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

Ψ : V ⟶ {(V^{*})}^{*}, v ⟼ (f \mapsto f (v)) .

If $V$ has finite dimension, then $Ψ$ is an

isomorphism.

Proof

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

(Ψ (v)) (f_{1} + f_{2}) = (f_{1} + f_{2}) (v) = f_{1} (v) + f_{2} (v) = (Ψ (v)) (f_{1}) + (Ψ (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

(Ψ (v)) (s f) = (s f) (v) = s (f (v)) = s ((Ψ (v)) (f)) .

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

Ψ (v + w) = Ψ (v) + Ψ (w) .

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

(Ψ (v + w)) (f) = f (v + w) = f (v) + f (w) = (Ψ (v)) (f) + (Ψ (w)) (f) .

The scalar compatibility follows from

(Ψ (s v)) (f) = f (s v) = s (f (v)) = s ((Ψ (v)) (f)) .

In order to prove injectivity, let $v \in V$ with $Ψ (v) = 0$ 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, $Ψ$ is injective.

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

◻

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