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In Nielsen and Chuang, the chapter about Schmidt decomposition, there is an interesting result states that for a bipartite pure state $|\psi\rangle_{AB}$, if part A is a pure state, then $|\psi\rangle_{AB}$ is a product state. But what if the bipartite state is not a pure state, instead, it's a mixed state, do we still have a similar result, i.e., if part A is pure, then $\rho_{AB}$ is a product state?

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Yes. Any mixed state $\rho$ is a convex combination of pure states, that is $$ \rho = \sum_i \lambda_i |\phi_i\rangle\langle\phi_i| $$ where $\lambda_i >0$, $\sum_i\lambda_i=1$. The partial trace is linear, so that $$ \rho_A = \sum_i \lambda_i \text{Tr}_B(|\phi_i\rangle\langle\phi_i|). $$

Every $\lambda_i\text{Tr}_B(|\phi_i\rangle\langle\phi_i|)$ is a positive semidefinite operator. The kernel of a sum of such operators is the intersection of kernels, that is $$ \text{Ker}(\rho_A) = \bigcap_i \text{Ker}(\lambda_i \text{Tr}_B(|\phi_i\rangle\langle\phi_i|)). $$

Consequently, the image of a sum is the sum of images $$ \text{Im}(\rho_A) = \sum_i \text{Im}(\lambda_i \text{Tr}_B(|\phi_i\rangle\langle\phi_i|)). $$

If $\rho_A$ is pure, i.e. $\rho_A = |\psi\rangle\langle\psi|$, then it's image is the subspace of dimension $1$ (it's the vectors collinear with $|\psi\rangle$). Hence every $\text{Im}(\lambda_i \text{Tr}_B(|\phi_i\rangle\langle\phi_i|))$ coincides with the $\text{Im}(|\psi\rangle\langle\psi|)$. But this can happen only if $\text{Tr}_B(|\phi_i\rangle\langle\phi_i|) = |\psi\rangle\langle\psi|$ for every $i$. For the pure case we know that this implies that $|\phi_i\rangle = |\psi\rangle|\psi_i\rangle$. Hence $$ \rho = |\psi\rangle\langle\psi| \otimes \sum_i \lambda_i |\psi_i\rangle\langle\psi_i| $$

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