Section 5.6 Natural isomorphism between a vector space and its double dual
Throughout this section we assume that \(V\) is a finite-dimensional vector space over a field \(F\text{.}\) Recall from Definition 4.1.6 that the dual space of \(V\text{,}\) \(V^*=\Hom_F(V,F)\) is an vector space over \(F\text{.}\) The double dual of \(V\) is a vector space over \(F\) defined by \(V^{**}=\Hom_F(V^*,F)\text{.}\)
Proof.
Let \(\mathfrak{B}=\{v_1,v_2,\ldots,v_n\}\) be a basis of \(V\) over \(F\text{.}\) We define \(v_i^*(v_j)=\delta_{ij}\text{,}\) and extend it linearly over \(V\text{.}\) Thus, we get a subset \(\mathfrak{B}^*=\{v_1^*,v_2^*,\ldots,v_n^*\}\) of \(V^*\text{.}\)
We claim that \(\mathfrak{B}^*\) is a basis for \(V^*\) over \(F\text{.}\) Let \(\sum\alpha_iv_i^*=0\text{,}\) i.e., for every \(v_j\) we have
\begin{equation*}
\sum\alpha_iv_i^*(v_j)=0.
\end{equation*}
Therefore, for every \(j\text{,}\) we get \(\alpha_j=0\text{.}\) Hence, \(\mathfrak{B}^*\) is \(F\)-linearly independent.
We now show that \(\mathfrak{B}^*\) spans \(V^*\text{.}\) Let \(f\in V^*\) be a linear functional. The map \(f\) is determined by its action on \(v_i\text{.}\) Therefore,
\begin{equation*}
f=\sum f(v_i)v_i^*.
\end{equation*}
\begin{equation*}
v_i\mapsto v_i^*.
\end{equation*}
Corollary 5.6.2.
We have \(\dim_FV=\dim_FV^*\text{.}\)Definition 5.6.3.
Let \(V\) be a finite-dimensional vector space over a field \(F\text{.}\) As in the proof of Proposition 5.6.1, let \(v_i^*(v_j)=\delta_{ij}\text{.}\) Let \(\{v_1,\ldots,v_n\}\) be a basis of \(V\text{.}\) The set \(\{v_1^*,\ldots,v_n^*\}\) is called a dual basis for \(V^*\) . The set \(\{v_1^*,\ldots,v_n^*\}\) is a basis of \(V^*\) (refer to the proof of Proposition 5.6.1).Example 5.6.4. (Matrix representation of dual basis).
\begin{equation*}
[T]_{\mathfrak{B}}=\begin{pmatrix}a_{11}\amp a_{12}\amp\cdots\amp a_{1n}\\a_{21}\amp a_{22}\amp\cdots\amp a_{2n}\\\vdots\amp \vdots\amp\ddots\amp \vdots\\a_{n1}\amp a_{n2}\amp\cdots\amp a_{nn}\\\end{pmatrix}.
\end{equation*}
We have a map
\begin{equation*}
T^*\colon V^*\to V^*\quad\text{given by}\quad f\mapsto f\circ T.
\end{equation*}
We have the following.
\begin{equation*}
T^*(v_j^*)(v_k)=v_j^*\circ T(v_k)=v_j^*\left(\sum_i a_{ik}v_i\right)=a_{jk}
\end{equation*}
Hence,
\begin{equation*}
T^*(v_j^*)=\sum_ka_{jk}v^*_k.
\end{equation*}
The matrix of \(T^*\) with respect to \(\mathfrak{B}^*\) is the transpose of \([T]_{\mathfrak{B}}\text{.}\)
\begin{equation*}
[T^*]_{\mathfrak{B}^*}=\begin{pmatrix}a_{11}\amp a_{21}\amp\cdots\amp a_{n1}\\a_{12}\amp a_{22}\amp\cdots\amp a_{n2}\\\vdots\amp \vdots\amp\ddots\amp \vdots\\a_{1n}\amp a_{2n}\amp\cdots\amp a_{nn}\\\end{pmatrix}
\end{equation*}
Remark 5.6.5.
The isomorphism defined in Proposition 5.6.1 depends on the choice of a basis.
Indeed, suppose that \(V\) is one-dimensional vector space over a field \(\R\text{.}\) Consider any nonzero vector \(v\text{.}\) Then, \(\{v\}\) is a basis for \(V\text{,}\) and its dual basis is \(\{v^*\}\text{.}\) Similarly, \(\{2v\}\) is a basis and \(\{v^*/2\}\) is a dual basis. Note that mappings \(v\mapsto v^*\) and \(2v\mapsto v^*/2\) are different.
We now define an isomorphism between a vector space and its double dual, which is independent of the choice of a basis. We call this isomorphism the canonical isomorphism or natural isomorphism.
We begin with the definition of evaluation map.
Definition 5.6.6.
Let \(v\in V\text{.}\) The evaluation map is an \(F\)-linear function on \(V^*\text{:}\)
\begin{equation*}
\ev_v\colon V^*\to F\quad\text{is defined by}\quad f\mapsto f(v).
\end{equation*}
We often denote \(f(v)\) by \(\langle v,f\rangle\text{.}\) In this notation the map \(\varphi^*\) as defined in Exercise 4.3.5, will assume the following form.
\begin{equation*}
\langle v,\varphi^*(f)\rangle=\langle \varphi(v),f\rangle.
\end{equation*}
Proposition 5.6.7.
Let \(V\) be a finite-dimensional vector space over a field \(F\text{.}\) We have the following.- A vector \(v\in V\) is zero if and only if \(f(v)=0\) for every \(f\in V^*\text{.}\)
- If \(v\in V\) is such that \(f(v)\neq 0\) for some linear function \(f\in V^*\) then,\begin{equation*} V=\langle v\rangle\bigoplus\ker(f). \end{equation*}
- Consider nonzero linear functions \(f,g\in V^*\text{.}\) Then, \(\ker(f)=\ker(g)\) if and only if there exists a scalar \(\alpha\) such that \(f=\alpha g\text{.}\)
We now prove the following natural isomorphism between a vector space and its double dual.
Theorem 5.6.8.
Let \(V\) be a finite-dimensional vector space over a field. The \(F\)-linear map \(\ev\colon V\to V^{**}\) defined by
\begin{equation*}
v\mapsto\ev_v
\end{equation*}
is an \(F\)-isomorphism.Proof.
We obtain the following useful consequence of the Theorem 5.6.8.
Corollary 5.6.9.
We keep notations of Theorem 5.6.8. If \(\varphi\in\Hom_F(V^*,F)\) then there exists a unique vector \(v\in V\) such that
\begin{equation*}
\varphi=\ev_v\text{.}
\end{equation*}
