# If $\text{tr}_B \rho \in A$, then $\rho \in A \otimes B$?

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.

• 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? Commented Apr 11 at 6:41
• @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. Commented Apr 11 at 17:10

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.