pervognsen, 3 months ago (edited 3 months ago)

@shachaf The trace of a rank 1 operator f = u v^T is Tr(u v^T) = v^T u. Since v^T is the "input part" and u is the "output part" of f this provides another way to think about how the trace connects the output of an operator back to its own input. Here I'm just using v^T as "abstract matrix notation" for a linear functional, so no inner product or basis is assumed.