Transpose of Linear Transformation is a Linear Transformation

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Theorem

Let $R$ be a commutative ring.

Let $G$ and $H$ be $R$-modules.

Let $G^*$ and $H^*$ be the algebraic duals of $G$ and $H$ respectively.


Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.

Let $u \in \map {\LL_R} {G, H}$.

Let $u^t: H^* \to G^*$ be the transpose of $u$.


Then $u^t: H^* \to G^*$ is itself a linear transformation.


Proof

By definition of evaluation linear transformation:

$\forall x \in G: y' \in H^*: \sequence {x, \map {u^t} {y'} } = \sequence {\map u x, y'}$

Since we have:

\(\ds \sequence {x, \map {u^t} {y' + z'} }\) \(=\) \(\ds \sequence {\map u x, y' + z'}\)
\(\ds \) \(=\) \(\ds \sequence {\map u x, y'} + \sequence {\map u x, z'}\)
\(\ds \) \(=\) \(\ds \sequence {x, \map {u^t} {y'} } + \sequence {x, \map {u^t} {z'} }\)
\(\ds \) \(=\) \(\ds \sequence {x, \map {u^t} {y'} + \map {u^t} {z'} }\)


and:

\(\ds \sequence {x, \map {u^t} {\lambda y'} }\) \(=\) \(\ds \sequence {\map u x, \lambda y'}\)
\(\ds \) \(=\) \(\ds \lambda \sequence {\map u x, y'}\)
\(\ds \) \(=\) \(\ds \lambda \sequence {x, \map {u^t} {y'} }\)
\(\ds \) \(=\) \(\ds \sequence {x, \lambda \map {u^t} {y'} }\)


it follows that $u^t: H^* \to G^*$ is a linear transformation.

$\blacksquare$


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