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In [[linear algebra]], given a vector space ''V'' with a basis ''B'' of vectors indexed by an [[index set]] ''I'' (the [[cardinality]] of ''I'' is the dimensionality of ''V''), its '''dual set''' is a set ''B''<sup>∗</sup> of vectors in the [[dual space]] ''V''<sup>∗</sup> with the same index set ''I'' such that ''B'' and ''B''<sup>∗</sup> form a [[biorthogonal system]]. The dual set is always [[linearly independent]] but does not necessarily span ''V''<sup>∗</sup>. If it does span ''V''<sup>∗</sup>, then ''B''<sup>∗</sup> is called the '''dual basis''' for the basis ''B''.
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Denoting the indexed vector sets as <math>B = \{v_i\}_{i\in I}</math> and <math>B^{*} = \{v^i\}_{i \in I},</math> being biorthogonal means that the elements pair to 1 if the indexes are equal, and to zero otherwise. Symbolically, evaluating a dual vector in ''V''<sup>∗</sup> on a vector in the original space ''V'':
:<math>
v^i(v_j) = \delta^i_j =
\begin{cases}
  1 & \text{if } i = j\\
  0 & \text{if } i \ne j\text{,}
\end{cases}
</math>
where <math>\delta^i_j</math> is the [[Kronecker delta]] symbol.
 
==A Categorical and Algebraic Construction of the Dual Space==
 
Another way to introduce the dual space of a vector space (module) is by introducing it in a categorical sense. To do this, let <math>A</math> be a module defined over the ring <math>R</math> (that is, <math>A</math> is an object in the category <math>R\text{-}\mathbf{Mod}</math>). Then we define the dual space of <math>A</math>, denoted <math>A^{\ast}</math>, to be <math>\text{Hom}_R(V,R)</math>, the module formed of all <math>R</math>-linear module homomorphisms from <math>A</math> into <math>R</math>. Note then that we may define a dual to the dual, referred to as the double dual of <math>A</math>, written as <math>A^{\ast\ast}</math>, and defined as <math>\text{Hom}_R(A^{\ast},R)</math>.
 
To formally construct a basis for the dual space, we shall now restrict our view to the case where <math>F</math> is a finite dimensional free (left) <math>R</math>-module, where <math>R</math> is a ring of unity. Then, we assume that the set <math>X</math> is a basis for <math>F</math>. From here, we define the Kronecker Delta function <math>\delta_{xy}</math> over the basis <math>X</math> by <math>\delta_{xy}=1</math> if <math>x=y</math> and <math>\delta_{xy}=0</math> if <math>x\ne y</math>. Then the set <math> S = \lbrace f_x:F \to R \; | \; f_x(y)=\delta_{xy} \rbrace </math> describes a linearly independent set with each <math>f_x \in \text{Hom}_R(F,R)</math>. Since <math>F</math> is finite dimensional, the basis <math>X</math> is of finite cardinality. Then, the set <math> S </math> is a basis to <math>F^\ast</math> and <math>F^\ast</math> is a free (right) <math>R</math>-module.
 
==Existence and uniqueness==
The dual set always exists and gives an injection from ''V'' into ''V''<sup>∗</sup>, namely the mapping that sends ''v<sub>i</sub>'' to ''v<sup>i</sup>''. This says, in particular, that the dual space has dimension greater or equal to that of ''V''.
 
However, the dual set of an infinite dimensional ''V'' does not span its dual space ''V''<sup>∗</sup>. For example, consider the map ''w'' in ''V''<sup>∗</sup> from ''V'' into the underlying scalars ''F'' given by {{nowrap|1=''w''(''v<sub>i</sub>'') = 1}} for all ''i''. This map is clearly nonzero on all ''v<sub>i</sub>''. If ''w'' were a finite linear combination of the dual basis vectors ''v<sup>i</sup>'', say <math>w=\sum_{i\in K}\alpha_iv^i</math> for a finite subset ''K'' of ''I'', then for any ''j'' not in ''K'', <math>w(v_j)=(\sum_{i\in K}\alpha_iv^i)(v_j)=0</math>, contradicting the definition of ''w''. So, this ''w'' does not lie in the span of the dual set.
 
The dual of an infinite-dimensional space has greater dimensionality (this being a greater infinite cardinality) than the original space has, and thus these cannot have a basis with the same indexing set. However, a dual set of vectors exists, which defines a subspace of the dual isomorphic to the original space. Further, for [[topological vector space]]s, a [[continuous dual space]] can be defined, in which case a dual basis may exist.
 
;Finite-dimensional vector spaces
 
In the case of finite-dimensional vector spaces, the dual set is always a dual basis and it is unique. These bases are denoted by {{nowrap|1=''B'' = { ''e''<sub>1</sub>, …, ''e''<sub>''n''</sub> }<nowiki/>}} and {{nowrap|1=''B''<sup>∗</sup> = { ''e''<sup>1</sup>, …, ''e''<sup>''n''</sup> }<nowiki/>}}. If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:
:<math>
\left\langle e^i, e_j \right\rangle = \delta^i_j.
</math>
 
The association of a dual basis with a basis gives a map from the space of bases of ''V'' to the space of bases of ''V''<sup>∗</sup>, and this is also an isomorphism. For [[topological field]]s such as the real numbers, the space of duals is a [[topological space]], and this gives a [[homeomorphism]] between the [[Stiefel manifold]]s of bases of these spaces.
 
==Introduction==
To perform operations with a vector, we must have a straightforward method of calculating its components. In a Cartesian frame the necessary operation is simple dot product by the base vector.{{sfn|Lebedev|Cloud|Eremeyev|2010|p=12}}
 
e.g. <math>\mathbf{x}=x^1 \mathbf{i}_1+x^2 \mathbf{i}_2+x^3 \mathbf{i}_3</math> 
 
where <math>\mathbf{i}_k</math> is the bases in Cartesian frame.The components of <math>\mathbf{x}</math> can be found by
 
<math>x^k=\mathbf{x}\cdot\mathbf{i}_k</math>
 
In non-Cartesian frame, we do not necessarily have <var>'''e'''</var><sub>i</sub>·<var>'''e'''</var><sub>j</sub>=0 for all i≠j. However, it may still be possible to find a vector <var>'''e'''</var><sup>i</sup> such that
 
<math>x^i=\mathbf{x}\cdot\mathbf{e}^i \qquad  (i=1,2,3)</math>.
 
the equality holds when <var>'''e'''</var><sup>i</sup> is the dual base of <var>'''e'''</var><sub>i</sub>
 
In Cartesian frame, we have <math>\mathbf{e}^k=\mathbf{e}_k=\mathbf{i}_k</math>
 
==Examples==
For example, the standard basis vectors of '''R'''<sup>2</sup> (the [[Cartesian plane]]) are
:<math>
\{\mathbf{e}_1, \mathbf{e}_2\} = \left\{
\begin{pmatrix}
  1 \\
  0
\end{pmatrix},
\begin{pmatrix}
  0 \\
  1
\end{pmatrix}
\right\}
</math>
and the standard basis vectors of its dual space '''R'''<sup>2</sup>* are
:<math>
\{\mathbf{e}^1, \mathbf{e}^2\} = \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 {'''e'''<sub>1</sub>, '''e'''<sub>2</sub>, '''e'''<sub>3</sub>}, you can find the biorthogonal (dual) basis {'''e'''<sup>1</sup>, '''e'''<sup>2</sup>, '''e'''<sup>3</sup>} by formulas below:
 
:<math>
\mathbf{e}^1 = \left(\frac{\mathbf{e}_2\times\mathbf{e}_3}{V}\right)^\text{T},\
\mathbf{e}^2 = \left(\frac{\mathbf{e}_3\times\mathbf{e}_1}{V}\right)^\text{T},\
\mathbf{e}^3 = \left(\frac{\mathbf{e}_1\times\mathbf{e}_2}{V}\right)^\text{T}.
</math>
<!-- Maybe, this formula can illustrate, why dual basis is also called biorthogonal... -->
 
where {{math|{{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>
 
==See also==
* [[Dual lattice]]
 
==References==
{{Reflist}}
{{refbegin}}
* {{cite isbn|9789814313124}}
{{refend}}
 
{{DEFAULTSORT:Dual Basis}}
[[Category:Linear algebra]]
 
[[he:מרחב דואלי#הבסיס הדואלי]]
[[zh:对偶基]]

Latest revision as of 18:26, 18 December 2014

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