I was just wondering if anyone knew of a neat proof of the fact that for and matrices.

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- Mar 25th 2010, 08:25 AMSwlabrTrace of a tensor product
I was just wondering if anyone knew of a neat proof of the fact that for and matrices.

- Mar 25th 2010, 12:30 PMOpalg
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. - Mar 25th 2010, 08:36 PMNonCommAlg
if then using the definition of we have:

- Mar 26th 2010, 03:42 AMSwlabr
- Mar 26th 2010, 04:15 AMOpalg
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.