Trace in Terms of Orthonormal Basis
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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 | |||||||||||
\(\ds \) | \(=\) | \(\ds c_{ii}\) |
$\blacksquare$