# Transpose of Linear Transformation is a Linear Transformation

## 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.

$\blacksquare$