I was just wondering if anyone knew of a neat proof of the fact that for and matrices.
The trace on the n×n matrices is the unique linear functional satisfying the conditions (1) and (2) .
For matrices (the set of m×m matrices) and , the functional is bilinear, and therefore extends to a linear functional on . 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:
Condition (2) is trivially satisfied, so f is the functional Tr on the mn×mn matrices.
Yes it should. (I'm used to dealing with the normalised trace, where you divide by the dimension so that the identity has trace 1.) For the usual trace, condition (2) in my previous comment is equally obvious, because both sides become mn when you put A and B equal to the appropriately-sized identity matrices.