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## Summary

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), 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 $B=\{v_{i}\}_{i\in I}$ 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.

## 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 ei · ej = 0 for all ij. 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 ei. Notice the difference in position of the index i.

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

## Existence and uniqueness

The dual set always exists and gives an injection from V into V, namely the mapping that sends vi 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(vi) = 1 for all i. This map is clearly nonzero on all vi. If w were a finite linear combination of the dual basis vectors vi, say ${\textstyle w=\sum _{i\in K}\alpha _{i}v^{i}}$  for a finite subset K of I, then for any j not in K, ${\textstyle w(v_{j})=\left(\sum _{i\in K}\alpha _{i}v^{i}\right)\left(v_{j}\right)=0}$ , 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, …, en } 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.

## 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.

## 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 $\mathbf {e} _{3}.$

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{aligned}F&={\begin{bmatrix}f_{1}&\cdots &f_{n}\end{bmatrix}}\\G&={\begin{bmatrix}f^{1}&\cdots &f^{n}\end{bmatrix}}\end{aligned}}

Then the defining property of the dual basis states that

$G^{\mathsf {T}}F=I$

Hence the matrix for the dual basis $G$  can be computed as

$G=\left(F^{-1}\right)^{\mathsf {T}}$