Evaluation Linear Transformation is Linear Transformation

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Theorem

Let $R$ 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 $G^*$.


Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$ defined as:

$\forall x \in G: \map J x = x^\wedge$

where for each $x \in G$, $x^\wedge: G^* \to R$ is defined as:

$\forall t \in G^*: \map {x^\wedge} t = \map t x$


Then $J$ is a linear transformation.


Proof

From Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual, we have that:

$x^\wedge \in G^{**}$

Hence $x^\wedge$ a fortiori is a linear transformation.


It remains to be shown that $J: G \to G^{**}$ is a linear transformation.

That is, that the following conditions are satisfied by $J$:

$(1): \quad \forall x, y \in G: \map J {x + y} = \map J x + \map J y$
$(2): \quad \forall x \in G: \forall \lambda \in R: \map J {\lambda \times x} = \lambda \times J x$


Hence:

\(\text {(1)}: \quad\) \(\ds \map J {x + y}\) \(=\) \(\ds \paren {x + y}^\wedge\) Definition of $J$
\(\ds \) \(=\) \(\ds x^\wedge + y^\wedge\) Definition of Pointwise Addition of Linear Transformations
\(\ds \) \(=\) \(\ds \map J x + \map J y\) Definition of $x^\wedge$


and:

\(\text {(2)}: \quad\) \(\ds \map J {\lambda \times x}\) \(=\) \(\ds \paren {\lambda \times x}^\wedge\) Definition of $x^\wedge$
\(\ds \) \(=\) \(\ds \lambda \times x^\wedge\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \lambda \times \map J x\) Definition of $x^\wedge$

Hence the result.

$\blacksquare$


Sources

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