Suppose we have a square matrix $A$ with singular value decomposition $A=U\Sigma V'$.
How can we show the equation $$\begin{bmatrix}0 & A^T \\ A & 0 \end{bmatrix}=\frac{1}{\sqrt{2}} \begin{bmatrix}V & V\\ U & -U\end{bmatrix} \begin{bmatrix}\Sigma&0\\0&\Sigma\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}V^T &U^T\\ V^T &-U^T\end{bmatrix} ?$$
Something is wrong with the identity.
Block matrices with blocks of the same size can be multiplied as if you had numbers: $$ \begin{split}\frac{1}{\sqrt{2}} \begin{bmatrix}V & V \\ U & -U\end{bmatrix} \begin{bmatrix}\Sigma&0\\0&\Sigma\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}V^T &U^T\\ V^T &-U^T\end{bmatrix} &=\frac12 \begin{bmatrix}V\Sigma & V\Sigma\\ U\Sigma & -U\Sigma\end{bmatrix} \begin{bmatrix}V^T &U^T\\ V^T &-U^T\end{bmatrix}\\ &= \begin{bmatrix}V\Sigma V^T & 0\\ 0 & U\Sigma U^T\end{bmatrix}. \end{split} $$
You probably want instead: $$ \begin{split}\frac{1}{\sqrt{2}} \begin{bmatrix}V & V \\ U & -U\end{bmatrix} \begin{bmatrix}\Sigma&0\\0&\Sigma\end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix}V^T &U^T\\ \color{red}{-V^T} &\color{red}{U^T}\end{bmatrix} &=\frac12 \begin{bmatrix}V\Sigma & V\Sigma\\ U\Sigma & -U\Sigma\end{bmatrix} \begin{bmatrix}V^T &U^T\\ -V^T &U^T\end{bmatrix}\\ &= \begin{bmatrix}0 & V\Sigma U^T\\ U\Sigma V^T & 0\end{bmatrix}\\ &= \begin{bmatrix}0 & A^T\\ A & 0\end{bmatrix}. \end{split} $$