1
$\begingroup$

Assume $V$ is a finite dimensional vectorspace and let $W$ be a subspace of $V$. Denote by $i \colon W \to V$ the injection of $W$ into $V$.

Then it is possible to define a dual map $\pi \colon V^* \to W^*$, the projection from the dualspace of $V$ to the dualspace of $W$.

My question is now, if there is a canonical way of defining a map $\lambda \colon W^* \to V^*$ as the embedding from $W^*$ to $V^*$

$\endgroup$
1
$\begingroup$

Not without extra data. There are many injective linear maps $\lambda \colon W^{*} \rightarrow V^{*}$. Even if you require compatability with $\pi$ in the sense that $\pi \circ \lambda = \operatorname{id}|_{W^{*}}$, then such an injective linear map provides a "linear scheme" of extending functionals that are defined on $W$ to functionals that are defined on $V$ and there are many possible ways to do this (by choosing a basis for $W$, extending it to a basis of $V$ and letting the functional act arbitrary on the extra basis elements).

If you choose a complement $W'$ to $W$ so that $V = W \oplus W'$, then you have a surjective projection $\pi_{W|W'} \colon V \rightarrow W$ to $W$ along $W'$ (that depends on both $W$ and $W'$!) and then the dual map $\pi_{W|W'}^{*} \colon W^{*} \rightarrow V^{*}$ will give you an injective linear map from $W^{*}$ to $V^{*}$. If you are working over $\mathbb{R}$ or $\mathbb{C}$ and have an inner product, you can take $W' = W^{\perp}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.