Trace in Terms of Orthonormal Basis

Theorem

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

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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:

\(\ds \map \tr f\)

\(=\)

\(\ds \map \tr A\)

Definition of Trace of Linear Operator

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^n A_{ii}\)

Definition of Trace of Matrix

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^n c_{ii}\)

From above

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

\(\ds \innerprod {\map f {e_i} } {e_i}\)

\(=\)

\(\ds \innerprod {\sum_{j \mathop = 1}^n c_{ij} e_j} {e_i}\)

From above assumption

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^n c_{ij} \innerprod {e_j} {e_i}\)

Axiom $(2)$ of Inner Product: Bilinearity

\(\ds \)

\(=\)

\(\ds c_{ii} (1)\)

Definition of Orthonormal Subset: other terms vanish

\(\ds \)

\(=\)

\(\ds c_{ii}\)

$\blacksquare$

Also see

• Trace in Terms of Dual Basis