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 $\tuple {e_1, \ldots, e_n}$ be a basis of $M$.

Let $\tuple {e_1^*,\ldots, e_n^*}$ be its dual basis

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

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

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

Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$.

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

Then:

Now it remains to show that $c_{ii} = e_i^* \paren {\map f {e_i} }$:

Also see

 * Trace in Terms of Orthonormal Basis