Linear functional

From Wikipedia, the free encyclopedia

This article deals with linear maps from a vector space to its field of scalars. These maps may be functionals in the traditional sense of functions of functions, but this is not necessarily the case.

1 Definition
2 Discussion using coordinates
3 See also

[edit] Definition

In linear algebra, a branch of mathematics, a linear functional or linear form (in older terminology, covariant vector or covector; also called a one-form) is a linear map from a vector space to its field of scalars. Specifically, if V is a vector space over a field k, then a linear functional is a linear function from V to k.

The set of all linear functionals from V to k, Hom_K(V,k), is itself a k-vector space. This space is called the dual space of V. If V is a topological vector space, the space of continuous linear functionals — the continuous dual — is often simply called the dual space. If V is a Banach space, then so is its (continuous) dual.

Any linear functional is either trivial (equal to 0 everywhere) or surjective onto the scalar field. (Because the image of a vector subspace under a linear transformation is a subspace, so is the image of V under L. But the only subspaces of k are {0} and k proper).

Linear functionals first appeared in functional analysis, the study of vector spaces of functions. A typical example of a linear functional is integration: the linear transformation defined by the Riemann integral

$f \mapsto \int_a^b f(x)\, dx$

is a linear functional from the space of continuous functions on the interval [a, b] to the real numbers.

Linear functionals are particularly important in quantum mechanics. Quantum mechanical systems are represented by Hilbert spaces, which are isomorphic to their own dual spaces. A state of a quantum mechanical system can be identified with a linear functional. For more information see bra-ket notation.

A generalized function is an example of a linear functional.

The reason for the use of the term "functional" instead of the traditional term "function" is to avoid potential confusion when a vector space is a space of functions, which is often the case. Hence, linear functionals are often, in practice, functionals in the traditional sense of functions of functions.

A linear functional is a tensor of type $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ . It is the simplest non-scalar tensor.

[edit] Discussion using coordinates

Let $\tilde{f}$ represent a linear functional which acts on vectors of space V, including vectors $\vec u$ and $\vec v$ . Then the linearity properties of $\tilde{f}$ are

$\tilde{f} (\vec u + \vec v) = \tilde{f} (\vec u) + \tilde{f} (\vec v)$

$\tilde{f} (\alpha \vec v) = \alpha \tilde{f} (\vec v)$

where α is a scalar.

The set of all linear functionals definable on the vector space V can also itself be a vector space if functionals can be added to each other or be multiplied by scalars in a pointwise linear manner. That is, if the vectors of the space V are position vectors of points, then for every point $\vec v$ in the space V, the following should hold true:

$(\tilde{f} + \tilde{g}) (\vec v) = \tilde{f}(\vec v) + \tilde{g}(\vec v)$

$(\alpha \tilde{f}) (\vec v) = \alpha \tilde{f}(\vec v).$

If these last two conditions are true for every $\vec v \isin V$ then the linear functionals constitute a vector space.

If V is an inner-product space with inner product 〈 , 〉 then every vector $\vec v$ can be mapped to a dual linear functional $\tilde{v}$ defined by

$\tilde{v} := \langle \vec v, \ \rangle$

(i.e. $\tilde{v} := \lambda x. \langle \vec v, x \rangle$ in lambda notation) so that the functional $\tilde{v}$ applied to a vector $\vec u$ yields

$\tilde{v} (\vec u) = \langle \vec v, \vec u\rangle.$

Thus the inner product provides a bijection of each vector in V to a functional in its dual vector space $\tilde{V}$ (Note that this mapping is not necessarily linear, but is conjugate linear for complex vector spaces).

[edit] Visualizing linear functionals

A vector is usually visualized as an arrow extending from the origin to a point in space. A linear functional can be visualized as a set of equally spaced parallel planes that partition the entire space. The magnitude of a linear functional is directly proportional to the density of parallel planes and inversely proportional to the spacing between pairs of neighboring planes. To find the result of applying a functional to a vector, basically count the number of planes which a vector cuts through. (Note: this visualization is discrete, whereas linear functionals and vectors have magnitudes which range continuously over the real numbers. The visualization can be interpolated linearly, as it were, to increase the precision.)

Unfortunately, the problem with visualizing a linear functional as a set of planes is that additional structure (a direction) needs to be included in order to define the negative of the functional. Also, adding functionals is not as straightforward as adding vectors. Because of this, such a visualization must be seen as only a rudimentary concept.

[edit] Basis of the dual space in finite dimensions

Let the vector space V have a basis ${\vec e}_1,\ {\vec e}_2$ , … , ${\vec e}_n$ , not necessarily orthonormal nor even orthogonal. Then the dual space $\tilde{V}$ has a basis $\tilde{\omega}^1, \ \tilde{\omega}^2$ , … , $\ \tilde{\omega}^n$ which in the three-dimensional case (n = 3) can be defined by

$\tilde{\omega}^i = {1 \over 2} \, \left\langle { \epsilon^{ijk} \, (\vec e_j \times \vec e_k) \over \vec e_1 \cdot \vec e_2 \times \vec e_3} , \qquad \right\rangle$

where $\epsilon\,\!$ is the Levi-Civita symbol . This definition has the special property that

$\tilde{\omega}^i (\vec e_j) = \delta^i {}_j$

where δ is the Kronecker delta. Thus, these two dual bases are mutually orthonormal even if each basis is not self-orthonormal.

N.B. The superscripts of the basis functionals are not exponents but are instead contravariant indices.

A linear functional $\tilde{u}$ belonging to the dual space $\tilde{V}$ can be expressed as a linear combination of basis functionals, with coefficients ("components") u_i ,

$\tilde{u} = u_i \, \tilde{\omega}^i$

Then, applying the functional $\tilde{u}$ to a basis vector e_j yields

$\tilde{u}(\vec e_j) = (u_i \, \tilde{\omega}^i) \vec e_j = u_i (\tilde{\omega}^i (\vec e_j))$

due to linearity of scalar multiples of functionals and pointwise linearity of sums of functionals. Then

$\tilde{u}({\vec e}_j) = u_i (\tilde{\omega}^i ({\vec e}_j)) = u_i \delta^i {}_j = u_j$

that is

$\tilde{u} (\vec e_j) = u_j.$

This last equation shows that an individual component of a linear functional can be extracted by applying the functional to a corresponding basis vector.

[edit] References

Bernard F. Schutz (1985, 2002). A first course in general relativity. Cambridge University Press: Cambridge, UK. Chapter 3. ISBN 0-521-27703-5.
Richard Bishop and Samuel Goldberg(1968,1980). "Tensor Analysis on Manifolds" Dover Publications. Chapter 4. ISBN 0-486-64039-6

Categories: Linear algebra | Functional analysis

Linear functional

From Wikipedia, the free encyclopedia

Contents

[edit] Definition

[edit] Discussion using coordinates

[edit] Visualizing linear functionals

[edit] Basis of the dual space in finite dimensions

[edit] References

[edit] See also

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