Editing Dual basis (section)

==Examples==
For example, the [[standard basis]] vectors of <math>\R^2</math> (the [[Cartesian plane]]) are
:<math>
  \left\{\mathbf{e}_1, \mathbf{e}_2\right\} = \left\{
    \begin{pmatrix}
      1 \\
      0 
    \end{pmatrix},
    \begin{pmatrix}
      0 \\
     1 
    \end{pmatrix}
  \right\}
</math>

and the standard basis vectors of its dual space <math>(\R^2)^*</math> are
:<math>
  \left\{\mathbf{e}^1, \mathbf{e}^2\right \} = \left\{
    \begin{pmatrix}
      1 & 0 
    \end{pmatrix},
    \begin{pmatrix}
      0 & 1 
    \end{pmatrix}
    \right\}\text{.}
</math>

In 3-dimensional [[Euclidean space]], for a given basis <math>\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}</math>, the biorthogonal (dual) basis <math>\{\mathbf{e}^1, \mathbf{e}^2, \mathbf{e}^3\}</math> can be found by formulas below:

:<math>
  \mathbf{e}^1 = \left(\frac{\mathbf{e}_2 \times \mathbf{e}_3}{V}\right)^\mathsf{T},\ 
  \mathbf{e}^2 = \left(\frac{\mathbf{e}_3 \times \mathbf{e}_1}{V}\right)^\mathsf{T},\ 
  \mathbf{e}^3 = \left(\frac{\mathbf{e}_1 \times \mathbf{e}_2}{V}\right)^\mathsf{T}.
</math>
<!-- Maybe, this formula can illustrate, why dual basis is also called biorthogonal... -->

where {{sup|T}} denotes the [[transpose]] and

:<math>
  V                                                   \,=\,
  \left(\mathbf{e}_1;\mathbf{e}_2;\mathbf{e}_3\right) \,=\,
  \mathbf{e}_1\cdot(\mathbf{e}_2\times\mathbf{e}_3)   \,=\,
  \mathbf{e}_2\cdot(\mathbf{e}_3\times\mathbf{e}_1)   \,=\,
  \mathbf{e}_3\cdot(\mathbf{e}_1\times\mathbf{e}_2)
</math>

is the volume of the [[parallelepiped]] formed by the basis vectors <math>\mathbf{e}_1,\,\mathbf{e}_2</math> and <math>\mathbf{e}_3.</math>

In general the dual basis of a basis in a finite-dimensional vector space can be readily computed as follows: given the basis <math>f_1,\ldots,f_n</math> and corresponding dual basis <math>f^1,\ldots,f^n</math> we can build matrices
:<math>
\begin{align}
F &= \begin{bmatrix}f_1 & \cdots & f_n \end{bmatrix} \\
G &= \begin{bmatrix}f^1 & \cdots & f^n \end{bmatrix}
\end{align}
</math>

Then the defining property of the dual basis states that
:<math>G^\mathsf{T}F = I</math>

Hence the matrix for the dual basis <math>G</math> can be computed as
:<math>G = \left(F^{-1}\right)^\mathsf{T}</math>