In linear algebra, given a vector space <math>V</math> with a basis <math>B</math> of vectors indexed by an index set <math>I</math> (the cardinality of <math>I</math> is the 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 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. 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 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 spaces, 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 fields such as the real numbers, the space of duals is a topological space, and this gives a homeomorphism between the Stiefel manifolds 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) 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 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
References
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