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 λ_{i} tensor x_{i}, with λ_{i} in V^{*} and x_{i} in V. (It maps a vector v in V to the sum of λ_{i}(v)x_{i}.) 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 λ_{i} tensor x_{i} and the sum of μ_{j} tensor y_{j} to the sum over i and j of λ_{i}(y_{j})μ_{j}(x_{i}).

If e_{1},...,e_{n} is a basis of V and e_{1}^{*},...,e_{n}^{*} is the dual basis of V^{*}, then the identity map is the sum of e_{i}^{*} tensor e_{i}. Applying it as above to the linear transformation T given by the sum of μ_{j} tensor y_{j} yields the sum of e_{i}^{*}(y_{j})μ_{j}(e_{i}), which is the sum over all i of the e_{i}-component of T(e_{i}). 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).