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Dual space

From Wikipedia, the free encyclopedia
(Redirected fromDual vector space)
In mathematics, vector space of linear forms
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Inmathematics, anyvector spaceV{\displaystyle V} has a correspondingdual vector space (or justdual space for short) consisting of alllinear forms onV,{\displaystyle V,} together with the vector space structure ofpointwise addition andscalar multiplication by constants.

The dual space as defined above is defined for all vector spaces, and to avoid ambiguity may also be called thealgebraic dual space.When defined for atopological vector space, there is a subspace of the dual space, corresponding tocontinuous linear functionals, called thecontinuous dual space.

Dual vector spaces find application in many branches of mathematics that use vector spaces, such as intensor analysis withfinite-dimensional vector spaces.When applied to vector spaces of functions (which are typically infinite-dimensional), dual spaces are used to describemeasures,distributions, andHilbert spaces. Consequently, the dual space is an important concept infunctional analysis.

Early terms fordual includepolarer Raum [Hahn 1927],espace conjugué,adjoint space [Alaoglu 1940], andtransponierter Raum [Schauder 1930] and [Banach 1932]. The termdual is due toBourbaki 1938.[1]

Algebraic dual space

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Given anyvector spaceV{\displaystyle V} over afieldF{\displaystyle F}, the(algebraic) dual spaceV{\displaystyle V^{*}}[2] (alternatively denoted byV{\displaystyle V^{\lor }}[3] orV{\displaystyle V'}[4][5])[nb 1] is defined as the set of alllinear mapsφ:VF{\displaystyle \varphi :V\to F} (linear functionals). Since linear maps are vector spacehomomorphisms, the dual space may be denotedhom(V,F){\displaystyle \hom(V,F)}.[3]The dual spaceV{\displaystyle V^{*}} itself becomes a vector space overF{\displaystyle F} when equipped with an addition and scalar multiplication satisfying:

(φ+ψ)(x)=φ(x)+ψ(x)(aφ)(x)=a(φ(x)){\displaystyle {\begin{aligned}(\varphi +\psi )(x)&=\varphi (x)+\psi (x)\\(a\varphi )(x)&=a\left(\varphi (x)\right)\end{aligned}}}

for allφ,ψV{\displaystyle \varphi ,\psi \in V^{*}},xV{\displaystyle x\in V}, andaF{\displaystyle a\in F}.

Elements of the algebraic dual spaceV{\displaystyle V^{*}} are sometimes calledcovectors,one-forms, orlinear forms.

The pairing of a functionalφ{\displaystyle \varphi } in the dual spaceV{\displaystyle V^{*}} and an elementx{\displaystyle x} ofV{\displaystyle V} is sometimes denoted by a bracket:φ(x)=[x,φ]{\displaystyle \varphi (x)=[x,\varphi ]}[6]orφ(x)=x,φ{\displaystyle \varphi (x)=\langle x,\varphi \rangle }.[7] This pairing defines a nondegeneratebilinear mapping[nb 2],:V×VF{\displaystyle \langle \cdot ,\cdot \rangle :V\times V^{*}\to F} called thenatural pairing.

Finite-dimensional case

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See also:Dual basis

IfV{\displaystyle V} is finite-dimensional, thenV{\displaystyle V^{*}} has the same dimension asV{\displaystyle V}. Given abasis{e1,,en}{\displaystyle \{\mathbf {e} _{1},\dots ,\mathbf {e} _{n}\}} inV{\displaystyle V}, it is possible to construct a specific basis inV{\displaystyle V^{*}}, called thedual basis. This dual basis is a set{e1,,en}{\displaystyle \{\mathbf {e} ^{1},\dots ,\mathbf {e} ^{n}\}} of linear functionals onV{\displaystyle V}, defined by the relation

ei(c1e1++cnen)=ci,i=1,,n{\displaystyle \mathbf {e} ^{i}(c^{1}\mathbf {e} _{1}+\cdots +c^{n}\mathbf {e} _{n})=c^{i},\quad i=1,\ldots ,n}

for any choice of coefficientsciF{\displaystyle c^{i}\in F}. In particular, letting in turn each one of those coefficients be equal to one and the other coefficients zero, gives the system of equations

ei(ej)=δji{\displaystyle \mathbf {e} ^{i}(\mathbf {e} _{j})=\delta _{j}^{i}}

whereδji{\displaystyle \delta _{j}^{i}} is theKronecker delta symbol. This property is referred to as thebi-orthogonality property.

Proof

Consider{e1,,en}{\displaystyle \{\mathbf {e} _{1},\dots ,\mathbf {e} _{n}\}} the basis of V. Let{e1,,en}{\displaystyle \{\mathbf {e} ^{1},\dots ,\mathbf {e} ^{n}\}} be defined as the following:

ei(c1e1++cnen)=ci,i=1,,n{\displaystyle \mathbf {e} ^{i}(c^{1}\mathbf {e} _{1}+\cdots +c^{n}\mathbf {e} _{n})=c^{i},\quad i=1,\ldots ,n}.

These are a basis ofV{\displaystyle V^{*}} because:

  1. Theei,i=1,2,,n,{\displaystyle \mathbf {e} ^{i},i=1,2,\dots ,n,} are linear functionals, which mapx,yV{\displaystyle x,y\in V} such asx=α1e1++αnen{\displaystyle x=\alpha _{1}\mathbf {e} _{1}+\dots +\alpha _{n}\mathbf {e} _{n}} andy=β1e1++βnen{\displaystyle y=\beta _{1}\mathbf {e} _{1}+\dots +\beta _{n}\mathbf {e} _{n}} to scalarsei(x)=αi{\displaystyle \mathbf {e} ^{i}(x)=\alpha _{i}} andei(y)=βi{\displaystyle \mathbf {e} ^{i}(y)=\beta _{i}}. Then also,x+λy=(α1+λβ1)e1++(αn+λβn)en{\displaystyle x+\lambda y=(\alpha _{1}+\lambda \beta _{1})\mathbf {e} _{1}+\dots +(\alpha _{n}+\lambda \beta _{n})\mathbf {e} _{n}} andei(x+λy)=αi+λβi=ei(x)+λei(y){\displaystyle \mathbf {e} ^{i}(x+\lambda y)=\alpha _{i}+\lambda \beta _{i}=\mathbf {e} ^{i}(x)+\lambda \mathbf {e} ^{i}(y)}. Therefore,eiV{\displaystyle \mathbf {e} ^{i}\in V^{*}} fori=1,2,,n{\displaystyle i=1,2,\dots ,n}.
  2. Supposeλ1e1++λnen=0V{\displaystyle \lambda _{1}\mathbf {e} ^{1}+\cdots +\lambda _{n}\mathbf {e} ^{n}=0\in V^{*}}. Applying this functional on the basis vectors ofV{\displaystyle V} successively, lead us toλ1=λ2==λn=0{\displaystyle \lambda _{1}=\lambda _{2}=\dots =\lambda _{n}=0} (The functional applied inei{\displaystyle \mathbf {e} _{i}} results inλi{\displaystyle \lambda _{i}}). Therefore,{e1,,en}{\displaystyle \{\mathbf {e} ^{1},\dots ,\mathbf {e} ^{n}\}} is linearly independent onV{\displaystyle V^{*}}.
  3. Lastly, considergV{\displaystyle g\in V^{*}}. Then
g(x)=g(α1e1++αnen)=α1g(e1)++αng(en)=e1(x)g(e1)++en(x)g(en){\displaystyle g(x)=g(\alpha _{1}\mathbf {e} _{1}+\dots +\alpha _{n}\mathbf {e} _{n})=\alpha _{1}g(\mathbf {e} _{1})+\dots +\alpha _{n}g(\mathbf {e} _{n})=\mathbf {e} ^{1}(x)g(\mathbf {e} _{1})+\dots +\mathbf {e} ^{n}(x)g(\mathbf {e} _{n})}

and{e1,,en}{\displaystyle \{\mathbf {e} ^{1},\dots ,\mathbf {e} ^{n}\}} generatesV{\displaystyle V^{*}}. Hence, it is a basis ofV{\displaystyle V^{*}}.

For example, ifV{\displaystyle V} isR2{\displaystyle \mathbb {R} ^{2}}, let its basis be chosen as{e1=(1/2,1/2),e2=(0,1)}{\displaystyle \{\mathbf {e} _{1}=(1/2,1/2),\mathbf {e} _{2}=(0,1)\}}. The basis vectors are not orthogonal to each other. Then,e1{\displaystyle \mathbf {e} ^{1}} ande2{\displaystyle \mathbf {e} ^{2}} areone-forms (functions that map a vector to a scalar) such thate1(e1)=1{\displaystyle \mathbf {e} ^{1}(\mathbf {e} _{1})=1},e1(e2)=0{\displaystyle \mathbf {e} ^{1}(\mathbf {e} _{2})=0},e2(e1)=0{\displaystyle \mathbf {e} ^{2}(\mathbf {e} _{1})=0}, ande2(e2)=1{\displaystyle \mathbf {e} ^{2}(\mathbf {e} _{2})=1}. (Note: The superscript here is the index, not an exponent.) This system of equations can be expressed using matrix notation as

[e11e12e21e22][e11e21e12e22]=[1001].{\displaystyle {\begin{bmatrix}e^{11}&e^{12}\\e^{21}&e^{22}\end{bmatrix}}{\begin{bmatrix}e_{11}&e_{21}\\e_{12}&e_{22}\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}.}

Solving for the unknown values in the first matrix shows the dual basis to be{e1=(2,0),e2=(1,1)}{\displaystyle \{\mathbf {e} ^{1}=(2,0),\mathbf {e} ^{2}=(-1,1)\}}. Becausee1{\displaystyle \mathbf {e} ^{1}} ande2{\displaystyle \mathbf {e} ^{2}} are functionals, they can be rewritten ase1(x,y)=2x{\displaystyle \mathbf {e} ^{1}(x,y)=2x} ande2(x,y)=x+y{\displaystyle \mathbf {e} ^{2}(x,y)=-x+y}.

In general, whenV{\displaystyle V} isRn{\displaystyle \mathbb {R} ^{n}}, ifE=[e1||en]{\displaystyle E=[\mathbf {e} _{1}|\cdots |\mathbf {e} _{n}]} is a matrix whose columns are the basis vectors andE^=[e1||en]{\displaystyle {\hat {E}}=[\mathbf {e} ^{1}|\cdots |\mathbf {e} ^{n}]} is a matrix whose columns are the dual basis vectors, then

E^TE=In,{\displaystyle {\hat {E}}^{\textrm {T}}\cdot E=I_{n},}

whereIn{\displaystyle I_{n}} is theidentity matrix of ordern{\displaystyle n}. The biorthogonality property of these two basis sets allows any pointxV{\displaystyle \mathbf {x} \in V} to be represented as

x=ix,eiei=ix,eiei,{\displaystyle \mathbf {x} =\sum _{i}\langle \mathbf {x} ,\mathbf {e} ^{i}\rangle \mathbf {e} _{i}=\sum _{i}\langle \mathbf {x} ,\mathbf {e} _{i}\rangle \mathbf {e} ^{i},}

even when the basis vectors are not orthogonal to each other. Strictly speaking, the above statement only makes sense once the inner product,{\displaystyle \langle \cdot ,\cdot \rangle } and the corresponding duality pairing are introduced, as described below in§ Bilinear products and dual spaces.

In particular,Rn{\displaystyle \mathbb {R} ^{n}} can be interpreted as the space of columns ofn{\displaystyle n}real numbers, its dual space is typically written as the space ofrows ofn{\displaystyle n} real numbers. Such a row acts onRn{\displaystyle \mathbb {R} ^{n}} as a linear functional by ordinarymatrix multiplication. This is because a functional maps everyn{\displaystyle n}-vectorx{\displaystyle x} into a real numbery{\displaystyle y}. Then, seeing this functional as a matrixM{\displaystyle M}, andx{\displaystyle x} as ann×1{\displaystyle n\times 1} matrix, andy{\displaystyle y} a1×1{\displaystyle 1\times 1} matrix (trivially, a real number) respectively, ifMx=y{\displaystyle Mx=y} then, by dimension reasons,M{\displaystyle M} must be a1×n{\displaystyle 1\times n} matrix; that is,M{\displaystyle M} must be a row vector.

IfV{\displaystyle V} consists of the space of geometricalvectors in the plane, then thelevel curves of an element ofV{\displaystyle V^{*}} form a family of parallel lines inV{\displaystyle V}, because the range is 1-dimensional, so that every point in the range is a multiple of any one nonzero element.So an element ofV{\displaystyle V^{*}} can be intuitively thought of as a particular family of parallel lines covering the plane. To compute the value of a functional on a given vector, it suffices to determine which of the lines the vector lies on. Informally, this "counts" how many lines the vector crosses.More generally, ifV{\displaystyle V} is a vector space of any dimension, then thelevel sets of a linear functional inV{\displaystyle V^{*}} are parallel hyperplanes inV{\displaystyle V}, and the action of a linear functional on a vector can be visualized in terms of these hyperplanes.[8]

Infinite-dimensional case

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IfV{\displaystyle V} is not finite-dimensional but has abasis[nb 3]eα{\displaystyle \mathbf {e} _{\alpha }} indexed by an infinite setA{\displaystyle A}, then the same construction as in the finite-dimensional case yieldslinearly independent elementseα{\displaystyle \mathbf {e} ^{\alpha }} (αA{\displaystyle \alpha \in A}) of the dual space, but they will not form a basis.

For instance, consider the spaceR{\displaystyle \mathbb {R} ^{\infty }}, whose elements are thosesequences of real numbers that contain only finitely many non-zero entries, which has a basis indexed by the natural numbersN{\displaystyle \mathbb {N} }. ForiN{\displaystyle i\in \mathbb {N} },ei{\displaystyle \mathbf {e} _{i}} is the sequence consisting of all zeroes except in thei{\displaystyle i}-th position, which is 1.The dual space ofR{\displaystyle \mathbb {R} ^{\infty }} is (isomorphic to)RN{\displaystyle \mathbb {R} ^{\mathbb {N} }}, the space ofall sequences of real numbers: each real sequence(an){\displaystyle (a_{n})} defines a function where the element(xn){\displaystyle (x_{n})} ofR{\displaystyle \mathbb {R} ^{\infty }} is sent to the number

nanxn,{\displaystyle \sum _{n}a_{n}x_{n},}

which is a finite sum because there are only finitely many nonzeroxn{\displaystyle x_{n}}. Thedimension ofR{\displaystyle \mathbb {R} ^{\infty }} iscountably infinite, whereasRN{\displaystyle \mathbb {R} ^{\mathbb {N} }} does not have a countable basis.

This observation generalizes to any[nb 3] infinite-dimensional vector spaceV{\displaystyle V} over any fieldF{\displaystyle F}: a choice of basis{eα:αA}{\displaystyle \{\mathbf {e} _{\alpha }:\alpha \in A\}} identifiesV{\displaystyle V} with the space(FA)0{\displaystyle (F^{A})_{0}} of functionsf:AF{\displaystyle f:A\to F} such thatfα=f(α){\displaystyle f_{\alpha }=f(\alpha )} is nonzero for only finitely manyαA{\displaystyle \alpha \in A}, where such a functionf{\displaystyle f} is identified with the vector

αAfαeα{\displaystyle \sum _{\alpha \in A}f_{\alpha }\mathbf {e} _{\alpha }}

inV{\displaystyle V} (the sum is finite by the assumption onf{\displaystyle f}, and anyvV{\displaystyle v\in V} may be written uniquely in this way by the definition of the basis).

The dual space ofV{\displaystyle V} may then be identified with the spaceFA{\displaystyle F^{A}} ofall functions fromA{\displaystyle A} toF{\displaystyle F}: a linear functionalT{\displaystyle T} onV{\displaystyle V} is uniquely determined by the valuesθα=T(eα){\displaystyle \theta _{\alpha }=T(\mathbf {e} _{\alpha })} it takes on the basis ofV{\displaystyle V}, and any functionθ:AF{\displaystyle \theta :A\to F} (withθ(α)=θα{\displaystyle \theta (\alpha )=\theta _{\alpha }}) defines a linear functionalT{\displaystyle T} onV{\displaystyle V} by

T(αAfαeα)=αAfαT(eα)=αAfαθα.{\displaystyle T\left(\sum _{\alpha \in A}f_{\alpha }\mathbf {e} _{\alpha }\right)=\sum _{\alpha \in A}f_{\alpha }T(e_{\alpha })=\sum _{\alpha \in A}f_{\alpha }\theta _{\alpha }.}

Again, the sum is finite becausefα{\displaystyle f_{\alpha }} is nonzero for only finitely manyα{\displaystyle \alpha }.

The set(FA)0{\displaystyle (F^{A})_{0}} may be identified (essentially by definition) with thedirect sum of infinitely many copies ofF{\displaystyle F} (viewed as a 1-dimensional vector space over itself) indexed byA{\displaystyle A}, i.e. there are linear isomorphisms

V(FA)0αAF.{\displaystyle V\cong (F^{A})_{0}\cong \bigoplus _{\alpha \in A}F.}

On the other hand,FA{\displaystyle F^{A}} is (again by definition), thedirect product of infinitely many copies ofF{\displaystyle F} indexed byA{\displaystyle A}, and so the identification

V(αAF)αAFαAFFA{\displaystyle V^{*}\cong \left(\bigoplus _{\alpha \in A}F\right)^{*}\cong \prod _{\alpha \in A}F^{*}\cong \prod _{\alpha \in A}F\cong F^{A}}

is a special case of ageneral result relating direct sums (ofmodules) to direct products.

If a vector space is not finite-dimensional, then its (algebraic) dual space isalways of larger dimension (as acardinal number) than the original vector space. This is in contrast to the case of the continuous dual space, discussed below, which may beisomorphic to the original vector space even if the latter is infinite-dimensional.

The proof of this inequality between dimensions results from the following.

IfV{\displaystyle V} is an infinite-dimensionalF{\displaystyle F}-vector space, the arithmetical properties ofcardinal numbers implies that

dim(V)=|A|<|F||A|=|V|=max(|dim(V)|,|F|),{\displaystyle \mathrm {dim} (V)=|A|<|F|^{|A|}=|V^{\ast }|=\mathrm {max} (|\mathrm {dim} (V^{\ast })|,|F|),}

where cardinalities are denoted asabsolute values. For proving thatdim(V)<dim(V),{\displaystyle \mathrm {dim} (V)<\mathrm {dim} (V^{*}),} it suffices to prove that|F||dim(V)|,{\displaystyle |F|\leq |\mathrm {dim} (V^{\ast })|,} which can be done with an argument similar toCantor's diagonal argument.[9] The exact dimension of the dual is given by theErdős–Kaplansky theorem.

Bilinear products and dual spaces

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IfV is finite-dimensional, thenV is isomorphic toV. But there is in general nonatural isomorphism between these two spaces.[10] Anybilinear form,{\displaystyle \langle \cdot ,\cdot \rangle } onV{\displaystyle V} gives a mapping ofV{\displaystyle V} into its dual space via

vv,{\displaystyle v\mapsto \langle v,\cdot \rangle }

where the right hand side is defined as the functional onV taking eachwV{\displaystyle w\in V} tov,w{\displaystyle \langle v,w\rangle }. In other words, the bilinear form determines a linear mapping

Φ,:VV{\displaystyle \Phi _{\langle \cdot ,\cdot \rangle }:V\to V^{*}}

defined by

[Φ,(v),w]=v,w.{\displaystyle \left[\Phi _{\langle \cdot ,\cdot \rangle }(v),w\right]=\langle v,w\rangle .}

If the bilinear form isnondegenerate, then this is an isomorphism onto a subspace ofV.IfV is finite-dimensional, then this is an isomorphism onto all ofV. Conversely, any isomorphismΦ{\displaystyle \Phi } fromV to a subspace ofV (resp., all ofV ifV is finite dimensional) defines a unique nondegenerate bilinear form,Φ{\displaystyle \langle \cdot ,\cdot \rangle _{\Phi }} onV by

v,wΦ=(Φ(v))(w)=[Φ(v),w].{\displaystyle \langle v,w\rangle _{\Phi }=(\Phi (v))(w)=[\Phi (v),w].\,}

Thus there is a one-to-one correspondence between isomorphisms ofV to a subspace of (resp., all of)V and nondegenerate bilinear forms onV.

If the vector spaceV is over thecomplex field, then sometimes it is more natural to considersesquilinear forms instead of bilinear forms.In that case, a given sesquilinear form⟨·,·⟩ determines an isomorphism ofV with thecomplex conjugate of the dual space

Φ,:VV¯.{\displaystyle \Phi _{\langle \cdot ,\cdot \rangle }:V\to {\overline {V^{*}}}.}

The conjugate of the dual spaceV¯{\displaystyle {\overline {V^{*}}}} can be identified with the set of all additive complex-valued functionalsf :VC such that

f(αv)=α¯f(v).{\displaystyle f(\alpha v)={\overline {\alpha }}f(v).}

Injection into the double-dual

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There is anaturalhomomorphismΨ{\displaystyle \Psi } fromV{\displaystyle V} into the double dualV=hom(V,F){\displaystyle V^{**}=\hom(V^{*},F)}, defined by(Ψ(v))(φ)=φ(v){\displaystyle (\Psi (v))(\varphi )=\varphi (v)} for allvV,φV{\displaystyle v\in V,\varphi \in V^{*}}. In other words, ifevv:VF{\displaystyle \mathrm {ev} _{v}:V^{*}\to F} is the evaluation map defined byφφ(v){\displaystyle \varphi \mapsto \varphi (v)}, thenΨ:VV{\displaystyle \Psi :V\to V^{**}} is defined as the mapvevv{\displaystyle v\mapsto \mathrm {ev} _{v}}. This mapΨ{\displaystyle \Psi } is alwaysinjective;[nb 3] and it is always anisomorphism ifV{\displaystyle V} is finite-dimensional.[11]Indeed, the isomorphism of a finite-dimensional vector space with its double dual is an archetypal example of anatural isomorphism.Infinite-dimensional Hilbert spaces are not isomorphic to their algebraic double duals, but instead to their continuous double duals.

Transpose of a linear map

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Main article:Transpose of a linear map

Iff :VW is alinear map, then thetranspose (ordual)f :WV is defined by

f(φ)=φf{\displaystyle f^{*}(\varphi )=\varphi \circ f\,}

for everyφW{\displaystyle \varphi \in W^{*}}. The resulting functionalf(φ){\displaystyle f^{*}(\varphi )} inV{\displaystyle V^{*}} is called thepullback ofφ{\displaystyle \varphi } alongf{\displaystyle f}.

The following identity holds for allφW{\displaystyle \varphi \in W^{*}} andvV{\displaystyle v\in V}:

[f(φ),v]=[φ,f(v)],{\displaystyle [f^{*}(\varphi ),\,v]=[\varphi ,\,f(v)],}

where the bracket [·,·] on the left is the natural pairing ofV with its dual space, and that on the right is the natural pairing ofW with its dual. This identity characterizes the transpose,[12] and is formally similar to the definition of theadjoint.

The assignmentff produces aninjective linear map between the space of linear operators fromV toW and the space of linear operators fromW toV; this homomorphism is anisomorphism if and only ifW is finite-dimensional.IfV =W then the space of linear maps is actually analgebra undercomposition of maps, and the assignment is then anantihomomorphism of algebras, meaning that(fg) =gf.In the language ofcategory theory, taking the dual of vector spaces and the transpose of linear maps is therefore acontravariant functor from the category of vector spaces overF to itself.It is possible to identify (f) withf using the natural injection into the double dual.

If the linear mapf is represented by thematrixA with respect to two bases ofV andW, thenf is represented by thetranspose matrixAT with respect to the dual bases ofW andV, hence the name.Alternatively, asf is represented byA acting on the left on column vectors,f is represented by the same matrix acting on the right on row vectors.These points of view are related by the canonical inner product onRn, which identifies the space of column vectors with the dual space of row vectors.

Quotient spaces and annihilators

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LetS{\displaystyle S} be a subset ofV{\displaystyle V}.Theannihilator ofS{\displaystyle S} inV{\displaystyle V^{*}}, denoted hereS0{\displaystyle S^{0}}, is the collection of linear functionalsfV{\displaystyle f\in V^{*}} such that[f,s]=0{\displaystyle [f,s]=0} for allsS{\displaystyle s\in S}.That is,S0{\displaystyle S^{0}} consists of all linear functionalsf:VF{\displaystyle f:V\to F} such that the restriction toS{\displaystyle S} vanishes:f|S=0{\displaystyle f|_{S}=0}.Within finite dimensional vector spaces, the annihilator is dual to (isomorphic to) theorthogonal complement.

The annihilator of a subset is itself a vector space.The annihilator of the zero vector is the whole dual space:{0}0=V{\displaystyle \{0\}^{0}=V^{*}}, and the annihilator of the whole space is just the zero covector:V0={0}V{\displaystyle V^{0}=\{0\}\subseteq V^{*}}.Furthermore, the assignment of an annihilator to a subset ofV{\displaystyle V} reverses inclusions, so that if{0}STV{\displaystyle \{0\}\subseteq S\subseteq T\subseteq V}, then

{0}T0S0V.{\displaystyle \{0\}\subseteq T^{0}\subseteq S^{0}\subseteq V^{*}.}

IfA{\displaystyle A} andB{\displaystyle B} are two subsets ofV{\displaystyle V} then

A0+B0(AB)0.{\displaystyle A^{0}+B^{0}\subseteq (A\cap B)^{0}.}

If(Ai)iI{\displaystyle (A_{i})_{i\in I}} is any family of subsets ofV{\displaystyle V} indexed byi{\displaystyle i} belonging to some index setI{\displaystyle I}, then

(iIAi)0=iIAi0.{\displaystyle \left(\bigcup _{i\in I}A_{i}\right)^{0}=\bigcap _{i\in I}A_{i}^{0}.}

In particular ifA{\displaystyle A} andB{\displaystyle B} are subspaces ofV{\displaystyle V} then

(A+B)0=A0B0{\displaystyle (A+B)^{0}=A^{0}\cap B^{0}}

and[nb 3]

(AB)0=A0+B0.{\displaystyle (A\cap B)^{0}=A^{0}+B^{0}.}

IfV{\displaystyle V} is finite-dimensional andW{\displaystyle W} is avector subspace, then

W00=W{\displaystyle W^{00}=W}

after identifyingW{\displaystyle W} with its image in the second dual space under the double duality isomorphismVV{\displaystyle V\approx V^{**}}. In particular, forming the annihilator is aGalois connection on the lattice of subsets of a finite-dimensional vector space.

IfW{\displaystyle W} is a subspace ofV{\displaystyle V} then thequotient spaceV/W{\displaystyle V/W} is a vector space in its own right, and so has a dual. By thefirst isomorphism theorem, a functionalf:VF{\displaystyle f:V\to F} factors throughV/W{\displaystyle V/W} if and only ifW{\displaystyle W} is in thekernel off{\displaystyle f}. There is thus an isomorphism

(V/W)W0.{\displaystyle (V/W)^{*}\cong W^{0}.}

As a particular consequence, ifV{\displaystyle V} is adirect sum of two subspacesA{\displaystyle A} andB{\displaystyle B}, thenV{\displaystyle V^{*}} is a direct sum ofA0{\displaystyle A^{0}} andB0{\displaystyle B^{0}}.

Dimensional analysis

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The dual space is analogous to a "negative"-dimensional space. Most simply, since a vectorvV{\displaystyle v\in V} can be paired with a covectorφV{\displaystyle \varphi \in V^{*}} by the natural pairingx,φ:=φ(x)F{\displaystyle \langle x,\varphi \rangle :=\varphi (x)\in F} to obtain a scalar, a covector can "cancel" the dimension of a vector, similar toreducing a fraction. Thus while the direct sumVV{\displaystyle V\oplus V^{*}} is a2n{\displaystyle 2n}-dimensional space (ifV{\displaystyle V} isn{\displaystyle n}-dimensional),V{\displaystyle V^{*}} behaves as an(n){\displaystyle (-n)}-dimensional space, in the sense that its dimensions can be canceled against the dimensions ofV{\displaystyle V}. This is formalized bytensor contraction.

This arises in physics viadimensional analysis, where the dual space has inverse units.[13] Under the natural pairing, these units cancel, and the resulting scalar value isdimensionless, as expected. For example, in (continuous)Fourier analysis, or more broadlytime–frequency analysis:[nb 4] given a one-dimensional vector space with aunit of timet{\displaystyle t}, the dual space has units offrequency: occurrencesper unit of time (units of1/t{\displaystyle 1/t}). For example, if time is measured inseconds, the corresponding dual unit is theinverse second: over the course of 3 seconds, an event that occurs 2 times per second occurs a total of 6 times, corresponding to3s2s1=6{\displaystyle 3s\cdot 2s^{-1}=6}. Similarly, if the primal space measures length, the dual space measuresinverse length.

Continuous dual space

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When dealing withtopological vector spaces, thecontinuous linear functionals from the space into the base fieldF=C{\displaystyle \mathbb {F} =\mathbb {C} } (orR{\displaystyle \mathbb {R} }) are particularly important.This gives rise to the notion of the "continuous dual space" or "topological dual" which is a linear subspace of the algebraic dual spaceV{\displaystyle V^{*}}, denoted byV{\displaystyle V'}.For anyfinite-dimensional normed vector space or topological vector space, such asEuclideann-space, the continuous dual and the algebraic dual coincide.This is however false for any infinite-dimensional normed space, as shown by the example ofdiscontinuous linear maps.Nevertheless, in the theory oftopological vector spaces the terms "continuous dual space" and "topological dual space" are often replaced by "dual space".

For atopological vector spaceV{\displaystyle V} itscontinuous dual space,[14] ortopological dual space,[15] or justdual space[14][15][16][17] (in the sense of the theory of topological vector spaces)V{\displaystyle V'} is defined as the space of all continuous linear functionalsφ:VF{\displaystyle \varphi :V\to {\mathbb {F} }}.

Important examples for continuous dual spaces are the space of compactly supportedtest functionsD{\displaystyle {\mathcal {D}}} and its dualD,{\displaystyle {\mathcal {D}}',} the space of arbitrarydistributions (generalized functions); the space of arbitrary test functionsE{\displaystyle {\mathcal {E}}} and its dualE,{\displaystyle {\mathcal {E}}',} the space of compactly supported distributions; and the space of rapidly decreasing test functionsS,{\displaystyle {\mathcal {S}},} theSchwartz space, and its dualS,{\displaystyle {\mathcal {S}}',} the space oftempered distributions (slowly growing distributions) in the theory ofgeneralized functions.

Properties

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IfX is aHausdorfftopological vector space (TVS), then the continuous dual space ofX is identical to the continuous dual space of thecompletion ofX.[1]

Topologies on the dual

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Main articles:Polar topology andDual system

There is a standard construction for introducing a topology on the continuous dualV{\displaystyle V'} of a topological vector spaceV{\displaystyle V}. Fix a collectionA{\displaystyle {\mathcal {A}}} ofbounded subsets ofV{\displaystyle V}.This gives the topology onV{\displaystyle V} ofuniform convergence on sets fromA,{\displaystyle {\mathcal {A}},} or what is the same thing, the topology generated byseminorms of the form

φA=supxA|φ(x)|,{\displaystyle \|\varphi \|_{A}=\sup _{x\in A}|\varphi (x)|,}

whereφ{\displaystyle \varphi } is a continuous linear functional onV{\displaystyle V}, andA{\displaystyle A} runs over the classA.{\displaystyle {\mathcal {A}}.}

This means that a net of functionalsφi{\displaystyle \varphi _{i}} tends to a functionalφ{\displaystyle \varphi } inV{\displaystyle V'} if and only if

 for all AAφiφA=supxA|φi(x)φ(x)|i0.{\displaystyle {\text{ for all }}A\in {\mathcal {A}}\qquad \|\varphi _{i}-\varphi \|_{A}=\sup _{x\in A}|\varphi _{i}(x)-\varphi (x)|{\underset {i\to \infty }{\longrightarrow }}0.}

Usually (but not necessarily) the classA{\displaystyle {\mathcal {A}}} is supposed to satisfy the following conditions:

If these requirements are fulfilled then the corresponding topology onV{\displaystyle V'} is Hausdorff and the sets

UA = {φV : φA<1}, for AA{\displaystyle U_{A}~=~\left\{\varphi \in V'~:~\quad \|\varphi \|_{A}<1\right\},\qquad {\text{ for }}A\in {\mathcal {A}}}

form its local base.

Here are the three most important special cases.

IfV{\displaystyle V} is anormed vector space (for example, aBanach space or aHilbert space) then the strong topology onV{\displaystyle V'} is normed (in fact a Banach space if the field of scalars is complete), with the norm

φ=supx1|φ(x)|.{\displaystyle \|\varphi \|=\sup _{\|x\|\leq 1}|\varphi (x)|.}

Each of these three choices of topology onV{\displaystyle V'} leads to a variant ofreflexivity property for topological vector spaces:

  • IfV{\displaystyle V'} is endowed with thestrong topology, then the corresponding notion of reflexivity is the standard one: the spaces reflexive in this sense are just calledreflexive.[18]
  • IfV{\displaystyle V'} is endowed with the stereotype dual topology, then the corresponding reflexivity is presented in the theory ofstereotype spaces: the spaces reflexive in this sense are calledstereotype.
  • IfV{\displaystyle V'} is endowed with theweak topology, then the corresponding reflexivity is presented in the theory ofdual pairs:[19] the spaces reflexive in this sense are arbitrary (Hausdorff) locally convex spaces with the weak topology.[20]

Examples

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Let 1 <p < ∞ be a real number and consider the Banach space p of allsequencesa = (an) for which

ap=(n=0|an|p)1p<.{\displaystyle \|\mathbf {a} \|_{p}=\left(\sum _{n=0}^{\infty }|a_{n}|^{p}\right)^{\frac {1}{p}}<\infty .}

Define the numberq by1/p + 1/q = 1. Then the continuous dual ofp is naturally identified withq: given an elementφ(p){\displaystyle \varphi \in (\ell ^{p})'}, the corresponding element ofq is the sequence(φ(en)){\displaystyle (\varphi (\mathbf {e} _{n}))} whereen{\displaystyle \mathbf {e} _{n}} denotes the sequence whosen-th term is 1 and all others are zero. Conversely, given an elementa = (an) ∈q, the corresponding continuous linear functionalφ{\displaystyle \varphi } onp is defined by

φ(b)=nanbn{\displaystyle \varphi (\mathbf {b} )=\sum _{n}a_{n}b_{n}}

for allb = (bn) ∈p (seeHölder's inequality).

In a similar manner, the continuous dual of 1 is naturally identified with ∞ (the space of bounded sequences).Furthermore, the continuous duals of the Banach spacesc (consisting of allconvergent sequences, with thesupremum norm) andc0 (the sequences converging to zero) are both naturally identified with 1.

By theRiesz representation theorem, the continuous dual of a Hilbert space is again a Hilbert space which isanti-isomorphic to the original space.This gives rise to thebra–ket notation used by physicists in the mathematical formulation ofquantum mechanics.

By theRiesz–Markov–Kakutani representation theorem, the continuous dual of certain spaces of continuous functions can be described using measures.

Transpose of a continuous linear map

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See also:Transpose of a linear map andDual system § Transposes

IfT :V → W is a continuous linear map between two topological vector spaces, then the (continuous) transposeT′ :W′ → V′ is defined by the same formula as before:

T(φ)=φT,φW.{\displaystyle T'(\varphi )=\varphi \circ T,\quad \varphi \in W'.}

The resulting functionalT′(φ) is inV′. The assignmentT → T′ produces a linear map between the space of continuous linear maps fromV toW and the space of linear maps fromW′ toV′.WhenT andU are composable continuous linear maps, then

(UT)=TU.{\displaystyle (U\circ T)'=T'\circ U'.}

WhenV andW are normed spaces, the norm of the transpose inL(W′,V′) is equal to that ofT inL(V,W).Several properties of transposition depend upon theHahn–Banach theorem.For example, the bounded linear mapT has dense rangeif and only if the transposeT′ is injective.

WhenT is acompact linear map between two Banach spacesV andW, then the transposeT′ is compact.This can be proved using theArzelà–Ascoli theorem.

WhenV is a Hilbert space, there is an antilinear isomorphismiV fromV onto its continuous dualV′.For every bounded linear mapT onV, the transpose and theadjoint operators are linked by

iVT=TiV.{\displaystyle i_{V}\circ T^{*}=T'\circ i_{V}.}

WhenT is a continuous linear map between two topological vector spacesV andW, then the transposeT′ is continuous whenW′ andV′ are equipped with "compatible" topologies: for example, when forX =V andX =W, both dualsX′ have thestrong topologyβ(X′,X) of uniform convergence on bounded sets ofX, or both have the weak-∗ topologyσ(X′,X) of pointwise convergence on X.The transposeT′ is continuous fromβ(W′,W) toβ(V′,V), or fromσ(W′,W) toσ(V′,V).

Annihilators

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Assume thatW is a closed linear subspace of a normed space V, and consider the annihilator ofW inV′,

W={φV:Wkerφ}.{\displaystyle W^{\perp }=\{\varphi \in V':W\subseteq \ker \varphi \}.}

Then, the dual of the quotientV / W can be identified withW, and the dual ofW can be identified with the quotientV′ / W.[21]Indeed, letP denote the canonicalsurjection fromV onto the quotientV / W; then, the transposeP′ is an isometric isomorphism from(V / W )′ intoV′, with range equal toW.Ifj denotes the injection map fromW intoV, then the kernel of the transposej′ is the annihilator ofW:

ker(j)=W{\displaystyle \ker(j')=W^{\perp }}

and it follows from theHahn–Banach theorem thatj′ induces an isometric isomorphismV′ / WW′.

Further properties

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If the dual of a normed spaceV isseparable, then so is the spaceV itself.The converse is not true: for example, the space 1 is separable, but its dual ∞ is not.

Double dual

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This is anatural transformation of vector addition from a vector space to its double dual.x1,x2 denotes theordered pair of two vectors. The addition + sendsx1 andx2 tox1 +x2. The addition +′ induced by the transformation can be defined as[Ψ(x1)+Ψ(x2)](φ)=φ(x1+x2)=φ(x){\displaystyle [\Psi (x_{1})+'\Psi (x_{2})](\varphi )=\varphi (x_{1}+x_{2})=\varphi (x)} for anyφ{\displaystyle \varphi } in the dual space.

In analogy with the case of the algebraic double dual, there is always a naturally defined continuous linear operatorΨ :VV′′ from a normed spaceV into its continuous double dualV′′, defined by

Ψ(x)(φ)=φ(x),xV, φV.{\displaystyle \Psi (x)(\varphi )=\varphi (x),\quad x\in V,\ \varphi \in V'.}

As a consequence of theHahn–Banach theorem, this map is in fact anisometry, meaning‖ Ψ(x) ‖ = ‖x for allxV.Normed spaces for which the map Ψ is abijection are calledreflexive.

WhenV is atopological vector space then Ψ(x) can still be defined by the same formula, for everyxV, however several difficulties arise.First, whenV is notlocally convex, the continuous dual may be equal to { 0 } and the map Ψ trivial.However, ifV isHausdorff and locally convex, the map Ψ is injective fromV to the algebraic dualV′ of the continuous dual, again as a consequence of the Hahn–Banach theorem.[nb 5]

Second, even in the locally convex setting, several natural vector space topologies can be defined on the continuous dualV′, so that the continuous double dualV′′ is not uniquely defined as a set. Saying that Ψ maps fromV toV′′, or in other words, that Ψ(x) is continuous onV′ for everyxV, is a reasonable minimal requirement on the topology ofV′, namely that the evaluation mappings

φVφ(x),xV,{\displaystyle \varphi \in V'\mapsto \varphi (x),\quad x\in V,}

be continuous for the chosen topology onV′. Further, there is still a choice of a topology onV′′, and continuity of Ψ depends upon this choice.As a consequence, definingreflexivity in this framework is more involved than in the normed case.

See also

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Notes

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  1. ^ForV{\displaystyle V^{\lor }} used in this way, seeAn Introduction to Manifolds (Tu 2011, p. 19).This notation is sometimes used when(){\displaystyle (\cdot )^{*}} is reserved for some other meaning.For instance, in the above text,F{\displaystyle F^{*}} is frequently used to denote the codifferential ofF{\displaystyle F}, so thatFω{\displaystyle F^{*}\omega } represents the pullback of the formω{\displaystyle \omega }.Halmos (1974, p. 20) usesV{\displaystyle V'} to denote the algebraic dual ofV{\displaystyle V}. However, other authors useV{\displaystyle V'} for the continuous dual, while reservingV{\displaystyle V^{*}} for the algebraic dual (Trèves 2006, p. 35).
  2. ^In many areas, such asquantum mechanics,,{\displaystyle \langle \cdot ,\cdot \rangle } is reserved for asesquilinear form defined onV×V{\displaystyle V\times V}.
  3. ^abcdSeveral assertions in this article require theaxiom of choice for their justification. The axiom of choice is needed to show that an arbitrary vector space has a basis: in particular it is needed to show thatRN{\displaystyle \mathbb {R} ^{\mathbb {N} }} has a basis.It is also needed to show that the dual of an infinite-dimensional vector spaceV{\displaystyle V} is nonzero, and hence that the natural map fromV{\displaystyle V} to its double dual is injective.
  4. ^To be precise, continuous Fourier analysis studies the space offunctionals with domain a vector space and the space of functionals on the dual vector space.
  5. ^IfV is locally convex but not Hausdorff, thekernel of Ψ is the smallest closed subspace containing {0}.

References

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  1. ^abNarici & Beckenstein 2011, pp. 225–273.
  2. ^Katznelson & Katznelson (2008) p. 37, §2.1.3
  3. ^abTu (2011) p. 19, §3.1
  4. ^Axler (2015) p. 101, §3.94
  5. ^Halmos (1974) p. 20, §13
  6. ^Halmos (1974) p. 21, §14
  7. ^Misner, Thorne & Wheeler 1973
  8. ^Misner, Thorne & Wheeler 1973, §2.5
  9. ^Nicolas Bourbaki (1974). Hermann (ed.).Elements of mathematics: Algebra I, Chapters 1 - 3. Addison-Wesley Publishing Company. p. 400.ISBN 0201006391.
  10. ^Mac Lane & Birkhoff 1999, §VI.4
  11. ^Halmos (1974) pp. 25, 28
  12. ^Halmos (1974) §44
  13. ^Tao, Terence (2012-12-29)."A mathematical formalisation of dimensional analysis".Similarly, one can defineVT1{\displaystyle V^{T^{-1}}} as the dual space toVT{\displaystyle V^{T}} ...
  14. ^abRobertson & Robertson 1964, II.2
  15. ^abSchaefer 1966, II.4
  16. ^Rudin 1973, 3.1
  17. ^Bourbaki 2003, II.42
  18. ^Schaefer 1966, IV.5.5
  19. ^Schaefer 1966, IV.1
  20. ^Schaefer 1966, IV.1.2
  21. ^Rudin 1991, chapter 4

Bibliography

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