TITLE: Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?
QUESTION [45 upvotes]: I am searching for a short coordinate-free  proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ for linear operators  $A$, $B$ between finite dimensional vector spaces of the same dimension. 

The usual proof is to represent the operators as matrices and then use matrix multiplication. I want a coordinate-free proof. That is, one that does not  make reference to an explicit matrix representation of the operator. I define trace as the sum of the eigenvalues of an operator.
Ideally, the proof the should be shorter and require fewer preliminary lemmas than the one given in this blog post.
I would be especially interested in a proof that generalizes to the trace class of operators on a Hilbert space.

REPLY [7 votes]: Here is a proof that shows more:
If $A$ and $B$ are such that both $AB$ and $BA$ are square matrices then 
$AB$ and $BA$ have the same non-zero eigenvalues.
Proof: Let $\lambda \ne 0$ be an eigenvalue and $e$ the eigenvector of $B$. Since $AB~e= \lambda e$ we have $B e \ne 0$. Hence
$$
(BA) (Be) = B (AB) e = B (\lambda e) = \lambda (Be) $$
Hence $\lambda $ is also an eigenvalue of $BA$ (with eigenvector $Be$).
One can extend this argument to repeated eigenvalues with generalized eigenvectors.
Since trace = sum of eigenvalues, the result follows.