Suppose you have 4 matrices with singular value decompositions
$A = U\_1 \Sigma\_A V\_1^{\dagger}$, $B = U\_2 \Sigma\_B V\_2^{\dagger}$, $C = U\_1 \Sigma\_C V\_1^{\dagger}$ and $D = U\_2 \Sigma\_D V\_2^{\dagger}$ such that $\Sigma\_A \Sigma\_C$ and $\Sigma\_B \Sigma\_D$ are both nonzero.


Are the singular value decompositions of $A+B$ and $C+D$ of the form
$A+B = U\_3 \Sigma\_{A+B} V\_3^{\dagger}$ and $C+D = U\_3 \Sigma\_{C+D} V\_3^{\dagger}$?



does the basis in the singular value decomposition of a sum depend on the singular values of the summands
no, just take as a counterexample: $V\_1=U\_1$, $V\_2=U\_2$, $\Sigma\_A=\mathbb{1}$, $\Sigma\_D=\mathbb{1}$, so $A+B=U\_2(\Sigma\_B+\mathbb{1})U\_2^{\dagger}$, $C+D=U\_1(\Sigma\_C+\mathbb{1})U\_1^{\dagger}$, so your assumption fails unless $U\_1=U\_2$.


