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


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


Proof

By definition of evaluation linear transformation:

$\forall x \in G: y \in H^*: \innerprod x {\map {u^\intercal} y} = \innerprod {\map u x} y$

Since we have:

\(\ds \innerprod x {\map {u^\intercal} {y + z} }\) \(=\) \(\ds \innerprod {\map u x} {y + z}\)
\(\ds \) \(=\) \(\ds \innerprod {\map u x} y + \innerprod {\map u x} z\)
\(\ds \) \(=\) \(\ds \innerprod x {\map {u^\intercal} y} + \innerprod x {\map {u^\intercal} z}\)
\(\ds \) \(=\) \(\ds \innerprod x {\map {u^\intercal} y + \map {u^\intercal} z}\)


and:

\(\ds \innerprod x {\map {u^\intercal} {\lambda y} }\) \(=\) \(\ds \innerprod {\map u x} {\lambda y}\)
\(\ds \) \(=\) \(\ds \lambda \innerprod {\map u x} y\)
\(\ds \) \(=\) \(\ds \lambda \innerprod x {\map {u^\intercal} y}\)
\(\ds \) \(=\) \(\ds \innerprod x {\lambda \map {u^\intercal} y}\)


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

$\blacksquare$


Sources

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