Is the partial trace identity $\mathrm{Tr}_2 [(A\otimes B)C]=A\cdot\mathrm{Tr}_2[(\mathbb{I}_1\otimes B)C]$ true?

Question

Suppose we have two (finite dimensional) Hilbert spaces $H_1$ and $H_2$. Let $A$ be an operator on $H_1$, and $B$ an operator on $H_2$, and $C$ an operator on $H_1 \otimes H_2$. Is the quantity (where $\mathrm{Tr}_2[\cdot]$ is the partial trace over the Hilbert space $H_2$) $$ \mathrm{LHS} := \mathrm{Tr}_2 \left[ ( A \otimes B ) C \right] $$ the same as $$ \mathrm{RHS} := A \cdot \mathrm{Tr}_2 \left[ ( \mathbb{I}_1 \otimes B ) C \right] \ ? $$ That is to say, is it true that $\mathrm{LHS} =\mathrm{RHS}$? It seems correct, but I am having difficulty proving it.

Answer 1

Yes, this identity is correct. The key is to note that $H_1 \otimes H_2$ is spanned by simple tensors, which is to say that $C$ can be expressed in the form $$ C = \sum_{j=1}^k P_j \otimes Q_j. $$ With that, we can write $$ \begin{align} \mathrm{Tr}_2 \left[ ( A \otimes B ) C \right] &= \mathrm{Tr}_2 \left[ ( A \otimes B ) \sum_{j=1}^k P_j \otimes Q_j \right] \\&= \mathrm{Tr}_2 \left[ \sum_{j=1}^k (AP_j) \otimes (BQ_j) \right] \\ &= \sum_{j=1}^k \mathrm{Tr}(BQ_j)(AP_j) \\ &= A \sum_{j=1}^k \mathrm{Tr}(BQ_j)(P_j) \\ &= A \cdot \mathrm{Tr}_2\left[ \sum_{j=1}^k P_j \otimes BQ_j\right] \\ &= A \cdot \mathrm{Tr}_2\left[ (I \otimes B)\sum_{j=1}^k P_j \otimes Q_j\right] = A \cdot \mathrm{Tr}_2\left[ (I \otimes B)C\right]. \end{align} $$