23 - Dual Map Isomorphism Theorem

Dual Map Isomorphism Theorem

Let $V,W$ be two finite dimensional vector space
Assignment $T\mapsto T^{\prime }$ is a natural isomorphism from

$$\mathrm{hom}(V,W)\text{ to }\mathrm{hom}(W^{\prime },V^{\prime })$$

Proof

We need to check that the assignment $T\mapsto T^{\prime }$ linear in $T$
Let $T,S\in \mathrm{hom}(V,W)$ and $\lambda \in \mathbf{F}$
For $f\in W^{\prime }$ then

$$\begin{array}{rl}((T+\lambda S{)}^{\prime })(f))(v)& =f((T+\lambda S)(v))\\ & =f(T(v)+\lambda S(v))\\ & =f(T(v))+\lambda f(S(v))\\ & =T^{\prime }(f)(v)+\lambda S^{\prime }(f)(v)\\ & =(T^{\prime }(f)+\lambda S^{\prime }(f))(v)\\ & =((T^{\prime }+\lambda S^{\prime })(f))(v)\end{array}$$

Hence

$$(T+\lambda S{)}^{\prime }=T^{\prime }+\lambda S^{\prime }$$

To prove injectivity assume $T^{\prime }=0$ then
For all $f\in W^{\prime }$ then

$$T^{\prime }(f)=0$$

which is an identity of functionals on $V$ so
For all $f\in W^{\prime }$ , and for all $v\in V$ we have $T^{\prime }(f)(v)=0$ so

$$T^{\prime }(f)(v):=f(Tv)=:E_{Tv}(f)=0$$

But then $E_{Tv}=0$ hence $Tv=0$ by Evaluation Map applied to $W$

As this holds for all $v\in V$ then $T=0$

When the vector spaces are finite dimensional, we have

$$\dim (\mathrm{hom}(V,W))=\dim (V)\dim (W)=\dim W^{\prime }\dim V^{\prime }=\dim (\mathrm{hom}(W^{\prime },V^{\prime }))$$

Hence the map is also surjective