Here's an entertaining characterization of the trace map from linear algebra. It's not original, but it doesn't seem to be as well known as it should be.
Let V be a finite-dimensional vector space. Then the vector space Hom(V,V) of linear transformations from V to V is canonically isomorphic to its own dual: Hom(V,V) = V* tensor V = V* tensor V** = (V tensor V*)* = (V* tensor V)* = Hom(V,V)*. Each equality denotes a canonical isomorphism. Which element of Hom(V,V)* corresponds to the identity map in Hom(V,V)? The answer is the trace.
To see why, consider an element of Hom(V,V). Viewing it as an element of V* tensor V amounts to viewing it as a sum over i of λi tensor xi, with λi in V* and xi in V. (It maps a vector v in V to the sum of λi(v)xi.) The isomorphism between Hom(V,V) and Hom(V,V)* amounts to a perfect pairing between elements of Hom(V,V). Unraveling the definitions shows that this pairing sends the sum of λi tensor xi and the sum of μj tensor yj to the sum over i and j of λi(yj)μj(xi).
If e1,...,en is a basis of V and e1*,...,en* is the dual basis of V*, then the identity map is the sum of ei* tensor ei. Applying it as above to the linear transformation T given by the sum of μj tensor yj yields the sum of ei*(yj)μj(ei), which is the sum over all i of the ei-component of T(ei). In other words, it is the trace of T.
Of course when V is infinite-dimensional this approach fails to define an analogue of the trace. It's no longer even true that Hom(V,V) is isomorphic to V* tensor V (the identity map is not in V* tensor V).