Trace in Terms of Orthonormal Basis

Theorem
Let $\mathbb K \subset \C$ be a field.

Let $\struct {V, \innerprod {\, \cdot \,} {\, \cdot \,} }$ be an inner product space over $\mathbb K$ of dimension $n$.

Let $\tuple {e_1, \ldots, e_n}$ be an orthonormal basis of $V$.

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

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

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

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 $c_{ii} = \innerprod {\map f {e_i} } {e_i}$:

Also see

 * Trace in Terms of Dual Basis