The bidual of a normed vector space is the continuous dual . Evaluation defines the canonical embedding into the bidual . The norm inequality gives , and a norming functional from the Hahn-Banach theorem gives equality. Thus is isometric, and its image is dense in its closure, a completion of a normed space.
For finite-dimensional vector spaces over a field , a bilinear pairing is perfect when the induced linear maps and are isomorphisms. Equivalently, both dimensions agree and the matrix of the pairing in any two bases is invertible. The evaluation pairing between a finite-dimensional vector space and its dual space is perfect. This notion allows two different spaces; a nondegenerate bilinear form is the case .
A linear subspace is separating if every nonzero has for some . For finite-dimensional , an evaluation map is injective, so the rank-nullity theorem forces and . For the infinite-dimensional polynomial space, the span of its coefficient functionals is a proper separating linear subspace of the dual space, which contains all sequences of coefficients, not just finitely supported ones.
For an algebraic dual space , the linear map sends to evaluation at . It is injective because linear functionals separate nonzero vectors: extend any nonzero vector to a basis and use its coordinate functional. If has a finite basis with dual basis , every equals , so is an isomorphism. For infinite-dimensional spaces, surjection does not follow from this finite sum argument.
For a nonzero Euclidean vector , the Cauchy-Schwarz inequality gives when . Equality is attained uniquely at . On a sphere of radius the minimizer is , and the minimum is . This turns many constrained vector problems into a normalization of one fixed vector. If , every point minimizes the functional.
If a linear functional on a normed vector space has for a sequence tending to zero in norm, it is discontinuous and unbounded. An explicit example is the algebraic span in of the coordinate vectors and . Define and . The tails have norm but image one. Thus the functional vanishes on the dense finite-support subspace while being nonzero elsewhere.
For a linear map between finite-dimensional vector spaces, the image of its dual map is the annihilator of a vector subspace given by its kernel. Inclusion follows by composition. Conversely a linear functional vanishing on the kernel descends to the image and extends to the target by basis extension. The identity and the rank-nullity theorem show that a linear map and its dual have equal rank.
For the map represented by a matrix , column rank is the rank of the map and row rank is the rank of its dual. Since the dual map has kernel equal to the annihilator of the original image, rank-nullity gives equal ranks.
If the columns of an by matrix are independent, some by row minor is nonzero. A nonzero term of that minor's determinant selects distinct nonzero coordinates , and replacing the corresponding standard basis vectors by the leaves a basis.
If are linearly independent, thenmaps onto . Otherwise a nonzero functional annihilating its proper image would give a nontrivial linear relation among the .
A polynomial moment functional evaluates a polynomial through a weighted integral, such as ; finitely many moments determine it on a bounded-degree polynomial space.
Articles by others on the same topic
In mathematics, particularly in functional analysis and linear algebra, the concept of the **dual space** is important in studying vector spaces and linear maps. ### Definition Given a vector space \( V \) over a field \( F \) (commonly the real numbers \( \mathbb{R} \) or complex numbers \( \mathbb{C} \)), the **dual space** \( V^* \) is defined as the set of all linear functionals on \( V \).
The dual space of a vector space , sometimes denoted , is the vector space of all linear forms over with the obvious addition and scalar multiplication operations defined.
Since a linear form is completely determined by how it acts on a basis, and since for each basis element it is specified by a scalar, at least in finite dimension, the dimension of the dual space is the same as the , and so they are isomorphic because all vector spaces of the same dimension on a given field are isomorphic, and so the dual is quite a boring concept in the context of finite dimension.
Infinite dimension seems more interesting however, see: en.wikipedia.org/w/index.php?title=Dual_space&oldid=1046421278#Infinite-dimensional_case