Reciprocal Basis

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 dimension of ''V''), the dual set of ''B'' is a set ''B''^∗ of vectors in the dual space ''V''^∗ with the same index set ''I'' such that ''B'' and ''B''^∗ form a biorthogonal system. The dual set is always linearly independent but does not necessarily span ''V''^∗. If it does span ''V''^∗, then ''B''^∗ is called the dual basis or reciprocal basis for the basis ''B''.

Denoting the indexed vector sets as $B = \_$ and $B^ = \_$, 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 ''V''^∗ on a vector in the original space ''V'':
:$v^i\cdot v_j = \delta^i_j = \begin 1 & \text i = j\\ 0 & \text i \ne j\text \end$
where $\delta^i_j$ 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. E.g.,

: $\mathbf = x^1 \mathbf_1 + x^2 \mathbf_2 + x^3 \mathbf_3$

where $\mathbf_k$ is the bases in a Cartesian frame. The components of $\mathbf$ can be found by

: $x^k = \mathbf \cdot \mathbf_k.$

In a non-Cartesian frame, we do not necessarily have e_''i'' · for all . However, it is always possible to find a vector e^''i'' such that

: $x^i = \mathbf\cdot\mathbf^i \qquad (i = 1, 2, 3).$

The equality holds when e^''i'' is the dual base of e_''i''. Notice the difference in position of the index i.

In a Cartesian frame, we have $\mathbf^k = \mathbf_k = \mathbf_k.$

Existence and uniqueness

The dual set always exists and gives an injection from ''V'' into ''V''^∗, namely the mapping that sends ''v_i'' to ''vⁱ''. 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''^∗. For example, consider the map ''w'' in ''V''^∗ from ''V'' into the underlying scalars ''F'' given by for all ''i''. This map is clearly nonzero on all ''v_i''. If ''w'' were a finite linear combination of the dual basis vectors ''vⁱ'', say $w=\sum_\alpha_iv^i$ for a finite subset ''K'' of ''I'', then for any ''j'' not in ''K'', $w(v_j)=\left(\sum_\alpha_iv^i\right)\left(v_j\right)=0$, 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 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 and . If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:
:$\left\langle e^i, e_j \right\rangle = \delta^i_j.$

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''^∗, 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 $A$ be a module defined over the ring $R$ (that is, $A$ is an object in the category $R\text\mathbf$). Then we define the dual space of $A$, denoted $A^$, to be $\text_R(A,R)$, the module formed of all $R$-linear module homomorphisms from $A$ into $R$. Note then that we may define a dual to the dual, referred to as the double dual of $A$, written as $A^$, and defined as $\text_R(A^,R)$.

To formally construct a basis for the dual space, we shall now restrict our view to the case where $F$ is a finite-dimensional free (left) $R$-module, where $R$ is a ring with unity. Then, we assume that the set $X$ is a basis for $F$. From here, we define the Kronecker Delta function $\delta_$ over the basis $X$ by $\delta_=1$ if $x=y$ and $\delta_=0$ if $x\ne y$. Then the set $S = \lbrace f_x:F \to R \; , \; f_x(y)=\delta_ \rbrace$ describes a linearly independent set with each $f_x \in \text_R(F,R)$. Since $F$ is finite-dimensional, the basis $X$ is of finite cardinality. Then, the set $S$ is a basis to $F^\ast$ and $F^\ast$ is a free (right) $R$-module.

Examples

For example, the standard basis vectors of R² (the Cartesian plane) are
:$\left\ = \left\$

and the standard basis vectors of its dual space R²* are
:$\left\ = \left\\text$

In 3-dimensional Euclidean space, for a given basis , you can find the biorthogonal (dual) basis by formulas below:

:$\mathbf^1 = \left(\frac\right)^\mathsf,\ \mathbf^2 = \left(\frac\right)^\mathsf,\ \mathbf^3 = \left(\frac\right)^\mathsf.$

where denotes the transpose and

:$V \,=\, \left(\mathbf_1;\mathbf_2;\mathbf_3\right) \,=\, \mathbf_1\cdot(\mathbf_2\times\mathbf_3) \,=\, \mathbf_2\cdot(\mathbf_3\times\mathbf_1) \,=\, \mathbf_3\cdot(\mathbf_1\times\mathbf_2)$

is the volume of the parallelepiped formed by the basis vectors $\mathbf_1,\,\mathbf_2$ and $\mathbf_3.$

In general the dual basis of a basis in a finite dimensional vector space can be readily computed as follows: given the basis $f_1,\ldots,f_n$ and corresponding dual basis $f^1,\ldots,f^n$ we can build matrices
:$\begin F &= \beginf_1 & \cdots & f_n \end \\ G &= \beginf^1 & \cdots & f^n \end \end$

Then the defining property of the dual basis states that
:$G^\mathsfF = I$

Hence the matrix for the dual basis $G$ can be computed as
:$G = \left(F^\right)^\mathsf$

See also

* Reciprocal lattice
* Miller index
* Zone axis

Notes

References

*
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{{DEFAULTSORT:Dual Basis
Linear algebra

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