Introduction
Dual spaces have meaningful applications in several areas in mathematics likes differential geometry and functional analysis. In this chapter I will introduce some important concepts regarding dual spaces.
Linear Form
A linear form, also linear functional, one form or covariant vector, is a linear map in \(\mathcal{L}(\mathcal{V}, \mathbb{F})\) where \(\mathcal V (\mathbb{F})\) is a vector space.
Example 1: Let \(\mathcal{M}_n(\mathbb{R})\) be the vector space of all n x n matrices over \(\mathbb{R}\).
Then the trace of a matrix is a linear form:
Then we have \(\forall \ A, B \in \mathcal{M}_n(\mathbb{R}) \land \forall \ \lambda \in \mathbb{R}\):
(1) $\quad \textbf{Tr}(A + B) = \textbf{Tr}(A) + \textbf{Tr}(B) $
(2) $\quad \lambda \textbf{Tr}(A) = \textbf{Tr}(\lambda A)$
(3) $\quad \textbf{Tr} \in \mathcal{L}(\mathcal{M}_n(\mathbb{R}), \mathbb{R})$
Example 2: Let \(M\) be an n-dimensional smooth manifold and \(\phi : U \subset M \rightarrow \mathbb{R}^{n}\) be a coordinate chart where \(U\) is open and contains \(p\). More over, denote \(T_pM\) as the tangent space of \(M\) at \(p\). Let \(x^{1}, ..., x^{n}\) be the components of \(\phi\). These components are real valued functions:
\[x^{i} : U \longrightarrow \mathbb{R} \ , \quad i = 1, ..., n\]Then their derivative at a point \(p \in U\) are linear functionals:
\[dx^{i}(p) : T_p M \longrightarrow \mathbb{R} \ , \quad i = 1, ..., n\]
(1) $\quad dx^{i}(p)(\mu v_1 + \lambda v_2) = \mu \ dx^{i}(p) \ v_1 + \lambda \ dx^{i}(p) \ v_2 $
(2) $\quad dx^{i}(p) \in \mathcal{L}(T_pM, \mathbb{R})$
Dual Space
A dual space of a vector space \(\mathcal{V}(\mathbb{F})\) is defined as \(\mathcal{V}^{*} := \mathcal{L}(\mathcal{V}, \mathbb{F})\). The dual space \(\mathcal{V}^{*}\) is by itself a vector space. That means its vectors are linear forms.
Representation Matrix:
Let \(\mathcal{B}_1 = \{ b_1, ..., b_n \}\) be a basis of \(\mathcal{V}(\mathbb{F})\) and \(\mathcal{B}_2 = \{ 1 \}\) a basis of \(\mathcal{V}^{*}\).
For \(f \in \mathcal{V}^{*}\) its representation matrix is :
Coordinate Mapping:
Let \(v = \sum_{i=1}^{n} \lambda_i b_i \in \mathcal{V}\) then we have due to (1) and because \(\mathbb{F}^{1, 1} \cong \mathbb{F}\):
...
Dual Basis
Let \(\mathcal V (\mathbb{F})\) be a vector space over a field \(\mathbb{F}\). Then for every basis \(\mathcal{B} = \{ b_1, ..., b_n \}\) of \(\mathcal V\) there exists a unique dual basis \(\mathcal{B}^{*} = \{ \varphi^1, ..., \varphi^n \}\) of \(\mathcal{V}^{*}\) that is defined by:
\[\varphi^{i}(v_j) = \delta_{i, j} \ ; \qquad i, j = 1, ..., n\]Furthermore, if \(f \in \mathcal{V}^{*}\) we can write this linear form in terms of its basis:
\[f = \sum_{i=1}^{n} f(b_i) \ \varphi^i\]
Let $ \mathbb{R}[t]_{\leq 2} $ be the vector space of polynomials with degree at most 2.
Then the dual basis of $\mathcal{B} = \{ b_1, b_2, b_3 \} $ given by:
$$ \begin{align}
&b_1 = 3 - 5t + 3t²/2 \ ;
\qquad b_2 = -3/2 + 4t - 3t²/2 \ ;
\qquad b_3 = -1/3 -t + t²/2
\end{align}$$
$$ \begin{align} \mathcal{B}^{*} = \left\{ \ \varphi^i \quad \bigg\vert \quad \varphi^i(p(t)) = \int_{0}^{i} p(t) dt \quad \land \quad p(t) \in \mathbb{R}[t]_{\leq 2} \ \right\}_{i=1}^{3}
\end{align} $$
For $j=1, 2, 3$ let $ b_j = \sum_{k=0}^{2} \alpha_{j, k} t^{k} $ where $\{ \alpha_{j, k} \}_{k=0}^{2}$ are the coordinates of $ b_j $ w.r.t. the standard basis of $ \mathbb{R}[t]_{\leq 2} $ then we get in total:
$$ \varphi^{i}(b_j) = \int_{0}^{i} \sum_{k=0}^{2} \alpha_{j, k} t^{k} dt = \sum_{k=0}^{2} \alpha_{j, k} \int_{0}^{i} t^{k} dt \ \ ; \qquad i = 1, 2, 3$$
In that way we can rewrite those integrals as a matrix product:
$$ \begin{align}
\begin{bmatrix}
\varphi^1(b_1) & \varphi^2(b_1) & \varphi^3(b_1) \\
\varphi^1(b_2) & \varphi^2(b_2) & \varphi^3(b_2) \\
\varphi^1(b_3) & \varphi^2(b_3) & \varphi^3(b_3)
\end{bmatrix}
&=
\begin{bmatrix}
\alpha_{1, 1} & \alpha_{1, 2} & \alpha_{1, 3} \\
\alpha_{2, 1} & \alpha_{2, 2} & \alpha_{2, 3} \\
\alpha_{3, 1} & \alpha_{3, 2} & \alpha_{3, 3}
\end{bmatrix}
\begin{bmatrix}
t \vert_{0}^{1} & t \vert_{0}^{2} & t \vert_{0}^{3} \\
\frac{t²}{2} \vert_{0}^{1} & \frac{t²}{2} \vert_{0}^{2} & \frac{t²}{2} \vert_{0}^{3} \\
\frac{t³}{3} \vert_{0}^{1} & \frac{t³}{3} \vert_{0}^{2} & \frac{t³}{3} \vert_{0}^{2}
\end{bmatrix} \\
\\
&=
\begin{bmatrix}
3 & -5 & 3/2 \\
-3/2 & 4 & -3/2 \\
-1/3 & -1 & 1/2
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 3 \\
1/2 & 2 & 9/2 \\
1/3 & 8/3 & 9
\end{bmatrix} \\
\\
&=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\end{align} $$
Therefore, $\mathcal{B}^{*}$ is dual to $\mathcal{B}$.
Dual Map
Let \(\mathcal{V}(\mathbb{F})\) and \(\mathcal{W}(\mathbb{F})\) be two vector spaces and \(f \in \mathcal{L}(\mathcal{V}, \mathcal{W})\) then the dual map of \(f\) is defined by:
\[f^{*} \ : \mathcal{W}^{*} \longrightarrow \mathcal{V}^{*} \ ; \quad h \longmapsto f^{*}(h) = h \circ f\]
Bilinear form
Definition
Let \(\mathcal{V}(\mathbb{F})\) and \(\mathcal{W}(\mathbb{F})\) be two vector spaces then a bilinear form \(\beta : \mathcal{V} \times \mathcal{W} \longrightarrow \mathbb{F}\) is defined by:
\(\forall \ \ v, v_1, v_2 \in \mathcal{V} \ \land \ w, w_1, w_2 \in \mathcal{W} \ \land \ \mu \in \mathbb{F}\) :
(1) $\quad \beta(v_1 + v_2, w) = \beta(v_1, w) + \beta(v_2, w)$
(2) $\quad \beta(v, w_1 + w_2) = \beta(v, w_1) + \beta(v, w_2)$
(3) $\quad \beta(\mu v, w) = \mu \beta(v, w) = \beta(v, \mu w) $
This definition gives rise to the following natural mappings:
(1) $\quad \beta_L : \mathcal{V} \rightarrow \mathcal{W}^{*} \ ; \quad v \mapsto \beta_v = \beta(v, \cdot)$
(2) $\quad \beta_R : \mathcal{W} \rightarrow \mathcal{V}^{*} \ ; \quad w \mapsto \beta_w = \beta(\cdot, w)$
...
Special bilinear forms
1) Symmetric bilinear form:
Let \(\beta : \mathcal{V} \times \mathcal{V} \rightarrow \mathbb{F}\) then a symmetric bilinear form on \(\mathcal{V}\) is defined by:
...
2) Non-degenerate bilinear form:
Let \(\beta : \mathcal{V} \times \mathcal{W} \longrightarrow \mathbb{F}\) be a bilinear form.
I) \(\beta\) is considered to be non-degenerate in its left variable iff:
II) \(\beta\) is considered to be non-degenerate in its right variable iff:
III) Iff I) and II) holds true for \(\beta\) then: a) \(\ \beta\) is non-degenerate.
and if \(\mathcal{V} , \ \mathcal{W}\) are finite dimensional: b) \(\ \textbf{dim}(\mathcal{V}) = \textbf{dim}(\mathcal{W})\)
To proof this statement I will show that:
$$ \begin{align}
\text{I.} \quad \textbf{dim}(\mathcal{V}) \leq \textbf{dim}(\mathcal{W}) \\
\text{II.} \quad \textbf{dim}(\mathcal{W}) \leq \textbf{dim}(\mathcal{V}) \end{align} $$
From this follows that $\textbf{dim}(\mathcal{V}) = \textbf{dim}(\mathcal{W})$.
Proof of $\text{I}$
Since $\beta$ is non-degenerate $ \textbf{kern}(\beta_L) = \{ 0_{\mathcal{V}} \} $ and therefore $\textbf{dim}(\textbf{kern}(\beta_L)) = 0$.
More over, note that the $\textbf{img}(\beta_L)$ is a vector subspace of $\mathcal{W}^{*}$. This implies that $ \textbf{dim}(\textbf{img}(\beta_L)) \leq \textbf{dim}(\mathcal{W}^{*})$. From this property and the dimensionality theorem for linear transformations the following applies:
$$ \begin{align}
\textbf{dim}(\mathcal{V}) = \textbf{dim}(\textbf{img}(\beta_L)) + \textbf{dim}(\textbf{kern}(\beta_L)) = \textbf{dim}(\textbf{img}(\beta_L)) + 0 \leq \textbf{dim}(\mathcal{W}^{*})
\end{align}$$
From the condition that the vector spaces are finite dimensional it holds true that $\mathcal{W}^{*} \cong \mathcal{W}$ and we obtain the desired result:
$$ \textbf{dim}(\mathcal{V}) \leq \textbf{dim}(\mathcal{W}^{*}) = \textbf{dim}(\mathcal{W}) $$
Proof of $\text{II}$
This proof is analogous to the previous proof just exchange $\mathcal{V}$ and $\mathcal{W}$ and replace $\beta_L$ with $\beta_R$. $\qquad \qquad \qquad \square$
3) Dual system:
Let $\mathcal{V}$ and $\mathcal{W}$ be two vector spaces and $ \beta: \mathcal{V} \times \mathcal{W} \rightarrow \mathbb{F} $ a bilinear form. A dual system or dual pair over a field $\mathbb{F}$ is defined by:
$\quad$A triple $(\mathcal{V}, \mathcal{W}, \beta)$| $\quad \beta$ is non-degenerate|
Properties
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