# 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 and , 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*:

where is the Kronecker delta symbol.

## Contents

## A categorical and algebraic construction of the dual space[edit]

Another way to introduce the dual space of a vector space (module) is by introducing it in a categorical sense. To do this, let be a module defined over the ring (that is, is an object in the category ). Then we define the dual space of , denoted , to be , the module formed of all -linear module homomorphisms from into . Note then that we may define a dual to the dual, referred to as the double dual of , written as , and defined as .

To formally construct a basis for the dual space, we shall now restrict our view to the case where is a finite-dimensional free (left) -module, where is a ring of unity. Then, we assume that the set is a basis for . From here, we define the Kronecker Delta function over the basis by if and if . Then the set describes a linearly independent set with each . Since is finite-dimensional, the basis is of finite cardinality. Then, the set is a basis to and is a free (right) -module.

## Existence and uniqueness[edit]

The dual set always exists and gives an injection from *V* into *V*^{∗}, namely the mapping that sends *v _{i}* to

*v*. This says, in particular, that the dual space has dimension greater or equal to that of

^{i}*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*. If

_{i}*w*were a finite linear combination of the dual basis vectors

*v*, say for a finite subset

^{i}*K*of

*I*, then for any

*j*not in

*K*, , 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[edit]

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* = { *e*_{1}, …, *e*_{n} } and *B*^{∗} = { *e*^{1}, …, *e*^{n} }. If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:

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[edit]

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.^{[1]} E.g.,

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

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 **e**^{i} such that

- .

the equality holds when **e**^{i} is the dual base of **e**_{i}

In a Cartesian frame, we have

## Examples[edit]

For example, the standard basis vectors of **R**^{2} (the Cartesian plane) are

and the standard basis vectors of its dual space **R**^{2}* are

In 3-dimensional Euclidean space, for a given basis {**e**_{1}, **e**_{2}, **e**_{3}}, you can find the biorthogonal (dual) basis {**e**^{1}, **e**^{2}, **e**^{3}} by formulas below:

where ^{T} denotes the transpose and

is the volume of the parallelepiped formed by the basis vectors and

## See also[edit]

## Notes[edit]

**^**Lebedev, Cloud & Eremeyev 2010, p. 12.

## References[edit]

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