The partial trace is characterized by the trace pairing and is basis-independent

By [F8], choose ordered orthonormal bases $(e_i{)}_{i<a}$ of $H_A$ and $(f_j{)}_{j<b}$ of $H_B$, and set $S={\mathrm{Tr}}_B(R)$. For any $T\in \mathrm{End}(H_A)$, [F1], [F6] and [F7] give $\mathrm{tr}(S\circ T)={\sum }_{i,i^{\mathrm{\prime }}<a}{\sum }_{j<b}{R}_{(i,j),(i^{\mathrm{\prime }},j)}{T}_{i^{\mathrm{\prime }}i}$.

The identity $I_{H_B}$ is linear by [F9], so [F2] defines the linear map $T\otimes I_{H_B}$; its composite with $R$ is linear by [F9]. In the product basis, [F2] gives the matrix entry $T_{i^{\mathrm{\prime }}i}{\delta }_{j^{\mathrm{\prime }}j}$ of $T\otimes I$. By [F6] and [F7], $\mathrm{tr}(R\circ (T\otimes I))={\sum }_{i,j}{\sum }_{i^{\mathrm{\prime }},j^{\mathrm{\prime }}}{R}_{(i,j),(i^{\mathrm{\prime }},j^{\mathrm{\prime }})}{T}_{i^{\mathrm{\prime }}i}{\delta }_{j^{\mathrm{\prime }}j}={\sum }_{i,i^{\mathrm{\prime }},j}{R}_{(i,j),(i^{\mathrm{\prime }},j)}{T}_{i^{\mathrm{\prime }}i}$, which is the same finite sum as in step 1.1. Thus $S$ satisfies the displayed trace identity.

If $S^{\mathrm{\prime }}$ is any other map value at $R$ satisfying the identity for every $T$, then $\mathrm{tr}((S-S^{\mathrm{\prime }})\circ T)=0$ for every $T$. Apply [F5] on $H_A$ to obtain $S=S^{\mathrm{\prime }}$. Since this holds for every $R$, the trace identity characterizes the partial-trace map uniquely.

Repeat steps 1.1 and 2.1 with any other ordered orthonormal basis of $H_B$. The right-hand side is the trace of the same endomorphism $R\circ (T\otimes I_{H_B})$, and [F4] makes that trace basis-independent. Step 3.1 then forces the resulting partial-trace operator to equal $S$.

If $H_B=0$, the product space and $R$ are zero, [F1] is the empty sum, and both sides of the trace identity are $0$. If $H_A=0$, the input and output endomorphism spaces contain only the zero map, so the same identity and uniqueness hold.