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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 $C+D = U_3 \Sigma_{C+D} V_3^{\dagger}$?

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  • $\begingroup$ no, just take as a counterexample $\Sigma_B=0=\Sigma_C$. Then $A+B=A$ and $C+D=D$, so your assumption fails unless $U_1=U_2$ and $V_1=V_2$. $\endgroup$
    – Carlo Beenakker
    Commented Aug 29, 2013 at 18:50
  • $\begingroup$ Thanks, I guess that is the obvious counterexample. I've edited the question to reflect some of the extra structure in my problem that avoids this. $\endgroup$
    – Joel Wallman
    Commented Aug 29, 2013 at 18:53

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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 $U_1=U_2$.

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