Originally Posted by
Opalg The trace on the n×n matrices is the unique linear functional satisfying the conditions (1) $\displaystyle \text{Tr}(XY) = \text{Tr}(YX) $ and (2) $\displaystyle \text{Tr}(I) =1$.
For matrices $\displaystyle A\in M_m$ (the set of m×m matrices) and $\displaystyle B\in M_n$, the functional $\displaystyle f(A\otimes B) = \text{Tr}(AB)$ is bilinear, and therefore extends to a linear functional on $\displaystyle M_m\otimes M_n\cong M_{mn}$. So we just have to check that f satisfies (1) and (2). Condition (1) is bilinear, so you only need to check it on simple tensors:
$\displaystyle \begin{aligned}f\bigl((A_1\otimes B_1)(A_2\otimes B_2)\bigr) &= f(A_1A_2\otimes B_1B_2) \\ &= \text{Tr}(A_1A_2)\text{Tr}(B_1B_2) \\ &= \text{Tr}(A_2A_1)\text{Tr}(B_2B_1) \\ &= f(A_2A_1\otimes B_2B_1) = f\bigl((A_2\otimes B_2)(A_1\otimes B_1)\bigr).\end{aligned}$
Condition (2) is trivially satisfied, so f is the functional Tr on the mn×mn matrices.