Summary
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 I 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. For example, where is the bases 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 a vector such that The equality holds when is the dual base of . Notice the difference in position of the index . In a Cartesian frame, we have 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 dimensionality (this being a greater infinite cardinality) than the original space has, and thus these cannot have a basis with the same indexing set.
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