Coordinate (lame) version

Given a square matrix $A$ the trace is the sum of the diagonal entries, the trace of $A$ is

tr (A) = i \sum A_{ii} .

In this form its a bit hard to see why it is important has has some interesting properties.

Coordinate free version

For finite dimensional vector spaces $V$ and $W$ , there is a canonical isomorphism using the tensor product

V^{*} \otimes W ε \otimes w ⟶ Hom (V, W) ⟼ (f : V v \to W \mapsto ε (v) w)

So we can use this idea on the set of linear maps $V \to V$ to make a new linear map $Hom (V, V) \to C$ .

Hom (V, V) ≅ V^{*} \otimes V f \mapsto ε \otimes v ⟶ C ⟼ ε (v)

Thus, the trace of a linear map $f \in Hom (V, V)$ is

t r (f) = ε (v)

as above.

How do these two definitions match?

Given a basis ${v_{1}, \dots, v_{n}}$ for $V$ , then

f (v) = f (i \sum c_{i} v_{i}) = i \sum a_{i} f (v_{i})

Then looking at each $f (v_{i})$ then we can see that

f (v_{i}) = k \sum n b_{k} v_{k}

So, all together, (combining coefficients) this is equivalent to

f (v) ≅ \sum c_{ij} (v_{i}^{*} \otimes v_{j})

where ${v_{1}^{*}, \dots, v_{n}^{*}}$ is the dual basis.

So then, using the definition of the trace, we get

tr (f) = tr (\sum c_{ij} (v_{i}^{*} \otimes v_{j})) = \sum c_{ij} v_{i}^{*} (v_{j}) = \sum c_{ii}

since $v_{i}^{*} (v_{j}) = 0$ for $i \neq = j$ . This coincides with the trace formula for a matrix form of a linear map.

@brocker2003 - Chapter 2 section 3