# 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.

$\blacksquare$

## Sources

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