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Section 5.6 Natural isomorphism between a vector space and its double dual

We describe the “natural isomorphism” between a vector space and its double dual. The term “natural” can be defined precisely. However, we will not define this term in this course. We refer the reader to the chapter on Category Theory in the book ‘Basic Algebra II’ by N. Jacobson.

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 \(V^{**}=\Hom_F(V^*,F)\text{.}\)

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*}
An \(F\)-isomorphism between \(V\) and \(V^*\) can now be defined by
\begin{equation*} v_i\mapsto v_i^*. \end{equation*}
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).
Let \(V\) be a finite-dimensional vector space over a field \(F\) with an ordered basis \(\mathfrak{B}=(v_1,\ldots,v_n)\text{.}\) Consider a dual basis \(\mathfrak{B}^*=(v_1^*,\ldots,v_n^*)\) as in Definition 5.6.3. Let \(T\colon V\to V\) be an \(F\)-linear map. Suppose that \(T(v_j)=\sum_i a_{ij}v_i\text{,}\) i.e.,
\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*}

We now prove the following natural isomorphism between a vector space and its double dual.

First verify that the map \(v\mapsto\ev_v\) is an \(F\)-linear map. Let \(\{v_1,\ldots,v_n\}\) be a basis for \(V\) over \(F\text{.}\) It is left to the reader to verify that \(\{\ev_{v_1},\ev_{v_2},\ldots,\ev_{v_n}\}\) is a dual basis for \(V^*\) (follow a similar argument as in the proof of Proposition 5.6.1). This will prove the theorem.

We obtain the following useful consequence of the Theorem 5.6.8.

Let \(V=M_n(\R)\) be a vector space over \(\R\) of \(n\times n\) matrices. Show that for any \(\R\)-linear functional \(f\in\Hom_F(V,F)\) there exists a unique \(A\in V\) such that \(f(X)=tr(AX)\) for every \(X\in V\text{.}\)