MA20216: Algebra 2A

Chapter 5 Duality

5.1 Dual spaces

Definition. Let \(V\) be a vector space over \(\F \). The dual space \(V^{*}\) of \(V\) is
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} V^{*}:=L(V,\F )=\set {\alpha :V\to \F \st \text {$\alpha $ is linear}}. \end{equation*}

Elements of \(V^{*}\) are called linear functionals or (less often) linear forms.

Proposition 5.1. Let \(V\) be a finite-dimensional vector space with basis \(\lst {v}1n\).

Define \(\dlst {v}1n\in V^{*}\) by setting
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} v_i^{*}(v_j)=\delta _{ij}\in \F \end{equation*}

and extending by linearity (thus applying Proposition 1.7).

Then \(\dlst {v}1n\) is a basis of \(V^{*}\) called the dual basis to \(\lst {v}1n\).

Corollary 5.2. If \(V\) is finite-dimensional then \(\dim V=\dim V^{*}\).

Theorem 5.3 (Sufficiency principle). Let \(V\) be a vector space and \(v\in V\). Then \(\alpha (v)=0\), for all \(\alpha \in V^{*}\), if and only if \(v=0\).

Proposition 5.4. Let \(V\) be a finite-dimensional vector space and \(\lst {\alpha }1n\) a basis of \(V^{*}\). Then there is a basis \(\lst {v}1n\) of \(V\) such that
\(\seteqnumber{0}{5.}{0}\)
\begin{equation*} \alpha _i(v_j)=\delta _{ij}. \end{equation*}

Thus \(\alpha _i=v^{*}_i\), for \(\bw 1{i}n\).

Theorem 5.5. If \(V\) is a finite-dimensional vector space then \(\ev :V\to V^{**}\) is an isomorphism.