Linear mapping/Dual mapping/Introduction/Section

Definition

Let $K$ denote a field, let $V$ and $W$ denote $K$-vector spaces, and let

$\phi :V⟶W$

denote a $K$-linear mapping. Then the mapping

${\phi }^{\ast }:{\mathrm{Hom}}_K(W,K)=W^{\ast }⟶{\mathrm{Hom}}_K(V,K)=V^{\ast },f⟼f\circ \phi ,$

is called the dual mapping of $\phi$.

This assignment arises from just considering the composition

$V\stackrel{\phi }{⟶}W\stackrel{f}{⟶}K.$

The dual mapping is a special case of the situation described in fact  (1). In particular, the dual mapping is again linear.

Lemma

Let $U,V,W$ denote vector spaces over a field $K$ and let

$\psi :U⟶V$

and

$\phi :V⟶W$

be

linear mappings. Then the following hold.

1. For the dual mapping, we have

   ${(\phi \circ \psi )}^{\ast }={\psi }^{\ast }\circ {\phi }^{\ast }.$

2. For the identity on $V$, we have

   ${{\mathrm{Id}}_V}^{\ast }={\mathrm{Id}}_{V^{\ast }}.$

3. If $\psi$ is surjective then ${\psi }^{\ast }$ is injective.
4. If $\psi$ is injective then ${\psi }^{\ast }$ is surjective.

Proof

1. For $f\in W^{\ast }$, we have

   ${(\phi \circ \psi )}^{\ast }(f)=f\circ (\phi \circ \psi )=(f\circ \phi )\circ \psi ={\phi }^{\ast }(f)\circ \psi ={\psi }^{\ast }({\phi }^{\ast }(f)).$

2. This follows directly from $f\circ {\mathrm{Id}}_V=f$.
3. Let $f\in V^{\ast }$ and

   ${\psi }^{\ast }(f)=0.$

   Because of the surjectivity of $\psi$, there exist for every $v\in V$ a $u\in U$ such that $\psi (u)=v$. Therefore

   $f(v)=f(\psi (u))=({\psi }^{\ast }(f))(u)=0,$

   and $f$ is itself the zero mapping. Due to fact, ${\psi }^{\ast }$ injective.

4. The condition means that we may consider $U\subseteq V$ as a linear subspace. Because of fact, we can write

   $V=U\oplus U^{\prime }$

   with another $K$-linear subspace $U^{\prime }\subseteq V$. A linear form

   $g:U⟶K$

   can always be extended to a linear form

   $\tilde{g}:V⟶K,$

   for example, by defining $\tilde{g}$ on $U^{\prime }$ to be the zero form. This means the surjectivity.

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