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 $\mathcal L_R \left({G, H}\right)$ be the set of all linear transformations from $G$ to $H$.

Let $u \in \mathcal L_R \left({G, H}\right)$.

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, $\forall x \in G: y' \in H^*: \left \langle {x, u^t \left({y'}\right)} \right \rangle = \left \langle {u \left({x}\right), y'} \right \rangle$.

Since we have:

\(\displaystyle \left \langle {x, u^t \left({y' + z'}\right)} \right \rangle\) \(=\) \(\displaystyle \left \langle {u \left({x}\right), y' + z'} \right \rangle\)
\(\displaystyle \) \(=\) \(\displaystyle \left \langle {u \left({x}\right), y'} \right \rangle + \left \langle {u \left({x}\right), z'} \right \rangle\)
\(\displaystyle \) \(=\) \(\displaystyle \left \langle {x, u^t \left({y'}\right)} \right \rangle + \left \langle {x, u^t \left({z'}\right)} \right \rangle\)
\(\displaystyle \) \(=\) \(\displaystyle \left \langle {x, u^t \left({y'}\right) + u^t \left({z'}\right)} \right \rangle\)


and:

\(\displaystyle \left \langle {x, u^t \left({\lambda y'}\right)} \right \rangle\) \(=\) \(\displaystyle \left \langle {u \left({x}\right), \lambda y'} \right \rangle\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \left \langle {u \left({x}\right), y'} \right \rangle\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \left \langle {x, u^t \left({y'}\right)} \right \rangle\)
\(\displaystyle \) \(=\) \(\displaystyle \left \langle {x, \lambda u^t \left({y'}\right)} \right \rangle\)


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

$\blacksquare$


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