My intuition tells me that the partial trace should be congruent under a change of basis. That is, if I have some matrix $A$ in the space of linear operators acting on a joint hilbert space: $A \in \mathbb{L}(\mathbb{H}_1 \otimes \mathbb{H}_2)$, then for every invertible matrix $U \in \mathbb{L}(\mathbb{H}_1 \otimes \mathbb{H}_2)$ there should exist some invertible matrix $G \in \mathbb{L}(\mathbb{H}_1)$ such that $G Tr_2(A) G^{-1}=Tr_2(U A U^{-1})$. The reason I believe it to be true is because the action of the operator should not change under a change of basis. Is this true? Is there a simple expression which relates $G$ to $U$? I have a physics education, so please forgive me if my notation is maybe too physicsy or assumes too much.
Thanks!
This is incorrect.
There is a $U$ in $\mathbb{L}(\mathbb{H}_1\otimes\mathbb{H}_2)$ which takes a tensor product state $e_1\otimes e_2$ to an entangled state $U(e_1\otimes e_2) = \frac{1}{\sqrt{2}} (e_1 \otimes e_2 + f_1 \otimes f_2)$. Now, let $A = e_1^\phantom{\dagger} e_1^\dagger \otimes e_2^\phantom{\dagger} e_2^\dagger.$ Then $G\,\mathrm{Tr}_2(A)G^{−1}$ has rank 1, but $\mathrm{Tr}_2(UAU^{−1})$ has rank 2.