Editing Linear form (section)

==Relationship to bases==
{{hatnote|Below, we assume that the dimension is finite.  For a discussion of analogous results in infinite dimensions, see [[Schauder basis]].}}

===Basis of the dual space===
Let the vector space {{mvar|V}} have a basis <math>\mathbf{e}_1, \mathbf{e}_2,\dots,\mathbf{e}_n</math>, not necessarily [[orthogonal]]. Then the [[dual space]] <math>V^*</math> has a basis <math>\tilde{\omega}^1,\tilde{\omega}^2,\dots,\tilde{\omega}^n</math> called the [[dual basis]] defined by the special property that
<math display="block"> \tilde{\omega}^i (\mathbf e_j) = \begin{cases}
1 &\text{if}\ i = j\\
0 &\text{if}\ i \neq j.
\end{cases} </math>

Or, more succinctly,
<math display="block"> \tilde{\omega}^i (\mathbf e_j) = \delta_{ij} </math>

where <math>\delta_{ij}</math> is the [[Kronecker delta]]. Here the superscripts of the basis functionals are not exponents but are instead [[Covariance and contravariance of vectors|contravariant]] indices.

A linear functional <math>\tilde{u}</math> belonging to the dual space <math>\tilde{V}</math> can be expressed as a [[linear combination]] of basis functionals, with coefficients ("components") {{math|''u<sub>i</sub>''}},
<math display="block">\tilde{u} = \sum_{i=1}^n u_i \, \tilde{\omega}^i. </math>

Then, applying the functional <math>\tilde{u}</math> to a basis vector <math>\mathbf{e}_j</math> yields
<math display="block">\tilde{u}(\mathbf e_j) = \sum_{i=1}^n \left(u_i \, \tilde{\omega}^i\right) \mathbf e_j = \sum_i u_i \left[\tilde{\omega}^i \left(\mathbf e_j\right)\right] </math>

due to linearity of scalar multiples of functionals and pointwise linearity of sums of functionals. Then
<math display="block">\begin{align}
  \tilde{u}({\mathbf e}_j)
&= \sum_i u_i \left[\tilde{\omega}^i \left({\mathbf e}_j\right)\right] \\&
= \sum_i u_i {\delta}_{ij} \\
&= u_j.
\end{align}</math>

So each component of a linear functional can be extracted by applying the functional to the corresponding basis vector.

=== The dual basis and inner product ===
When the space {{mvar|V}} carries an [[inner product]], then it is possible to write explicitly a formula for the dual basis of a given basis. Let {{mvar|V}} have (not necessarily orthogonal) basis <math>\mathbf{e}_1,\dots, \mathbf{e}_n.</math> In three dimensions ({{math|1=''n'' = 3}}), the dual basis can be written explicitly
<math display="block"> \tilde{\omega}^i(\mathbf{v}) = \frac{1}{2} \left\langle \frac { \sum_{j=1}^3\sum_{k=1}^3\varepsilon^{ijk} \, (\mathbf e_j \times \mathbf e_k)} {\mathbf e_1 \cdot \mathbf e_2 \times \mathbf e_3} , \mathbf{v} \right\rangle ,</math>
for <math>i = 1, 2, 3,</math> where ''ε'' is the [[Levi-Civita symbol]] and <math>\langle \cdot , \cdot \rangle</math> the inner product (or [[dot product]]) on {{mvar|V}}.

In higher dimensions, this generalizes as follows
<math display="block"> \tilde{\omega}^i(\mathbf{v}) = \left\langle \frac{\sum_{1 \le i_2 < i_3 < \dots < i_n \le n} \varepsilon^{ii_2\dots i_n}(\star \mathbf{e}_{i_2} \wedge \cdots \wedge \mathbf{e}_{i_n})}{\star(\mathbf{e}_1\wedge\cdots\wedge\mathbf{e}_n)}, \mathbf{v} \right\rangle ,</math>
where <math>\star</math> is the [[Hodge star operator]].