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Transpose of a linear map

From Wikipedia, the free encyclopedia
Induced map between the dual spaces of the two vector spaces
See also:Transpose,Dual system § Transposes, andTranspose § Transposes of linear maps and bilinear forms

Inlinear algebra, the transpose of alinear map between two vector spaces, defined over the samefield, is an induced map between thedual spaces of the two vector spaces. Thetranspose oralgebraic adjoint of a linear map is often used to study the original linear map. This concept is generalised byadjoint functors.

Definition

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See also:Dual system § Transposes, andTranspose § Transposes of linear maps and bilinear forms

LetX#{\displaystyle X^{\#}} denote thealgebraic dual space of a vector spaceX{\displaystyle X}. LetX{\displaystyle X} andY{\displaystyle Y} be vector spaces over the same fieldK{\displaystyle {\mathcal {K}}}. Ifu:XY{\displaystyle u:X\to Y} is alinear map, then itsalgebraic adjoint ordual,[1] is the map#u:Y#X#{\displaystyle {}^{\#\!}u:Y^{\#}\to X^{\#}} defined byffu{\displaystyle f\mapsto f\circ u}. The resulting functional#u(f):=fu{\displaystyle {}^{\#\!}u(f):=f\circ u} is called thepullback off{\displaystyle f} byu{\displaystyle u}.

Thecontinuous dual space of atopological vector space (TVS)X{\displaystyle X} is denoted byX{\displaystyle X^{\prime }}. IfX{\displaystyle X} andY{\displaystyle Y} are TVSs then a linear mapu:XY{\displaystyle u:X\to Y} isweakly continuous if and only if#u(Y)X{\displaystyle {}^{\#\!}u\left(Y^{\prime }\right)\subseteq X^{\prime }}, in which case we lettu:YX{\displaystyle {}^{\text{t}}\!u:Y^{\prime }\to X^{\prime }} denote the restriction of#u{\displaystyle {}^{\#\!}u} toY{\displaystyle Y^{\prime }}. The maptu{\displaystyle {}^{\text{t}}\!u} is called thetranspose[2] oralgebraic adjoint ofu{\displaystyle u}. The following identity characterizes the transpose ofu{\displaystyle u}:[3]tu(f),x=f,u(x) for all fY and xX,{\displaystyle \left\langle {}^{\text{t}}\!u(f),x\right\rangle =\left\langle f,u(x)\right\rangle \quad {\text{ for all }}f\in Y^{\prime }{\text{ and }}x\in X,}where,{\displaystyle \left\langle \cdot ,\cdot \right\rangle } is thenatural pairing defined by1{\displaystyle {1}}.

Properties

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The assignmentutu{\displaystyle u\mapsto {}^{\text{t}}\!u} produces aninjective linear map between the space of linear operators fromX{\displaystyle X} toY{\displaystyle Y} and the space of linear operators fromY#{\displaystyle Y^{\#}} toX#{\displaystyle X^{\#}}. IfX=Y{\displaystyle X=Y} then the space of linear maps is analgebra undercomposition of maps, and the assignment is then anantihomomorphism of algebras, meaning that1{\displaystyle {1}}. 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 overK{\displaystyle {\mathcal {K}}} to itself. One can identifyt(tu){\displaystyle {}^{\text{t}}\!\!\left({}^{\text{t}}\!u\right)} withu{\displaystyle u} using the natural injection into the double dual.

x=supx1|x(x)| for each xX{\displaystyle \|x\|=\sup _{\|x^{\prime }\|\leq 1}\left|x^{\prime }(x)\right|\quad {\text{ for each }}x\in X}and if the linear operatoru:XY{\displaystyle u:X\to Y} is bounded then theoperator norm oftu{\displaystyle {}^{\text{t}}\!u} is equal to the norm ofu{\displaystyle u}; that is[5][6]u=tu,{\displaystyle \|u\|=\left\|{}^{\text{t}}\!u\right\|,}and moreover,u=sup{|y(ux)|:x1,y1 where xX,yY}.{\displaystyle \|u\|=\sup \left\{\left|y^{\prime }(ux)\right|:\|x\|\leq 1,\left\|y^{*}\right\|\leq 1{\text{ where }}x\in X,y^{\prime }\in Y^{\prime }\right\}.}

Polars

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Suppose now thatu:XY{\displaystyle u:X\to Y} is a weakly continuous linear operator betweentopological vector spacesX{\displaystyle X} andY{\displaystyle Y} with continuous dual spacesX{\displaystyle X^{\prime }} andY{\displaystyle Y^{\prime }}, respectively. Let,:X×XC{\displaystyle \langle \cdot ,\cdot \rangle :X\times X^{\prime }\to \mathbb {C} } denote the canonicaldual system, defined byx,x=xx{\displaystyle \left\langle x,x^{\prime }\right\rangle =x^{\prime }x} wherex{\displaystyle x} andx{\displaystyle x^{\prime }} are said to beorthogonal ifx,x=xx=0{\displaystyle \left\langle x,x^{\prime }\right\rangle =x^{\prime }x=0}. For any subsetsAX{\displaystyle A\subseteq X} andSX{\displaystyle S^{\prime }\subseteq X^{\prime }}, letA={xX:supaA|x(a)|1} and S={xX:supsS|s(x)|1}{\displaystyle A^{\circ }=\left\{x^{\prime }\in X^{\prime }:\sup _{a\in A}\left|x^{\prime }(a)\right|\leq 1\right\}\qquad {\text{ and }}\qquad S^{\circ }=\left\{x\in X:\sup _{s^{\prime }\in S^{\prime }}\left|s^{\prime }(x)\right|\leq 1\right\}} denote the (absolute)polar ofA{\displaystyle A} inX{\displaystyle X^{\prime }} (resp.ofS{\displaystyle S^{\prime }} inX{\displaystyle X}).

[u(A)]=(tu)1(A){\displaystyle [u(A)]^{\circ }=\left({}^{\text{t}}\!u\right)^{-1}\left(A^{\circ }\right)}andu(A)B implies tu(B)A.{\displaystyle u(A)\subseteq B\quad {\text{ implies }}\quad {}^{\text{t}}\!u\left(B^{\circ }\right)\subseteq A^{\circ }.}

kertu=(Imu).{\displaystyle \operatorname {ker} {}^{\text{t}}\!u=\left(\operatorname {Im} u\right)^{\circ }.}

Annihilators

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SupposeX{\displaystyle X} andY{\displaystyle Y} aretopological vector spaces andu:XY{\displaystyle u:X\to Y} is a weakly continuous linear operator (so(tu)(Y)X{\displaystyle \left({}^{\text{t}}\!u\right)\left(Y^{\prime }\right)\subseteq X^{\prime }}). Given subsetsMX{\displaystyle M\subseteq X} andNX{\displaystyle N\subseteq X^{\prime }}, define theirannihilators (with respect to the canonical dual system) by[6]

M:={xX:m,x=0 for all mM}={xX:x(M)={0}} where x(M):={x(m):mM}{\displaystyle {\begin{alignedat}{4}M^{\bot }:&=\left\{x^{\prime }\in X^{\prime }:\left\langle m,x^{\prime }\right\rangle =0{\text{ for all }}m\in M\right\}\\&=\left\{x^{\prime }\in X^{\prime }:x^{\prime }(M)=\{0\}\right\}\qquad {\text{ where }}x^{\prime }(M):=\left\{x^{\prime }(m):m\in M\right\}\end{alignedat}}}

and

N:={xX:x,n=0 for all nN}={xX:N(x)={0}} where N(x):={n(x):nN}{\displaystyle {\begin{alignedat}{4}{}^{\bot }N:&=\left\{x\in X:\left\langle x,n^{\prime }\right\rangle =0{\text{ for all }}n^{\prime }\in N\right\}\\&=\left\{x\in X:N(x)=\{0\}\right\}\qquad {\text{ where }}N(x):=\left\{n^{\prime }(x):n^{\prime }\in N\right\}\\\end{alignedat}}}

kertu=(Imu){\displaystyle \ker {}^{\text{t}}\!u=(\operatorname {Im} u)^{\bot }}

Duals of quotient spaces

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LetM{\displaystyle M} be a closed vector subspace of a Hausdorff locally convex spaceX{\displaystyle X} and denote the canonical quotient map byπ:XX/M where π(x):=x+M.{\displaystyle \pi :X\to X/M\quad {\text{ where }}\quad \pi (x):=x+M.}AssumeX/M{\displaystyle X/M} is endowed with thequotient topology induced by the quotient mapπ:XX/M{\displaystyle \pi :X\to X/M}. Then the transpose of the quotient map is valued inM{\displaystyle M^{\bot }} andtπ:(X/M)MX{\displaystyle {}^{\text{t}}\!\pi :(X/M)^{\prime }\to M^{\bot }\subseteq X^{\prime }}is a TVS-isomorphism ontoM{\displaystyle M^{\bot }}. IfX{\displaystyle X} is aBanach space thentπ:(X/M)M{\displaystyle {}^{\text{t}}\!\pi :(X/M)^{\prime }\to M^{\bot }} is also anisometry.[6] Using this transpose, every continuous linear functional on the quotient spaceX/M{\displaystyle X/M} is canonically identified with a continuous linear functional in the annihilatorM{\displaystyle M^{\bot }} ofM{\displaystyle M}.

Duals of vector subspaces

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LetM{\displaystyle M} be a closed vector subspace of a Hausdorff locally convex spaceX{\displaystyle X}. IfmM{\displaystyle m^{\prime }\in M^{\prime }} and ifxX{\displaystyle x^{\prime }\in X^{\prime }} is a continuous linear extension ofm{\displaystyle m^{\prime }} toX{\displaystyle X} then the assignmentmx+M{\displaystyle m^{\prime }\mapsto x^{\prime }+M^{\bot }} induces a vector space isomorphismMX/(M),{\displaystyle M^{\prime }\to X^{\prime }/\left(M^{\bot }\right),}which is an isometry ifX{\displaystyle X} is a Banach space.[6]

Denote theinclusion map byIn:MX where In(m):=m for all mM.{\displaystyle \operatorname {In} :M\to X\quad {\text{ where }}\quad \operatorname {In} (m):=m\quad {\text{ for all }}m\in M.}The transpose of the inclusion map istIn:XM{\displaystyle {}^{\text{t}}\!\operatorname {In} :X^{\prime }\to M^{\prime }}whose kernel is the annihilatorM={xX:m,x=0 for all mM}{\displaystyle M^{\bot }=\left\{x^{\prime }\in X^{\prime }:\left\langle m,x^{\prime }\right\rangle =0{\text{ for all }}m\in M\right\}} and which is surjective by theHahn–Banach theorem. This map induces an isomorphism of vector spacesX/(M)M.{\displaystyle X^{\prime }/\left(M^{\bot }\right)\to M^{\prime }.}

Representation as a matrix

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If the linear mapu{\displaystyle u} is represented by thematrixA{\displaystyle A} with respect to two bases ofX{\displaystyle X} andY{\displaystyle Y}, thentu{\displaystyle {}^{\text{t}}\!u} is represented by thetranspose matrixAT{\displaystyle A^{\text{T}}} with respect to the dual bases ofY{\displaystyle Y^{\prime }} andX{\displaystyle X^{\prime }}, hence the name. Alternatively, asu{\displaystyle u} is represented byA{\displaystyle A} acting to the right on column vectors,tu{\displaystyle {}^{\text{t}}\!u} is represented by the same matrix acting to the left on row vectors. These points of view are related by the canonical inner product onRn{\displaystyle \mathbb {R} ^{n}}, which identifies the space of column vectors with the dual space of row vectors.

Relation to the Hermitian adjoint

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Main article:Hermitian adjoint
See also:Riesz representation theorem

The identity that characterizes the transpose, that is,1{\displaystyle {1}}, is formally similar to the definition of theHermitian adjoint, however, the transpose and the Hermitian adjoint are not the same map. The transpose is a mapYX{\displaystyle Y^{\prime }\to X^{\prime }} and is defined for linear maps between any vector spacesX{\displaystyle X} andY{\displaystyle Y}, without requiring any additional structure. The Hermitian adjoint mapsYX{\displaystyle Y\to X} and is only defined for linear maps between Hilbert spaces, as it is defined in terms of theinner product on the Hilbert space. The Hermitian adjoint therefore requires more mathematical structure than the transpose.

However, the transpose is often used in contexts where the vector spaces are both equipped with anondegenerate bilinear form such as the Euclideandot product or anotherrealinner product. In this case, the nondegenerate bilinear form is oftenused implicitly to map between the vector spaces and their duals, to express the transposed map as a mapYX{\displaystyle Y\to X}. For a complex Hilbert space, the inner product is sesquilinear and not bilinear, and these conversions change the transpose into the adjoint map.

More precisely: ifX{\displaystyle X} andY{\displaystyle Y} are Hilbert spaces andu:XY{\displaystyle u:X\to Y} is a linear map then the transpose ofu{\displaystyle u} and the Hermitian adjoint ofu{\displaystyle u}, which we will denote respectively bytu{\displaystyle {}^{\text{t}}\!u} andu{\displaystyle u^{*}}, are related. Denote byI:XX{\displaystyle I:X\to X^{*}} andJ:YY{\displaystyle J:Y\to Y^{*}} the canonical antilinear isometries of the Hilbert spacesX{\displaystyle X} andY{\displaystyle Y} onto their duals. Thenu{\displaystyle u^{*}} is the following composition of maps:[10]

YJYtuXI1X{\displaystyle Y{\overset {J}{\longrightarrow }}Y^{*}{\overset {{}^{{\text{t}}\!}u}{\longrightarrow }}X^{*}{\overset {I^{-1}}{\longrightarrow }}X}

Applications to functional analysis

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Suppose thatX{\displaystyle X} andY{\displaystyle Y} aretopological vector spaces and thatu:XY{\displaystyle u:X\to Y} is a linear map, then many ofu{\displaystyle u}'s properties are reflected intu{\displaystyle {}^{\text{t}}\!u}.

See also

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References

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  1. ^Schaefer & Wolff 1999, p. 128
  2. ^Trèves 2006, p. 240
  3. ^Halmos (1974, §44)
  4. ^abcdeSchaefer & Wolff 1999, pp. 129–130
  5. ^abTrèves 2006, pp. 240–252
  6. ^abcdRudin 1991, pp. 92–115
  7. ^abcSchaefer & Wolff 1999, pp. 128–130
  8. ^Trèves 2006, pp. 199–200
  9. ^Trèves 2006, pp. 382–383
  10. ^Trèves 2006, p. 488

Bibliography

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Basic concepts
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Vector space constructions
Numerical
Duality and spaces oflinear maps
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Main results
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