Dual basis

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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 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:

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

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 ${\text{Hom}}_{R}(V,R)$ , 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, $w(v_{j})=(\sum _{{i\in K}}\alpha _{i}v^{i})(v_{j})=0$ , 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:

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

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 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 <var>e</var>_i·<var>e</var>_j=0 for all i≠j. However, it may still be possible to find a vector <var>e</var>ⁱ such that

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

the equality holds when <var>e</var>ⁱ is the dual base of <var>e</var>_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 ${\mathbf {e}}_{3}.$

References

↑ 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.

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