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:

$\map \tr f = \ds \sum_{i \mathop = 1}^n e_i^* \paren {\map f {e_i} }$

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

\(\ds \)

\(=\)

\(\ds c_{ii}\)

$\blacksquare$

Also see

• Trace in Terms of Orthonormal Basis