Linear mapping/Dual mapping/Introduction/Section

Definition

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

$\varphi \colon V\longrightarrow W$

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

$\varphi ^{*}\colon \operatorname {Hom} _{K}{\left(W,K\right)}={W}^{*}\longrightarrow \operatorname {Hom} _{K}{\left(V,K\right)}={V}^{*},f\longmapsto f\circ \varphi ,$

is called the dual mapping of ${}\varphi$.

This assignment arises from just considering the composition

$V{\stackrel {\varphi }{\longrightarrow }}W{\stackrel {f}{\longrightarrow }}K.$

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

$\psi \colon U\longrightarrow V$

and

$\varphi \colon V\longrightarrow W$

be

linear mappings. Then the following hold.

1. For the dual mapping, we have

   ${}{\left(\varphi \circ \psi \right)}^{*}=\psi ^{*}\circ \varphi ^{*}\,.$

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

   ${}{\operatorname {Id} _{V}}^{*}=\operatorname {Id} _{{V}^{*}}\,.$

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

Proof

(1). For ${}f\in {W}^{*}$, we have

${}{\left(\varphi \circ \psi \right)}^{*}(f)=f\circ {\left(\varphi \circ \psi \right)}={\left(f\circ \varphi \right)}\circ \psi =\varphi ^{*}(f)\circ \psi =\psi ^{*}(\varphi ^{*}(f))\,.$

(2) follows directly from ${}f\circ \operatorname {Id} _{V}=f$.

(3). Let ${}f\in {V}^{*}$ and

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

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

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

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

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

${}V=U\oplus U'\,$

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

$g\colon U\longrightarrow K$

can always be extended to a linear form

${\tilde {g}}\colon V\longrightarrow K,$

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

$\Box$