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.
 A: 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.
A: I would rather leave a comment but I'm only able to answer, sorry.
If I take a closer look at the dimensions of the matrices involved it seems to me that your question might be ill-posed.
Since $A\in \mathbb{R}^{m\times n}$ the block matrix in the middle of the lhs is of dimension $(m+n)\times(m+n)$. Thus $Q$ as an orthogonal matrix must be of the same dimension which implies this is the dimension of the matrix product on the lhs.
This should be equal to a quadratic matrix of dimension $(2n+j)\times(2n+j)$ since on the rhs $\Sigma$ appears twice as a block matrix plus a block matrix of zeros of unspecified size $j$.
You may easily find examples like $m=1$ and $n=2$ for which this cannot work.
