Let $H_A, H_B, H_C$ be arbitrary Hilbert spaces. Let $ρ_{AB} ∈ D(H_A \otimes H_B)$ such that $ρ_{A}$ is pure. Prove that $ρ_{AB} = ρ_{A} \otimes ρ_{B}$ ( Hint: One way could be to prove it before for when $ρ_{AB}$ is pure and then use a purification to reduce to this case)
I haven't been able to go very far with this. If $\rho_A$ is pure $\exists$ unit vector $|\psi_A\rangle$ s.t. $ \rho_A =|\psi_A\rangle \langle|\psi_A|$. But I don't know what to do next. How should I go about it?
Since $\rho_A$ is pure, there exists a unit vector $|\psi_A\rangle \in H_A$ such that $\rho_A = |\psi_A\rangle\langle\psi_A|$.
Now consider a purification of $\rho_{AB}$, which is a vector $|\psi_{AB C}\rangle \in H_A \otimes H_B \otimes H_C$ such that:
$\text{Tr}_C(|\psi_{AB C}\rangle\langle\psi_{AB C}|) = \rho_{AB}$
By the properties of purification, $|\psi_{AB C}\rangle$ has the form:
$|\psi_{AB C}\rangle = |\psi_A\rangle \otimes |\phi_{B C}\rangle$
Where $|\phi_{B C}\rangle \in H_B \otimes H_C$.
Now we can calculate:
$\rho_{AB} = \text{Tr}_C(|\psi_{AB C}\rangle\langle\psi_{AB C}|) = (|\psi_A\rangle \otimes |\phi_{BC}\rangle)(\langle\psi_A|\otimes\langle\phi_{BC}|) = |\psi_A\rangle\langle\psi_A| \otimes \text{Tr}_C(|\phi_{BC}\rangle\langle\phi_{BC}|) = \rho_A \otimes \rho_B$
Then we conclude $\rho_{AB} = \rho_A \otimes \rho_B$.