Given a matrix $F$ and its SVD: $F=UDV^T$, then is it possible to calculate the SVD of the block matrix A using U,D,$V^T$ of F:$$A=\begin{bmatrix} F & F &F &F \\ F & F&F&F \end{bmatrix}$$
I tried to calculate some matrix and noticed SVD of A is also a block matrix. What's the connection between the SVD of A and the SVD of F.
If $F = UDV^T$ then, we have
\begin{align*} A &= \begin{bmatrix}F & F & F & F \\ F & F & F & F \end{bmatrix} \\ &= \begin{bmatrix}UDV^T & UDV^T & UDV^T & UDV^T \\ UDV^T & UDV^T & UDV^T & UDV^T \end{bmatrix} \\ &= \begin{bmatrix}U \\ U\end{bmatrix}D\begin{bmatrix}V^T & V^T & V^T & V^T\end{bmatrix} \\ &= \dfrac{1}{\sqrt{2}}\begin{bmatrix}U \\ U\end{bmatrix} \cdot 2\sqrt{2}D \cdot \dfrac{1}{2}\begin{bmatrix}V^T & V^T & V^T & V^T\end{bmatrix} \end{align*}
You can easily check that $\dfrac{1}{\sqrt{2}}\begin{bmatrix}U \\ U\end{bmatrix}$ and $\dfrac{1}{2}\begin{bmatrix}V^T & V^T & V^T & V^T\end{bmatrix}$ are orthonormal and $2\sqrt{2}D$ is diagonal. So that expression is the SVD of $A$.