Vector space/Bidual/Introduction/Section

Definition

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. Then the dual space of the dual space ${}{V}^{*}$, that is,

${({V}^{*})}^{*},$

is called the Bidual of ${}V$.

Lemma

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. Then there exists a natural injective linear mapping

$\Psi \colon V\longrightarrow {({V}^{*})}^{*},v\longmapsto {\left(f\mapsto f(v)\right)}.$

If ${}V$ has finite dimension, then ${}\Psi$ is an

isomorphism.

Proof

Let ${}v\in V$ be fixed. First of all, we show that ${}\Psi (v)$ is a linear form on the dual space ${}{V}^{*}$. Obviously, ${}\Psi (v)$ is a mapping from ${}{V}^{*}$ to ${}K$. The additivity follows from

${}(\Psi (v))(f_{1}+f_{2})=(f_{1}+f_{2})(v)=f_{1}(v)+f_{2}(v)=(\Psi (v))(f_{1})+(\Psi (v))(f_{2})\,,$

where we have used the definition of the addition on the dual space. The compatibility with the scalar multiplication follows similarly from

${}(\Psi (v))(sf)=(sf)(v)=s(f(v))=s((\Psi (v))(f))\,.$

In order to prove the additivity of ${}\Psi$, let ${}v,w\in V$ be given. We have to show the equality

${}\Psi (v+w)=\Psi (v)+\Psi (w)\,.$

This is an equality inside of ${}{\left({V}^{*}\right)}^{*}$, in particular, it is an equality of mappings. So let ${}f\in {V}^{*}$ be given. Then, the additivity follows from

${}(\Psi (v+w))(f)=f(v+w)=f(v)+f(w)=(\Psi (v))(f)+(\Psi (w))(f)\,.$

The scalar compatibility follows from

${}(\Psi (sv))(f)=f(sv)=s(f(v))=s((\Psi (v))(f))\,.$

In order to prove injectivity, let ${}v\in V$ with ${}\Psi (v)=0$ be given. this means that for all linear forms ${}f\in {V}^{*}$, we have ${}f(v)=0$. But then, due to fact, we have

${}v=0\,.$

By the criterion for injectiviy, ${}\Psi$ is injective.

In the finite-dimensional case, the bijectivity follows from injectivity and from fact.

$\Box$

Thus, the mapping ${}\Psi$ sends a vector ${}v$ to the evaluation (or evaluation mapping) which evaluates a linear form ${}f$ an the point ${}v$.