Orthogonal decomposition of block matrix in terms of SVD Let $A\in \mathbb{R}^{m\times n}$, $m\ge n$ and have an SVD $U \Sigma V^\top$ with $\Sigma\in \mathbb{R}^{n\times n}$. Now how do I find orthogonal $Q\in\mathbb{R}^{m+n\times m+n}$ from this decomposition such that
$Q^\top \begin{bmatrix}0 & A^\top\\A & 0 \end{bmatrix}Q= \begin{bmatrix}\Sigma & 0 & 0\\0 & -\Sigma & 0 \\ 0 & 0 & 0 \end{bmatrix}$?
I am not comfortable with block matrices and am lost here.

You have to be careful with dimensions, but matrix multiplication with block matrices looks very much like the usual matrix multiplication.
Let $ Q = \left[ \begin{array}{cc}B&C\\D & E \end{array} \right] $ where $B$ is $n \times n$, $C$ is $n\times m$, $D$ is $m\times n$ and E is $m \times m$.
$$ \begin{align*} Q^\top \left[ \begin{array}{cc}0&A^\top\\A & 0 \end{array} \right] Q &= \left[ \begin{array}{cc}B^\top&D^\top\\C^\top & E^\top \end{array} \right] \left[ \begin{array}{cc}0&A^\top\\A & 0 \end{array} \right]  \left[ \begin{array}{cc}B&C\\D & E \end{array} \right] \\
&=\left[ \begin{array}{cc}B^\top&D^\top\\C^\top & E^\top \end{array} \right] \left[ \begin{array}{cc}A^\top D &A^\top E \\AB & AC \end{array} \right] \\
&= \left[ \begin{array}{cc}B^\top A^\top D + D^\top AB & B^\top A^\top E + D^\top A C \\ C^\top A^\top D+ E^\top AB & C^\top A^\top E + E^\top A C \end{array} \right]
\end{align*} $$ 
And we'd like this to be equal to: $\left[\begin{array}{cc} \Sigma & 0 & 0\\ 0 & -\Sigma & 0 \\ 0 & 0 & 0 \end{array}\right]$.
Applying the SVD to $A$, we basically want: $$
\begin{align*} B^\top V \Sigma U^\top D + D^\top U \Sigma V^\top B &= \Sigma \\
C^\top V \Sigma U^\top E+ E^\top U \Sigma V^\top C &= -\Sigma\\
 C^\top  V \Sigma U^\top D+ E^\top U \Sigma V^\top B &= 0 \\ B^\top  V \Sigma U^\top E + D^\top U \Sigma V^\top C &= 0
 \end{align*}$$
This looks like a disaster, but it's actually simple! Remember that: $U^\top U = I$ and $V^\top V = I$. This suggests something like:
$$\begin{align*} B = \frac{1}{\sqrt{2}} V \quad
                 D = \frac{1}{\sqrt{2}} U \quad
                 C = -B \quad
                 E = D \end{align*}$$
 You need to add some extra zero columns to $E$ and $C$ so that $Q$ has the proper dimension and $Q$ is square.
