# For bipartite mixed state, if one part is pure, then the global mixed state is a product state?

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?

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 $$\rho = |\psi\rangle\langle\psi| \otimes \sum_i \lambda_i |\psi_i\rangle\langle\psi_i|$$