Dual 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 dimensionality of V), its dual set 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 $B=\{v_{i}\}_{i\in I}$ and $B^{*}=\{v^{i}\}_{i\in I}$, 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:

$v^{i}\cdot v_{j}=\delta _{j}^{i}={\begin{cases}1&{\text{if }}i=j\\0&{\text{if }}i\neq j{\text{,}}\end{cases}}$

where $\delta _{j}^{i}$ 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 $A$ be a module defined over the ring $R$ (that is, $A$ is an object in the category $R{\text{-}}\mathbf {Mod}$). Then we define the dual space of $A$, denoted $A^{\ast }$, to be ${\text{Hom}}_{R}(A,R)$, 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 $A^{\ast \ast }$, and defined as ${\text{Hom}}_{R}(A^{\ast },R)$.

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 of unity. Then, we assume that the set $X$ is a basis for $F$. From here, we define the Kronecker Delta function $\delta _{xy}$ over the basis $X$ by $\delta _{xy}=1$ if $x=y$ and $\delta _{xy}=0$ if $x\neq y$. Then the set $S=\lbrace f_{x}:F\to R\;|\;f_{x}(y)=\delta _{xy}\rbrace$ describes a linearly independent set with each $f_{x}\in {\text{Hom}}_{R}(F,R)$. 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.

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 w(v_i) = 1 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 $w=\sum _{i\in K}\alpha _{i}v^{i}$ for a finite subset K of I, then for any j not in K, $w(v_{j})=\left(\sum _{i\in K}\alpha _{i}v^{i}\right)\left(v_{j}\right)=0$, 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 B = { e₁, …, e_n } and B^∗ = { e¹, …, eⁿ }. If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:

$\left\langle e^{i},e_{j}\right\rangle =\delta _{j}^{i}.$

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.

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.,

$\mathbf {x} =x^{1}\mathbf {i} _{1}+x^{2}\mathbf {i} _{2}+x^{3}\mathbf {i} _{3}$

where $\mathbf {i} _{k}$ is the bases in a Cartesian frame.The components of $\mathbf {x}$ can be found by

$x^{k}=\mathbf {x} \cdot \mathbf {i} _{k}$

In a non-Cartesian frame, we do not necessarily have e_i · e_j = 0 for all i ≠ j. However, it is always possible to find a vector eⁱ such that

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

the equality holds when eⁱ is the dual base of e_i

In a Cartesian frame, we have $\mathbf {e} ^{k}=\mathbf {e} _{k}=\mathbf {i} _{k}$

Examples

For example, the standard basis vectors of R² (the Cartesian plane) are

$\left\{\mathbf {e} _{1},\mathbf {e} _{2}\right\}=\left\{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\end{pmatrix}}\right\}$

and the standard basis vectors of its dual space R²* are

$\left\{\mathbf {e} ^{1},\mathbf {e} ^{2}\right\}=\left\{{\begin{pmatrix}1&0\end{pmatrix}},{\begin{pmatrix}0&1\end{pmatrix}}\right\}{\text{.}}$

In 3-dimensional Euclidean space, for a given basis {e₁, e₂, e₃}, you can find the biorthogonal (dual) basis {e¹, e², e³} by formulas below:

$\mathbf {e} ^{1}=\left({\frac {\mathbf {e} _{2}\times \mathbf {e} _{3}}{V}}\right)^{\mathsf {T}},\ \mathbf {e} ^{2}=\left({\frac {\mathbf {e} _{3}\times \mathbf {e} _{1}}{V}}\right)^{\mathsf {T}},\ \mathbf {e} ^{3}=\left({\frac {\mathbf {e} _{1}\times \mathbf {e} _{2}}{V}}\right)^{\mathsf {T}}.$

where ^T denotes the transpose and

$V\,=\,\left(\mathbf {e} _{1};\mathbf {e} _{2};\mathbf {e} _{3}\right)\,=\,\mathbf {e} _{1}\cdot (\mathbf {e} _{2}\times \mathbf {e} _{3})\,=\,\mathbf {e} _{2}\cdot (\mathbf {e} _{3}\times \mathbf {e} _{1})\,=\,\mathbf {e} _{3}\cdot (\mathbf {e} _{1}\times \mathbf {e} _{2})$

is the volume of the parallelepiped formed by the basis vectors $\mathbf {e} _{1},\,\mathbf {e} _{2}$ and $\mathbf {e} _{3}.$

See also

• Reciprocal lattice
• Miller index
• Zone axis

References

• Lebedev, Leonid P.; Cloud, Michael J.; Eremeyev, Victor A. (2010). Tensor Analysis With Applications to Mechanics. World Scientific. ISBN 978-981431312-4.