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Linear form

IfV is a vector space over a field k, the set of all linear functionals from Vtok is itself a vector space over k with addition and scalar multiplication defined pointwise. This space is called the dual spaceofV, or sometimes the algebraic dual space, when a topological dual space is also considered. It is often denoted Hom(V, k),^[2] or, when the field k is understood, ${\displaystyle V^{*}}$;^[3] other notations are also used, such as ${\displaystyle V'}$,^[4][5] ${\displaystyle V^{\#}}$or${\displaystyle V^{\vee }.}$^[2] When vectors are represented by column vectors (as is common when a basis is fixed), then linear functionals are represented as row vectors, and their values on specific vectors are given by matrix products (with the row vector on the left).

The constant zero function, mapping every vector to zero, is trivially a linear functional. Every other linear functional (such as the ones below) is surjective (that is, its range is all of k).

• Indexing into a vector: The second element of a three-vector is given by the one-form ${\displaystyle [0,1,0].}$ That is, the second element of ${\displaystyle [x,y,z]}$is$${\displaystyle [0,1,0]\cdot [x,y,z]=y.}$$
• Mean: The mean element of an ${\displaystyle n}$-vector is given by the one-form ${\displaystyle \left[1/n,1/n,\ldots ,1/n\right].}$ That is, $${\displaystyle \operatorname {mean} ($$v)=\left[1/n,1/n,\ldots ,1/n\right]\cdot v.}
• Sampling: Sampling with a kernel can be considered a one-form, where the one-form is the kernel shifted to the appropriate location.
• Net present value of a net cash flow, ${\displaystyle R($t),} is given by the one-form ${\displaystyle w($t)=(1+i)^{-t}} where ${\displaystyle i}$ is the discount rate. That is, $${\displaystyle \mathrm {NPV} (R($$t))=\langle w,R\rangle =\int _{t=0}^{\infty }{\frac {R(t)}{(1+i)^{t}}}\,dt.}

Suppose that vectors in the real coordinate space ${\displaystyle \mathbb {R} ^{n}}$ are represented as column vectors $${\displaystyle \mathbf {x} ={\begin{bmatrix}x_{1}\\\vdots \\x_{n}\end{bmatrix}}.}$$

For each row vector ${\displaystyle \mathbf {a} ={\begin{bmatrix}a_{1}&\cdots &a_{n}\end{bmatrix}}}$ there is a linear functional ${\displaystyle f_{\mathbf {a} }}$ defined by $${\displaystyle f_{\mathbf {a} }(\mathbf {x} )=a_{1}x_{1}+\cdots +a_{n}x_{n},}$$ and each linear functional can be expressed in this form.

This can be interpreted as either the matrix product or the dot product of the row vector ${\displaystyle \mathbf {a} }$ and the column vector ${\displaystyle \mathbf {x} }$: $${\displaystyle f_{\mathbf {a} }(\mathbf {x} )=\mathbf {a} \cdot \mathbf {x} ={\begin{bmatrix}a_{1}&\cdots &a_{n}\end{bmatrix}}{\begin{bmatrix}x_{1}\\\vdots \\x_{n}\end{bmatrix}}.}$$

The trace ${\displaystyle \operatorname {tr} ($A)} of a square matrix ${\displaystyle A}$ is the sum of all elements on its main diagonal. Matrices can be multiplied by scalars and two matrices of the same dimension can be added together; these operations make a vector space from the set of all ${\displaystyle n\times n}$ matrices. The trace is a linear functional on this space because ${\displaystyle \operatorname {tr} ($sA)=s\operatorname {tr} (A)} and ${\displaystyle \operatorname {tr} (A+B)=\operatorname {tr} ($A)+\operatorname {tr} (B)} for all scalars ${\displaystyle s}$ and all ${\displaystyle n\times n}$ matrices ${\displaystyle A{\text{ and }}B.}$

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 $${\displaystyle I($$f)=\int _{a}^{b}f(x)\,dx} is a linear functional from the vector space ${\displaystyle C[a,b]}$ of continuous functions on the interval ${\displaystyle [a,b]}$ to the real numbers. The linearity of ${\displaystyle I}$ follows from the standard facts about the integral: $${\displaystyle {\begin{aligned}I(f+g)&=\int _{a}^{b}[f($$x)+g(x)]\,dx=\int _{a}^{b}f(x)\,dx+\int _{a}^{b}g(x)\,dx=I(f)+I(g)\\I(\alpha f)&=\int _{a}^{b}\alpha f(x)\,dx=\alpha \int _{a}^{b}f(x)\,dx=\alpha I(f).\end{aligned}}}

Let ${\displaystyle P_{n}}$ denote the vector space of real-valued polynomial functions of degree ${\displaystyle \leq n}$ defined on an interval ${\displaystyle [a,b].}$If${\displaystyle c\in [a,b],}$ then let ${\displaystyle \operatorname {ev} _{c}:P_{n}\to \mathbb {R} }$ be the evaluation functional $${\displaystyle \operatorname {ev} _{c}f=f($$c).} The mapping ${\displaystyle f\mapsto f($c)} is linear since $${\displaystyle {\begin{aligned}(f+g)($$c)&=f(c)+g(c)\\(\alpha f)(c)&=\alpha f(c).\end{aligned}}}

If${\displaystyle x_{0},\ldots ,x_{n}}$ are ${\displaystyle n+1}$ distinct points in ${\displaystyle [a,b],}$ then the evaluation functionals ${\displaystyle \operatorname {ev} _{x_{i}},}$ ${\displaystyle i=0,\ldots ,n}$ form a basis of the dual space of ${\displaystyle P_{n}}$ (Lax (1996) proves this last fact using Lagrange interpolation).

A function ${\displaystyle f}$ having the equation of a line ${\displaystyle f($x)=a+rx} with ${\displaystyle a\neq 0}$ (for example, ${\displaystyle f($x)=1+2x} ) is not a linear functional on ${\displaystyle \mathbb {R} }$, since it is not linear.^{[nb 2]} It is, however, affine-linear.

In finite dimensions, a linear functional can be visualized in terms of its level sets, the sets of vectors which map to a given value. In three dimensions, the level sets of a linear functional are a family of mutually parallel planes; in higher dimensions, they are parallel hyperplanes. This method of visualizing linear functionals is sometimes introduced in general relativity texts, such as GravitationbyMisner, Thorne & Wheeler (1973).

If${\displaystyle x_{0},\ldots ,x_{n}}$ are ${\displaystyle n+1}$ distinct points in [a, b], then the linear functionals ${\displaystyle \operatorname {ev} _{x_{i}}:f\mapsto f\left(x_{i}\right)}$ defined above form a basis of the dual space of P_n, the space of polynomials of degree ${\displaystyle \leq n.}$ The integration functional I is also a linear functional on P_n, and so can be expressed as a linear combination of these basis elements. In symbols, there are coefficients ${\displaystyle a_{0},\ldots ,a_{n}}$ for which $${\displaystyle I($$f)=a_{0}f(x_{0})+a_{1}f(x_{1})+\dots +a_{n}f(x_{n})} for all ${\displaystyle f\in P_{n}.}$ This forms the foundation of the theory of numerical quadrature.^[6]

Linear functionals are particularly important in quantum mechanics. Quantum mechanical systems are represented by Hilbert spaces, which are anti–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.

In the theory of generalized functions, certain kinds of generalized functions called distributions can be realized as linear functionals on spaces of test functions.

Every non-degenerate bilinear form on a finite-dimensional vector space V induces an isomorphism V → V^∗ : v ↦ v^∗ such that $${\displaystyle v^{*}($$w):=\langle v,w\rangle \quad \forall w\in V,}

where the bilinear form on V is denoted ${\displaystyle \langle \,\cdot \,,\,\cdot \,\rangle }$ (for instance, in Euclidean space, ${\displaystyle \langle v,w\rangle =v\cdot w}$ is the dot productofv and w).

The inverse isomorphism is V^∗ → V : v^∗ ↦ v, where v is the unique element of V such that $${\displaystyle \langle v,w\rangle =v^{*}($$w)} for all ${\displaystyle w\in V.}$

The above defined vector v^∗ ∈ V^∗ is said to be the dual vectorof${\displaystyle v\in V.}$

In an infinite dimensional Hilbert space, analogous results hold by the Riesz representation theorem. There is a mapping V ↦ V^∗ from V into its continuous dual space V^∗.

Let the vector space V have a basis ${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\dots ,\mathbf {e} _{n}}$, not necessarily orthogonal. Then the dual space ${\displaystyle V^{*}}$ has a basis ${\displaystyle {\tilde {\omega }}^{1},{\tilde {\omega }}^{2},\dots ,{\tilde {\omega }}^{n}}$ called the dual basis defined by the special property that $${\displaystyle {\tilde {\omega }}^{i}(\mathbf {e} _{j})={\begin{cases}1&{\text{if}}\ i=j\\0&{\text{if}}\ i\neq j.\end{cases}}}$$

Or, more succinctly, $${\displaystyle {\tilde {\omega }}^{i}(\mathbf {e} _{j})=\delta _{ij}}$$

where ${\displaystyle \delta _{ij}}$ is the Kronecker delta. Here the superscripts of the basis functionals are not exponents but are instead contravariant indices.

A linear functional ${\displaystyle {\tilde {u}}}$ belonging to the dual space ${\displaystyle {\tilde {V}}}$ can be expressed as a linear combination of basis functionals, with coefficients ("components") u_i, $${\displaystyle {\tilde {u}}=\sum _{i=1}^{n}u_{i}\,{\tilde {\omega }}^{i}.}$$

Then, applying the functional ${\displaystyle {\tilde {u}}}$ to a basis vector ${\displaystyle \mathbf {e} _{j}}$ yields $${\displaystyle {\tilde {u}}(\mathbf {e} _{j})=\sum _{i=1}^{n}\left(u_{i}\,{\tilde {\omega }}^{i}\right)\mathbf {e} _{j}=\sum _{i}u_{i}\left[{\tilde {\omega }}^{i}\left(\mathbf {e} _{j}\right)\right]}$$

due to linearity of scalar multiples of functionals and pointwise linearity of sums of functionals. Then $${\displaystyle {\begin{aligned}{\tilde {u}}({\mathbf {e} }_{j})&=\sum _{i}u_{i}\left[{\tilde {\omega }}^{i}\left({\mathbf {e} }_{j}\right)\right]\\&=\sum _{i}u_{i}{\delta }_{ij}\\&=u_{j}.\end{aligned}}}$$

So each component of a linear functional can be extracted by applying the functional to the corresponding basis vector.

When the space V carries an inner product, then it is possible to write explicitly a formula for the dual basis of a given basis. Let V have (not necessarily orthogonal) basis ${\displaystyle \mathbf {e} _{1},\dots ,\mathbf {e} _{n}.}$ In three dimensions (n = 3), the dual basis can be written explicitly $${\displaystyle {\tilde {\omega }}^{i}(\mathbf {v} )={\frac {1}{2}}\left\langle {\frac {\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsilon ^{ijk}\,(\mathbf {e} _{j}\times \mathbf {e} _{k})}{\mathbf {e} _{1}\cdot \mathbf {e} _{2}\times \mathbf {e} _{3}}},\mathbf {v} \right\rangle ,}$$ for ${\displaystyle i=1,2,3,}$ where ε is the Levi-Civita symbol and ${\displaystyle \langle \cdot ,\cdot \rangle }$ the inner product (ordot product) on V.

In higher dimensions, this generalizes as follows $${\displaystyle {\tilde {\omega }}^{i}(\mathbf {v} )=\left\langle {\frac {\sum _{1\leq i_{2}<i_{3}<\dots <i_{n}\leq n}\varepsilon ^{ii_{2}\dots i_{n}}(\star \mathbf {e} _{i_{2}}\wedge \cdots \wedge \mathbf {e} _{i_{n}})}{\star (\mathbf {e} _{1}\wedge \cdots \wedge \mathbf {e} _{n})}},\mathbf {v} \right\rangle ,}$$ where ${\displaystyle \star }$ is the Hodge star operator.

Modules over a ring are generalizations of vector spaces, which removes the restriction that coefficients belong to a field. Given a module M over a ring R, a linear form on M is a linear map from MtoR, where the latter is considered as a module over itself. The space of linear forms is always denoted Hom_k(V, k), whether k is a field or not. It is a right moduleifV is a left module.

The existence of "enough" linear forms on a module is equivalent to projectivity.^[8]

Suppose that ${\displaystyle X}$ is a vector space over ${\displaystyle \mathbb {C} .}$ Restricting scalar multiplication to ${\displaystyle \mathbb {R} }$ gives rise to a real vector space^[9] ${\displaystyle X_{\mathbb {R} }}$ called the realificationof${\displaystyle X.}$ Any vector space ${\displaystyle X}$ over ${\displaystyle \mathbb {C} }$ is also a vector space over ${\displaystyle \mathbb {R} ,}$ endowed with a complex structure; that is, there exists a real vector subspace ${\displaystyle X_{\mathbb {R} }}$ such that we can (formally) write ${\displaystyle X=X_{\mathbb {R} }\oplus X_{\mathbb {R} }i}$as${\displaystyle \mathbb {R} }$-vector spaces.

Every linear functional on ${\displaystyle X}$ is complex-valued while every linear functional on ${\displaystyle X_{\mathbb {R} }}$ is real-valued. If ${\displaystyle \dim X\neq 0}$ then a linear functional on either one of ${\displaystyle X}$or${\displaystyle X_{\mathbb {R} }}$ is non-trivial (meaning not identically ${\displaystyle 0}$) if and only if it is surjective (because if ${\displaystyle \varphi ($x)\neq 0} then for any scalar ${\displaystyle s,}$ ${\displaystyle \varphi \left((s/\varphi ($x))x\right)=s} ), where the image of a linear functional on ${\displaystyle X}$is${\displaystyle \mathbb {C} }$ while the image of a linear functional on ${\displaystyle X_{\mathbb {R} }}$is${\displaystyle \mathbb {R} .}$ Consequently, the only function on ${\displaystyle X}$ that is both a linear functional on ${\displaystyle X}$ and a linear function on ${\displaystyle X_{\mathbb {R} }}$ is the trivial functional; in other words, ${\displaystyle X^{\#}\cap X_{\mathbb {R} }^{\#}=\{0\},}$ where ${\displaystyle \,{\cdot }^{\#}}$ denotes the space's algebraic dual space. However, every ${\displaystyle \mathbb {C} }$-linear functional on ${\displaystyle X}$ is an ${\displaystyle \mathbb {R} }$-linear operator (meaning that it is additive and homogeneous over ${\displaystyle \mathbb {R} }$), but unless it is identically ${\displaystyle 0,}$ it is not an ${\displaystyle \mathbb {R} }$-linear functionalon${\displaystyle X}$ because its range (which is ${\displaystyle \mathbb {C} }$) is 2-dimensional over ${\displaystyle \mathbb {R} .}$ Conversely, a non-zero ${\displaystyle \mathbb {R} }$-linear functional has range too small to be a ${\displaystyle \mathbb {C} }$-linear functional as well.

If${\displaystyle \varphi \in X^{\#}}$ then denote its real partby${\displaystyle \varphi _{\mathbb {R} }:=\operatorname {Re} \varphi }$ and its imaginary partby${\displaystyle \varphi _{i}:=\operatorname {Im} \varphi .}$ Then ${\displaystyle \varphi _{\mathbb {R} }:X\to \mathbb {R} }$ and ${\displaystyle \varphi _{i}:X\to \mathbb {R} }$ are linear functionals on ${\displaystyle X_{\mathbb {R} }}$ and ${\displaystyle \varphi =\varphi _{\mathbb {R} }+i\varphi _{i}.}$ The fact that ${\displaystyle z=\operatorname {Re} z-i\operatorname {Re} ($iz)=\operatorname {Im} (iz)+i\operatorname {Im} z} for all ${\displaystyle z\in \mathbb {C} }$ implies that for all ${\displaystyle x\in X,}$^[9] $${\displaystyle {\begin{alignedat}{4}\varphi ($$x)&=\varphi _{\mathbb {R} }(x)-i\varphi _{\mathbb {R} }(ix)\\&=\varphi _{i}(ix)+i\varphi _{i}(x)\\\end{alignedat}}} and consequently, that ${\displaystyle \varphi _{i}($x)=-\varphi _{\mathbb {R} }(ix)} and ${\displaystyle \varphi _{\mathbb {R} }($x)=\varphi _{i}(ix).} ^[10]

The assignment ${\displaystyle \varphi \mapsto \varphi _{\mathbb {R} }}$ defines a bijective^[10] ${\displaystyle \mathbb {R} }$-linear operator ${\displaystyle X^{\#}\to X_{\mathbb {R} }^{\#}}$ whose inverse is the map ${\displaystyle L_{\bullet }:X_{\mathbb {R} }^{\#}\to X^{\#}}$ defined by the assignment ${\displaystyle g\mapsto L_{g}}$ that sends ${\displaystyle g:X_{\mathbb {R} }\to \mathbb {R} }$ to the linear functional ${\displaystyle L_{g}:X\to \mathbb {C} }$ defined by $${\displaystyle L_{g}($$x):=g(x)-ig(ix)\quad {\text{ for all }}x\in X.} The real part of ${\displaystyle L_{g}}$is${\displaystyle g}$ and the bijection ${\displaystyle L_{\bullet }:X_{\mathbb {R} }^{\#}\to X^{\#}}$ is an ${\displaystyle \mathbb {R} }$-linear operator, meaning that ${\displaystyle L_{g+h}=L_{g}+L_{h}}$ and ${\displaystyle L_{rg}=rL_{g}}$ for all ${\displaystyle r\in \mathbb {R} }$ and ${\displaystyle g,h\in X_{\mathbb {R} }^{\#}.}$^[10] Similarly for the imaginary part, the assignment ${\displaystyle \varphi \mapsto \varphi _{i}}$ induces an ${\displaystyle \mathbb {R} }$-linear bijection ${\displaystyle X^{\#}\to X_{\mathbb {R} }^{\#}}$ whose inverse is the map ${\displaystyle X_{\mathbb {R} }^{\#}\to X^{\#}}$ defined by sending ${\displaystyle I\in X_{\mathbb {R} }^{\#}}$ to the linear functional on ${\displaystyle X}$ defined by ${\displaystyle x\mapsto I($ix)+iI(x).}

This relationship was discovered by Henry Löwig in 1934 (although it is usually credited to F. Murray),^[11] and can be generalized to arbitrary finite extensions of a field in the natural way. It has many important consequences, some of which will now be described.

Suppose ${\displaystyle \varphi :X\to \mathbb {C} }$ is a linear functional on ${\displaystyle X}$ with real part ${\displaystyle \varphi _{\mathbb {R} }:=\operatorname {Re} \varphi }$ and imaginary part ${\displaystyle \varphi _{i}:=\operatorname {Im} \varphi .}$

Then ${\displaystyle \varphi =0}$ if and only if ${\displaystyle \varphi _{\mathbb {R} }=0}$ if and only if ${\displaystyle \varphi _{i}=0.}$

Assume that ${\displaystyle X}$ is a topological vector space. Then ${\displaystyle \varphi }$ is continuous if and only if its real part ${\displaystyle \varphi _{\mathbb {R} }}$ is continuous, if and only if ${\displaystyle \varphi }$'s imaginary part ${\displaystyle \varphi _{i}}$ is continuous. That is, either all three of ${\displaystyle \varphi ,\varphi _{\mathbb {R} },}$ and ${\displaystyle \varphi _{i}}$ are continuous or none are continuous. This remains true if the word "continuous" is replaced with the word "bounded". In particular, ${\displaystyle \varphi \in X^{\prime }}$ if and only if ${\displaystyle \varphi _{\mathbb {R} }\in X_{\mathbb {R} }^{\prime }}$ where the prime denotes the space's continuous dual space.^[9]

Let ${\displaystyle B\subseteq X.}$If${\displaystyle uB\subseteq B}$ for all scalars ${\displaystyle u\in \mathbb {C} }$ofunit length (meaning ${\displaystyle |u|=1}$) then^{[proof 1][12]} $${\displaystyle \sup _{b\in B}|\varphi ($$b)|=\sup _{b\in B}\left|\varphi _{\mathbb {R} }(b)\right|.} Similarly, if ${\displaystyle \varphi _{i}:=\operatorname {Im} \varphi :X\to \mathbb {R} }$ denotes the complex part of ${\displaystyle \varphi }$ then ${\displaystyle iB\subseteq B}$ implies $${\displaystyle \sup _{b\in B}\left|\varphi _{\mathbb {R} }($$b)\right|=\sup _{b\in B}\left|\varphi _{i}(b)\right|.} If${\displaystyle X}$ is a normed space with norm ${\displaystyle \|\cdot \|}$ and if ${\displaystyle B=\{x\in X:\|x\|\leq 1\}}$ is the closed unit ball then the supremums above are the operator norms (defined in the usual way) of ${\displaystyle \varphi ,\varphi _{\mathbb {R} },}$ and ${\displaystyle \varphi _{i}}$ so that ^[12] $${\displaystyle \|\varphi \|=\left\|\varphi _{\mathbb {R} }\right\|=\left\|\varphi _{i}\right\|.}$$ This conclusion extends to the analogous statement for polarsofbalanced sets in general topological vector spaces.