# Evaluation Linear Transformation is Linear Transformation

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## Theorem

Let $R$ be a commutative ring.

Let $G$ be an $R$-module.

Let $G^*$ be the algebraic dual of $G$.

Let $G^{**}$ be the algebraic dual of $G^*$.

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

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

- $\forall t' \in G^*: \map {x^\wedge} {t'} = \map {t'} x$

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:

- $(1): \quad x^\wedge \in G^{**}$
- $(2): \quad J$ is a linear transformation.

## Proof

$(1):$ First we show that $x^\wedge \in G^{**}$:

$(2):$ Then we show that $J: G \to G^{**}$ is a linear transformation:

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 28$