18 - Evaluation Map

Evaluation Map

Let $V$ be a finite dimensional vector space
Then

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

is defined by $v\mapsto E_v$ which is a natural linear isomorphism
where $E_v$ is the evaluation map at $v$ defined by

$$E_v(f):=f(v)\text{ for }f\in V^{\prime }$$

Note: Natural means independent of a choice of basis

Proof

First $E_v$ is a linear map from $V^{\prime }$ to $\mathbf{F}$ since

$$E_v(f+\lambda g):=(f+\lambda g)(v):=f(v)+\lambda g(v)=:E_v(f)+\lambda E_v(g)$$

for all $f,g\in V^{\prime }$, $v\in V$ and $\lambda \in \mathbf{F}$
Hence assignment $v\mapsto E_v$ is well-defined

Map $v\mapsto E_v$ itself is linear as

$$E_{\lambda v+w}=\lambda E_v+E_w\text{ for all }v,w\in V\text{ and }\lambda in\mathbf{F}$$

as each functional $f$ is linear

Map is also injective, since

$$E_v=0⟹E_v(f)=f(v)=0\text{ for all }f\in V^{\prime }$$

If $v\ne 0$ then extend $\{e_1=v\}$ to a basis $\mathcal{B}$ of $V$
For $f=e_1^{\prime }$ then

$$E_v(e_1^{\prime })=e_1^{\prime }(e_1)=1$$

This contradicts the fact that $E_v(e_1)=0$
Hence $v=0$ which shows that the assignment $v\mapsto E_v$ is injective

By Dual Basis Theorem then

$$\dim V=\dim (V^{\prime }))=\dim (V^{\prime }{)}^{\prime }$$

So by injectivity and the Rank-Nullity Theorem then assignment is surjective