# 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*), 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 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

## IntroductionEdit

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

## Existence and uniquenessEdit

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 spacesEdit

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.

## A categorical and algebraic construction of the dual spaceEdit

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.

## ExamplesEdit

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

In general the dual basis of a basis in a finite dimensional vector space can be readily computed as follows: given the basis and corresponding dual basis we can build matrices

Then the defining property of the dual basis states that

Hence the matrix for the dual basis can be computed as

## See alsoEdit

## NotesEdit

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

## ReferencesEdit

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