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Let our Hilbert space be $H = (A \otimes B) \oplus (A \otimes B)^{\perp}$.

If $\rho \in A \otimes B$, then we have $\text{tr}_B \rho \in A$. Is the converse true: if $\text{tr}_B \rho \in A$, then $\rho \in A \otimes B$? Or, is it possible that $\rho$ can be in a subspace of $H$ other than $A \otimes B$?

I know this is a very basic question and my intuition says that $\rho \in A \otimes B$ and cannot be in any other subspace, but I am not sure how to mathematically prove it.

Edit: as my set-up is unclear, I specified an instance of what I'm thinking about in the comment.

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    $\begingroup$ As $H$ is not exclusively of tensor product form, how do you define ${\rm tr}_B$ on it? Like if you have a "generic" product element $X:=(a\otimes b,c\otimes d)\in H$ what is ${\rm tr}_B$ supposed to be? $\endgroup$ Commented Apr 11 at 6:41
  • $\begingroup$ @FrederikvomEnde Consider a 2 qubit system in the computational basis. I can define my $A$ and $B$ to be first and second qubit systems, respectively. So, in this case $H=A \otimes B$ so that $(A \otimes B)^{\perp}$ is a $0$ subspace. Then, for any given state $\rho$, I can express it in terms of the computational basis and trace out the second qubit (which is $B$). We can generalize this to higher dimensions. So, you're definitely right, and it might be better to restrict the set-up to the Hilbert space in the computational basis with system that we're tracing out should be a set of qubits. $\endgroup$
    – karavan
    Commented Apr 11 at 17:10

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It's true in an appropriate formulation.

First of all, don't confuse pure states $|\psi\rangle$ from $H$ and density matrices $\rho$ on $H$ (which are $|\psi\rangle\langle\psi|$ for pure states).

Let $H = H_A \otimes H_B$, and $S_A \subset H_A$, $S_B \subset H_B$ are some subspaces. By ${\rm Im}(\rho)$ denote the image of $\rho$, which coincides with the support of $\rho$ since it's Hermitian. E.g. ${\rm Im}(|\psi\rangle\langle\psi|) = \{c|\psi\rangle\}$, a one-dimensional subspace.

Indeed, if ${\rm Im}(\rho) \subset S_A \otimes S_B$ then ${\rm Im}({\rm Tr}_B\rho) \subset S_A$ and ${\rm Im}({\rm Tr}_A\rho) \subset S_B$. It's not hard to see from linearity.

Conversely, we have $${\rm Im}(\rho) \subset {\rm Im}({\rm Tr}_B\rho) \otimes {\rm Im}({\rm Tr}_A\rho).$$ For pure $\rho$ it can be seen from the Schmidt decomposition, and for mixed $\rho$ from linearity.

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