Trace in Terms of Dual Basis

Theorem
Let $R$ be a ring with unity.

Let $M$ be a free $R$-module of dimension $n$.

Let $(e_1,\ldots, e_n)$ be a basis of $M$.

Let $(e_1^*,\ldots, e_n^*)$ be its dual basis

Let $f:M\to M$ be a linear operator.

Then its trace equals:
 * $\operatorname{tr}(f) = \displaystyle \sum_{i\mathop =1}^n e_i^*(f(e_i))$

Proof
Let $\displaystyle f(e_i) = \sum_{j \mathop = 1}^n c_{ij} e_j$

Let $A$ be the matrix relative to the basis $\left({e_1, \ldots, e_n}\right)$.

Then by the above assumption, $A_{ij} = c_{ij}$.

Then:

Now it remains to show that $c_{ii} = e_i^*(f(e_i))$ :

Also see

 * Trace in Terms of Orthonormal Basis