Consider a matrix $A \in \mathbb{C}^{m\times n}$. The SVD of $A$ reads: \begin{equation} A = U\Sigma V^H \end{equation} where $U \in \mathbb{C}^{m\times m},V \in\mathbb{C}^{n\times n}, \Sigma \in\mathbb{R}^{m\times n}$. The same matrix $A$ can however be represented using only real values either as: \begin{equation} A_1 = \begin{bmatrix}\text{Re}(A) & -\text{Im}(A) \\ \text{Im}(A) & \text{Re}(A) \end{bmatrix} \in\mathbb{R}^{2m\times 2n}\end{equation} The SVD of matrix $A_1$ will similarly read: \begin{equation} A_1 = U_1 \Sigma_1 V_1^T \end{equation} where $U_1\in\mathbb{R}^{2m\times2m}, \Sigma_1\in\mathbb{R}^{2m\times2n}, V_1\in\mathbb{R}^{2n\times2n}$ have the attractive property of being closely related to $U,V$ as: \begin{equation} U_1 = \begin{bmatrix}\text{Re}(U) & -\text{Im}(U) \\ \text{Im}(U) & \text{Re}(U) \end{bmatrix} \end{equation} \begin{equation} \Sigma_1 = \begin{bmatrix} \Sigma & 0 \\ 0 & \Sigma \end{bmatrix} \end{equation} \begin{equation} V_1 = \begin{bmatrix}\text{Re}(V) & -\text{Im}(V) \\ \text{Im}(V) & \text{Re}(V) \end{bmatrix} \end{equation}
However, matrix $A$ can also be represented using only real values as: \begin{equation} A_2 = \begin{bmatrix}\text{Re}(A) \\ \text{Im}(A) \end{bmatrix}\in\mathbb{R}^{2m\times n} \end{equation} with SVD $A_2 = U_2 \Sigma_2 V_2^T$ with $U_2 \in\mathbb{R}^{2m\times2m}, \Sigma_2 \in\mathbb{R}^{2m\times n}, V_2 \in\mathbb{R}^{n\times n}$. In this case, however, I can not find a similarly nice relation between $U_2, \Sigma_2, V_2$ and $U, \Sigma, V$ (not even the singular values seem to match, even when numerical errors are ruled out). I tried block-partitioning matrix $U_2$ similarly to the case of $U_1$, which yields:
\begin{equation} A_2 = \begin{bmatrix}\text{Re}(A) \\ \text{Im}(A) \end{bmatrix} = \begin{bmatrix}U_{21} \\ U_{22} \end{bmatrix}\Sigma_2V_2^T = \begin{bmatrix} U_{21}\Sigma_2V_2^T\\ U_{22}\Sigma_2V_2^T\end{bmatrix} \end{equation} From the decomposition of matrix $A_1$ I know that the real and imaginary part of $A$ are related to the real and imaginary parts of $U,V$ as: \begin{cases} \text{Re}(A) = \text{Re}(U)\cdot \Sigma\cdot \text{Re}(V)^T + \text{Im}(U)\cdot \Sigma\cdot \text{Im}(V)^T \\ \text{Im}(A) = \text{Im}(U)\cdot \Sigma\cdot \text{Re}(V)^T - \text{Re}(U)\cdot \Sigma\cdot \text{Im}(V)^T \\ \end{cases} which would thus give the system: \begin{cases} U_{21}\Sigma_2V_2^T = \text{Re}(U)\cdot \Sigma\cdot \text{Re}(V)^T + \text{Im}(U)\cdot \Sigma\cdot \text{Im}(V)^T \\ U_{22}\Sigma_2V_2^T= \text{Im}(U)\cdot \Sigma\cdot \text{Re}(V)^T - \text{Re}(U)\cdot \Sigma\cdot \text{Im}(V)^T \\ \end{cases}
Is this the end? Is there inherently no way to go back to the real and imaginary parts of $U,V$ knowing the SVD of $A_2$ (and thus, only the SVD of $A_1$ can be used to compute $U,V$ using only real values)? Or am I missing something? (My goal is to retrieve the SVD of $A$ without working with complex numbers, i.e. only real values can be used in the computing system.)
If you want to retrieve the SVD of $A$ using only real numbers then $A_1$ is the way to go. $A_2$ corresponds to treating the domain as $\mathbb{C}^n$ but the range as $\mathbb{R}^{2m}$ which means you only recover the SVD this way by writing $A_2 = U_2 \Sigma_2 V_2^T$ where $U_2$ is an orthogonal matrix but $V_2$ is a unitary matrix, so $V_2$ still needs to be complex; if you try to compute the real SVD of $A_2$ you are forcing $V_2$ to be real which just gets you a different thing.