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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'':
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| :<math>
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| v^i(v_j) = \delta^i_j =
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| \begin{cases}
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| 1 & \text{if } i = j\\
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| 0 & \text{if } i \ne j\text{,}
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| \end{cases}
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| </math>
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| where <math>\delta^i_j</math> is the [[Kronecker delta]] symbol.
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| ==A Categorical and Algebraic Construction of the Dual Space==
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| 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>.
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| 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.
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| ==Existence and uniqueness==
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| 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''. | |
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| 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.
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| 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.
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| ;Finite-dimensional vector spaces
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| 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:
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| :<math>
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| \left\langle e^i, e_j \right\rangle = \delta^i_j.
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| </math>
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| 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.
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| ==Introduction==
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| 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}}
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| e.g. <math>\mathbf{x}=x^1 \mathbf{i}_1+x^2 \mathbf{i}_2+x^3 \mathbf{i}_3</math>
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| where <math>\mathbf{i}_k</math> is the bases in Cartesian frame.The components of <math>\mathbf{x}</math> can be found by
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| <math>x^k=\mathbf{x}\cdot\mathbf{i}_k</math>
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| 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
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| <math>x^i=\mathbf{x}\cdot\mathbf{e}^i \qquad (i=1,2,3)</math>.
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| the equality holds when <var>'''e'''</var><sup>i</sup> is the dual base of <var>'''e'''</var><sub>i</sub>
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| In Cartesian frame, we have <math>\mathbf{e}^k=\mathbf{e}_k=\mathbf{i}_k</math>
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| ==Examples==
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| For example, the standard basis vectors of '''R'''<sup>2</sup> (the [[Cartesian plane]]) are
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| :<math>
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| \{\mathbf{e}_1, \mathbf{e}_2\} = \left\{
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| \begin{pmatrix}
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| 1 \\
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| 0
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| \end{pmatrix},
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| \begin{pmatrix}
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| 0 \\
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| 1
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| \end{pmatrix}
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| \right\}
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| </math>
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| and the standard basis vectors of its dual space '''R'''<sup>2</sup>* are | |
| :<math>
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| \{\mathbf{e}^1, \mathbf{e}^2\} = \left\{
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| \begin{pmatrix}
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| 1 & 0
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| \end{pmatrix},
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| \begin{pmatrix}
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| 0 & 1
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| \end{pmatrix}
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| \right\}\text{.}
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| </math>
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| 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:
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| :<math>
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| \mathbf{e}^1 = \left(\frac{\mathbf{e}_2\times\mathbf{e}_3}{V}\right)^\text{T},\
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| \mathbf{e}^2 = \left(\frac{\mathbf{e}_3\times\mathbf{e}_1}{V}\right)^\text{T},\
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| \mathbf{e}^3 = \left(\frac{\mathbf{e}_1\times\mathbf{e}_2}{V}\right)^\text{T}.
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| </math>
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| <!-- Maybe, this formula can illustrate, why dual basis is also called biorthogonal... -->
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| where {{math|{{sup|T}}}} denotes the [[transpose]] and
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| :<math>V \,=\,
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| \left(\mathbf{e}_1;\mathbf{e}_2;\mathbf{e}_3\right)\,=\,
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| \mathbf{e}_1\cdot(\mathbf{e}_2\times\mathbf{e}_3) \,=\,
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| \mathbf{e}_2\cdot(\mathbf{e}_3\times\mathbf{e}_1) \,=\,
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| \mathbf{e}_3\cdot(\mathbf{e}_1\times\mathbf{e}_2)
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| </math>
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| 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>
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| ==See also==
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| * [[Dual lattice]]
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| ==References==
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| {{Reflist}}
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| {{refbegin}}
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| * {{cite isbn|9789814313124}}
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| {{refend}}
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| {{DEFAULTSORT:Dual Basis}}
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| [[Category:Linear algebra]]
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| [[he:מרחב דואלי#הבסיס הדואלי]]
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| [[zh:对偶基]]
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