Trace of tensor product identity

Question

Let $A: V\to V$ and $B: W\to W$ be linear operators on vector spaces $V$ and $W$. I know how to prove $$\operatorname{tr}(A\otimes B) = \operatorname{tr}(A)\operatorname{tr}(B)$$ by appealing to a basis $\{a_i\}$ of $V$ and $\{b_j\}$ of $W$, and using the product basis $\{a_i\otimes b_j\}$ of $A\otimes B$, but I'm wondering if there is a way to prove this without talking about bases at all. I know there is a coordinate free characterization of the trace, but I'm not familiar enough with it to put it into use.

Answer 1

Mechanically, the proof using the coordinate-free characterization won't be all that different from what you likely did using the product basis.

Here's one approach. We claim without proof that that there exist rank-$1$ operators $A_1,\dots,A_p$ and $B_1,\dots,B_q$ such that $A = \sum_j A_j$ and $B = \sum_k B_k$. This statement is usually proved using the existence of a rank factorization or a singular value decomposition.

With that in mind: for any rank-1 operator $A$, there exists a vector $x$ and a functional $f$ such that $A(y) = f(y)\cdot x$ (which I will write as $A = xf(\cdot)$). For any two rank-1 operators $A = xf(\cdot)$ and $B = yg(\cdot)$, we have $$ \begin{align} \operatorname{tr}(A \otimes B) &= \operatorname{tr}([xf(\cdot)] \otimes [yg(\cdot)]) \\ &= \operatorname{tr}([x \otimes y][f(\cdot) \otimes g(\cdot)]) \\ &= [f(\cdot) \otimes g(\cdot)](x \otimes y) \\ & = f(x)g(y) = \operatorname{tr}(xf(\cdot)) \operatorname{tr}(yg(\cdot)) = \operatorname{tr}(A)\operatorname{tr}(B). \end{align} $$ From there, simply use the linearity of the trace function and tensor product. For our original operators $A,B$, we have $$ \begin{align} \operatorname{tr}(A \otimes B) &= \operatorname{tr}\left(\left[\sum_{j}A_j \right] \otimes \left[ \sum_k B_k \right] \right) \\ & = \sum_{j} \sum_k \operatorname{tr}(A_j \otimes B_k) \\ &= \sum_{j} \sum_k \operatorname{tr}(A_j) \operatorname{tr}(B_k) \\ & = \left( \sum_j \operatorname{tr}(A_j) \right) \left(\sum_k \operatorname{tr}(B_k) \right) = \operatorname{tr}(A) \operatorname{tr}(B). \end{align} $$