^2 = PDP^T$$ then $$B^2P = PD$$ and nothing that the columns of $P$ must be eigenvectors of $B^2$ we have that 
$$\lambda_i^2=\sigma_i^2 \quad, i\in[1,2q]$$
immediately. using the same logic as the latter part of the first answer, we obtain the same result.
