Editing Dual basis

{{Short description|Linear algebra concept}}
In [[linear algebra]], given a [[vector space]] <math>V</math> with a [[Basis (linear algebra)|basis]] <math>B</math> of [[Vector (mathematics and physics)|vectors]] indexed by an [[index set]] <math>I</math> (the [[cardinality]] of <math>I</math> is the [[dimension (vector space)|dimension]] of <math>V</math>), the '''dual set''' of <math>B</math> is a set <math>B^*</math> of vectors in the [[dual space]] <math>V^*</math> with the same index set <math>I</math> such that <math>B</math> and <math>B^*</math> form a [[biorthogonal system]]. The dual set is always [[linearly independent]] but does not necessarily [[Linear span|span]] <math>V^*</math>. If it does span <math>V^*</math>, then <math>B^*</math> is called the '''dual basis''' or '''reciprocal basis''' for the basis <math>B</math>.

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 have an [[inner product]] equal to 1 if the indexes are equal, and equal to 0 otherwise. Symbolically, evaluating a dual vector in <math>V^*</math> on a vector in the original space <math>V</math>:
:<math>
v^i\cdot 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.

==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 the [[dot product]] of the vector and the base vector.{{sfn|Lebedev|Cloud|Eremeyev|2010|p=12}} For example,

: <math>\mathbf{x} = x^1 \mathbf{i}_1 + x^2 \mathbf{i}_2 + x^3 \mathbf{i}_3</math>

where <math>\{\mathbf{i}_1, \mathbf{i}_2, \mathbf{i}_3\}</math> is the basis in a Cartesian frame. The components of <math>\mathbf{x}</math> can be found by

: <math>x^k = \mathbf{x} \cdot \mathbf{i}_k.</math>

However, in a non-Cartesian frame, we do not necessarily have <math>\mathbf{e}_i\cdot\mathbf{e}_j=0</math> for all <math>i\neq j</math>. However, it is always possible to find vectors <math>\mathbf{e}^i</math> in the dual space such that

: <math>x^i = \mathbf{e}^i(\mathbf{x}) \qquad  (i = 1, 2, 3).</math>

The equality holds when the <math>\mathbf{e}^i</math>s are the dual basis of <math>\mathbf{e}_i</math>s. Notice the difference in position of the index <math>i</math>.

==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 display="inline">w=\sum_{i\in K}\alpha_iv^i</math> for a finite subset ''K'' of ''I'', then for any ''j'' not in ''K'', <math display="inline">w(v_j)=\left(\sum_{i\in K}\alpha_iv^i\right)\left(v_j\right)=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 dimension (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 <math>B=\{e_1,\dots,e_n\}</math> and <math>B^*=\{e^1,\dots,e^n\}</math>. 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.

==A categorical and algebraic construction of the dual space==

Another way to introduce the dual space of a vector space ([[Module (mathematics)|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(A,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 with 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.

==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>

==See also==

* [[Reciprocal lattice]]
* [[Miller index]]
* [[Zone axis]]

==Notes==
{{reflist}}

==References==
* {{cite book |title=Tensor Analysis With Applications to Mechanics |last1=Lebedev |first1=Leonid P. |last2=Cloud |first2=Michael J. |last3=Eremeyev |first3=Victor A. |year=2010 |publisher=World Scientific |isbn=978-981431312-4 }}
* {{cite web |title=Finding the Dual Basis |work=[[Stack Exchange]] |date=May 27, 2012 |url=https://math.stackexchange.com/q/150526 }}

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[[Category:Linear algebra]]

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