Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual
Jump to navigation
Jump to search
Theorem
Let $\struct {R, +, \times}$ be a commutative ring with unity.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the double dual of $G$.
For each $x \in G$, let $x^\wedge: G^* \to R$ be defined as:
- $\forall t \in G^*: \map {x^\wedge} t = \map t x$
Then:
- $x^\wedge \in G^{**}$
Proof
We have that $x^\wedge$ is a mapping from $G^* \to R$.
It remains to be demonstrates that $x^\wedge$ is in fact a linear transformation.
Hence we need to show that:
- $(1): \quad \forall u, v \in G^*: \map {x^\wedge} {u + v} = \map {x^\wedge} u + \map {x^\wedge} v$
- $(2): \quad \forall u \in G^*: \forall \lambda \in R: \map {x^\wedge} {\lambda \times u} = \lambda \times \map {x^\wedge} u$
Hence:
\(\text {(1)}: \quad\) | \(\ds \map {x^\wedge} {u + v}\) | \(=\) | \(\ds \map {\paren {u + v} } t\) | Definition of $x^\wedge$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \map u t + \map v t\) | Definition of Pointwise Addition of Linear Transformations | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {x^\wedge} u + \map {x^\wedge} v\) | Definition of $x^\wedge$ |
and:
\(\text {(2)}: \quad\) | \(\ds \map {x^\wedge} {\lambda \times u}\) | \(=\) | \(\ds \map {\paren {\lambda \times u} } t\) | Definition of $x^\wedge$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \times \map u t\) | Definition of Linear Transformation | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \times \map {x^\wedge} u\) | Definition of $x^\wedge$ |
Hence the result.
$\blacksquare$
Also see
- Definition:Evaluation Linear Transformation/Module Theory, which uses $x^\wedge$ in its definition
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations