21 - Isomorphism of the Natural Mapping

Isomorphism of the Natural Mapping

Let $U$ be a subspace of a finite dimensional vector space $V$
Under the natural map

$$V\to V^{\prime \prime }(:=(V^{\prime }{)}^{\prime })$$

which is given by

$$v\mapsto E_V$$

Then $U$ is mapped isomorphically to

$$U^{00}:=(U^0{)}^0$$

Proof

Let us here write $E:V\to V^{\prime \prime }$ for the natural isomorphism $v\mapsto E_v$

For $v\in V$ the functional $E_v$ is in $U^{00}$ if and only if

$$\forall f\in U^0\text{ we have }{E}_v(f)=f(v)=0$$

Hence if $v\in U$ then $E_v\in U^{00}$ thus

$$U\cong E(U)\subseteq U^{00}$$

When $V$ is finite dimensional then by Dimension of Annihilators

$$\begin{array}{rl}\dim (U^{00})& =\dim (V^{\prime })-\dim (U^0)\\ & =\dim (V)-(\dim (V)-\dim (U))\\ & =\dim (U)\end{array}$$

Hence

$$U\cong E(U)=U^{00}$$