In Quantum Processes Systems, and Information by Schumacher and Westmoreland we are given this property of the partial trace
$$ Tr_R G^{RQ} = Tr_R \left( |\alpha^R,\phi^Q \rangle\langle\beta^R,\psi^Q| \right) =\\ =Tr_R \left(|\alpha^R \rangle \langle \beta^R|\otimes|\phi^Q\rangle\langle \psi^Q| \right)=\\ =\langle\beta^R|\alpha^R\rangle |\phi^Q\rangle\langle\psi^Q|. $$
Can the partial trace still be found if $G^{RQ}$ can't be expressed as a tensor product of two operators? If we can instead express it as a sum of tensor products and do this:
$$ Tr_R G^{RQ} = Tr_R \left( |\alpha^R,\phi^Q \rangle\langle\beta^R,\psi^Q| + |\kappa^R,\zeta^Q \rangle\langle\lambda^R,\xi^Q|\right) =\\ =Tr_R \left(|\alpha^R \rangle \langle \beta^R|\otimes|\phi^Q\rangle\langle \psi^Q| + |\kappa^R \rangle \langle \lambda^R|\otimes|\zeta^Q\rangle\langle \xi^Q|\right)=\\ =\langle\beta^R|\alpha^R\rangle |\phi^Q\rangle\langle\psi^Q| + \langle\lambda^R|\kappa^R\rangle |\zeta^Q\rangle\langle\xi^Q| ? $$
If it has this linearity property, how to prove it? What is the physical interpretation of there not being a tensor product that gives $G^{RQ}$. And if we can find a sum of tensor products then what's so special about them that makes this linearity true?
I didn't want to post it on math exchange because I'd probably not understand an answer written by a mathematician and I am also interested in the physical interpretation of what is happening here.
