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Let $\rho_A$ be a mixed state (density operator). If we wish, we can purify it and get a $\rho_{AB}$ that is pure on some composite Hilbert space $H_{AB}$. And by the question I just ask, How to prove that tensor products of two pure density operators is again pure, we know that tensor product of pure states is again pure. Observing the results above, I just come up with a natural question: how about tensor product of all but exactly one mixed states? Is such composite state necessarily mixed? (I'm not a English speaker, which might have used a weird grammar; to be clear, I mean $\otimes_{i\in I} \rho_i$ is mixed whenever $\exists !~i\in I,~\rho_i$ is mixed) If we cannot make any conclusion, how about composite of all but exactly one pure state?

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A state is pure iff it is rank one and $$ \mathrm{rank}\left(\bigotimes_i \rho_i \right) = \prod_i \mathrm{rank}(\rho_i). $$ This can be proven by considering the spectral decompositions of the $\rho_i$. So in answer to your question, your state is pure iff all of the $\rho_i$ are pure.

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The tensor product of a non-pure state with any other state gives a non-pure state.

One way to see it is noticing that the von Neumann entropy is additive with respect to tensor products: $S(\rho\otimes\sigma)=S(\rho)+S(\sigma)$ for any $\rho,\sigma$. Therefore if there is any $\rho_i$ that is not pure, then $S(\rho_i)>0$ and thus $S(\bigotimes_j\rho_j)>0$.

You can also just observe that a state being pure means that it is a rank-one projector, i.e. a projector with one-dimensional support. If any factor in a tensor product is not pure, then its support is at least two-dimensional, and thus so must be the support of the tensor product.

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