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 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 inner product equal to) 1 if the indexes are equal, and to zero otherwise. Symbolically, evaluating a dual vector in V^∗ on a vector in the original space V:

$v^i(v_j) = \delta^i_j = \begin{cases} 1 & \text{if } i = j\\ 0 & \text{if } i \ne j\text{,} \end{cases}$

where $\delta^i_j$ 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 $x\ne y$. 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 $w=\sum_{i\in K}\alpha_iv^i$ for a finite subset K of I, then for any j not in K, $w(v_j)=(\sum_{i\in K}\alpha_iv^i)(v_j)=0$, 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:

$\left\langle e^i, e_j \right\rangle = \delta^i_j.$

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.

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 simple dot product by the base vector.^[1]

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 Cartesian frame.The components of $\mathbf{x}$ can be found by

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

In 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 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

$\{\mathbf{e}_1, \mathbf{e}_2\} = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$

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

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

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:

$\mathbf{e}^1 = \left(\frac{\mathbf{e}_2\times\mathbf{e}_3}{V}\right)^\text{T},\ \mathbf{e}^2 = \left(\frac{\mathbf{e}_3\times\mathbf{e}_1}{V}\right)^\text{T},\ \mathbf{e}^3 = \left(\frac{\mathbf{e}_1\times\mathbf{e}_2}{V}\right)^\text{T}.$

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

1. Lebedev, Cloud & Eremeyev 2010, p. 12.

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