In linear algebra, given a vector space with a basis of vectors indexed by an index set (the cardinality of is the dimension of ), the dual set of is a set of vectors in the dual space with the same index set such that and form a biorthogonal system. The dual set is always linearly independent but does not necessarily span . If it does span , then is called the dual basis or reciprocal basis for the basis .
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 on a vector in the original space :
where is the Kronecker delta symbol.
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] For example,
where is the basis in a Cartesian frame. The components of can be found by
However, in a non-Cartesian frame, we do not necessarily have for all . However, it is always possible to find vectors in the dual space such that
The equality holds when the s are the dual basis of s. Notice the difference in position of the index .
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 for a finite subset 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 dimension (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.
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 and . 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.
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 with 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.
For example, the standard basis vectors of (the Cartesian plane) are
and the standard basis vectors of its dual space are
In 3-dimensional Euclidean space, for a given basis , the biorthogonal (dual) basis can be found 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