Trace in Terms of Dual Basis
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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:
- $\map \tr f = \ds \sum_{i \mathop = 1}^n e_i^* \paren {\map f {e_i} }$
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 $c_{ii} = e_i^* \paren {\map f {e_i} }$:
\(\ds e_i^* \paren {\map f {e_i} }\) | \(=\) | \(\ds e_i^* \paren {\sum_{j \mathop = 1}^n c_{ij} e_j}\) | from above assumption | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n c_{ij} e_i^* \paren {e_j}\) | $e_i$ is a linear form | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {c_{ii} } 1\) | Definition of Ordered Dual Basis: other terms vanish | |||||||||||
\(\ds \) | \(=\) | \(\ds c_{ii}\) |
$\blacksquare$