Linear map

"Linear transformation" redirects here. For fractional linear transformations, seeMöbius transformation.

"Linear Operators" redirects here. For the textbook by Dunford and Schwarz, seeLinear Operators (book).

Not to be confused withlinear function.

Inmathematics,and more specifically inlinear algebra,alinear map(also called alinear mapping,linear transformation,vector space homomorphism,or in some contextslinear function) is amapping$V\to W$between twovector spacesthat preserves the operations ofvector additionandscalar multiplication.The same names and the same definition are also used for the more general case ofmodulesover aring;seeModule homomorphism.

If a linear map is abijectionthen it is called alinear isomorphism.In the case where$V=W$,a linear map is called alinear endomorphism.Sometimes the termlinear operatorrefers to this case,but the term "linear operator" can have different meanings for different conventions: for example, it can be used to emphasize that$V$and$W$arerealvector spaces (not necessarily with$V=W$),or it can be used to emphasize that$V$is afunction space,which is a common convention infunctional analysis.Sometimes the termlinear functionhas the same meaning aslinear map,while inanalysisit does not.

A linear map from$V$to$W$always maps the origin of$V$to the origin of$W$.Moreover, it mapslinear subspacesin$V$onto linear subspaces in$W$(possibly of a lowerdimension);for example, it maps aplanethrough theoriginin$V$to either a plane through the origin in$W$,alinethrough the origin in$W$,or just the origin in$W$.Linear maps can often be represented asmatrices,and simple examples includerotation and reflection linear transformations.

In the language ofcategory theory,linear maps are themorphismsof vector spaces, and they form a categoryequivalenttothe one of matrices.

Definition and first consequences

Let$V$and$W$be vector spaces over the samefield$K$. A function$f:V\to W$is said to be alinear mapif for any two vectors${\textstyle \mathbf {u},\mathbf {v} \in V}$and any scalar$c\in K$the following two conditions are satisfied:

• Additivity/ operation of addition$$f(\mathbf {u} +\mathbf {v} )=f(\mathbf {u} )+f(\mathbf {v} )$$
• Homogeneityof degree 1 / operation of scalar multiplication$$f(c\mathbf {u} )=cf(\mathbf {u} )$$

Thus, a linear map is said to beoperation preserving.In other words, it does not matter whether the linear map is applied before (the right hand sides of the above examples) or after (the left hand sides of the examples) the operations of addition and scalar multiplication.

Bythe associativity of the addition operationdenoted as +, for any vectors${\textstyle \mathbf {u} _{1},\ldots,\mathbf {u} _{n}\in V}$and scalars${\textstyle c_{1},\ldots,c_{n}\in K,}$the following equality holds: $$f(c_{1}\mathbf {u} _{1}+\cdots +c_{n}\mathbf {u} _{n})=c_{1}f(\mathbf {u} _{1})+\cdots +c_{n}f(\mathbf {u} _{n}).$$Thus a linear map is one which preserveslinear combinations.

Denoting the zero elements of the vector spaces$V$and$W$by${\textstyle \mathbf {0} _{V}}$and${\textstyle \mathbf {0} _{W}}$respectively, it follows that${\textstyle f(\mathbf {0} _{V})=\mathbf {0} _{W}.}$Let$c=0$and${\textstyle \mathbf {v} \in V}$in the equation for homogeneity of degree 1: $$f(\mathbf {0} _{V})=f(0\mathbf {v} )=0f(\mathbf {v} )=\mathbf {0} _{W}.$$

A linear map$V\to K$with$K$viewed as a one-dimensional vector space over itself is called alinear functional.

These statements generalize to any left-module${\textstyle {}_{R}M}$over a ring$R$without modification, and to any right-module upon reversing of the scalar multiplication.

Examples

• A prototypical example that gives linear maps their name is a function$f:\mathbb {R} \to \mathbb {R}:x\mapsto cx$,of which thegraphis a line through the origin.
• More generally, anyhomothety${\textstyle \mathbf {v} \mapsto c\mathbf {v} }$centered in the origin of a vector space is a linear map (herecis a scalar).
• The zero map${\textstyle \mathbf {x} \mapsto \mathbf {0} }$between two vector spaces (over the samefield) is linear.
• Theidentity mapon any module is a linear operator.
• For real numbers, the map${\textstyle x\mapsto x^{2}}$is not linear.
• For real numbers, the map${\textstyle x\mapsto x+1}$is not linear (but is anaffine transformation).
• If$A$is a$m\times n$real matrix,then$A$defines a linear map from$\mathbb {R} ^{n}$to$\mathbb {R} ^{m}$by sending acolumn vector$\mathbf {x} \in \mathbb {R} ^{n}$to the column vector$A\mathbf {x} \in \mathbb {R} ^{m}$.Conversely, any linear map betweenfinite-dimensionalvector spaces can be represented in this manner; see the§ Matrices,below.
• If${\textstyle f:V\to W}$is anisometrybetween realnormed spacessuch that${\textstyle f(0)=0}$then$f$is a linear map. This result is not necessarily true for complex normed space.
• Differentiationdefines a linear map from the space of all differentiable functions to the space of all functions. It also defines alinear operatoron the space of allsmooth functions(a linear operator is alinear endomorphism,that is, a linear map with the samedomainandcodomain). Indeed,$${\frac {d}{dx}}\left(af(x)+bg(x)\right)=a{\frac {df(x)}{dx}}+b{\frac {dg(x)}{dx}}.$$
• A definiteintegralover someintervalIis a linear map from the space of all real-valued integrable functions onIto$\mathbb {R}$.Indeed,$$\int _{u}^{v}\left(af(x)+bg(x)\right)dx=a\int _{u}^{v}f(x)dx+b\int _{u}^{v}g(x)dx.$$
• An indefiniteintegral(orantiderivative) with a fixed integration starting point defines a linear map from the space of all real-valued integrable functions on$\mathbb {R}$to the space of all real-valued, differentiable functions on$\mathbb {R}$.Without a fixed starting point, the antiderivative maps to thequotient spaceof the differentiable functions by the linear space of constant functions.
• If$V$and$W$are finite-dimensional vector spaces over a fieldF,of respective dimensionsmandn,then the function that maps linear maps${\textstyle f:V\to W}$ton×mmatrices in the way described in§ Matrices(below) is a linear map, and even alinear isomorphism.
• Theexpected valueof arandom variable(which is in fact a function, and as such an element of a vector space) is linear, as for random variables$X$and$Y$we have$E[X+Y]=E[X]+E[Y]$and$E[aX]=aE[X]$,but thevarianceof a random variable is not linear.

• 
  The function${\textstyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}$with${\textstyle f(x,y)=(2x,y)}$is a linear map. This function scales the${\textstyle x}$component of a vector by the factor${\textstyle 2}$.
• 
  The function${\textstyle f(x,y)=(2x,y)}$is additive: It does not matter whether vectors are first added and then mapped or whether they are mapped and finally added:${\textstyle f(\mathbf {a} +\mathbf {b} )=f(\mathbf {a} )+f(\mathbf {b} )}$
• 
  The function${\textstyle f(x,y)=(2x,y)}$is homogeneous: It does not matter whether a vector is first scaled and then mapped or first mapped and then scaled:${\textstyle f(\lambda \mathbf {a} )=\lambda f(\mathbf {a} )}$

Linear extensions

Often, a linear map is constructed by defining it on a subset of a vector space and thenextending by linearityto thelinear spanof the domain. Suppose$X$and$Y$are vector spaces and$f:S\to Y$is afunctiondefined on some subset$S\subseteq X.$ Then alinear extensionof$f$to$X,$if it exists, is a linear map$F:X\to Y$defined on$X$thatextends$f$(meaning that$F(s)=f(s)$for all$s\in S$) and takes its values from the codomain of$f.$ When the subset$S$is a vector subspace of$X$then a ($Y$-valued) linear extension of$f$to all of$X$is guaranteed to exist if (and only if)$f:S\to Y$is a linear map.In particular, if$f$has a linear extension to$\operatorname {span} S,$then it has a linear extension to all of$X.$

The map$f:S\to Y$can be extended to a linear map$F:\operatorname {span} S\to Y$if and only if whenever$n>0$is an integer,$c_{1},\ldots,c_{n}$are scalars, and$s_{1},\ldots,s_{n}\in S$are vectors such that$0=c_{1}s_{1}+\cdots +c_{n}s_{n},$then necessarily$0=c_{1}f\left(s_{1}\right)+\cdots +c_{n}f\left(s_{n}\right).$ If a linear extension of$f:S\to Y$exists then the linear extension$F:\operatorname {span} S\to Y$is unique and $$F\left(c_{1}s_{1}+\cdots c_{n}s_{n}\right)=c_{1}f\left(s_{1}\right)+\cdots +c_{n}f\left(s_{n}\right)$$ holds for all$n,c_{1},\ldots,c_{n},$and$s_{1},\ldots,s_{n}$as above. If$S$is linearly independent then every function$f:S\to Y$into any vector space has a linear extension to a (linear) map$\;\operatorname {span} S\to Y$(the converse is also true).

For example, if$X=\mathbb {R} ^{2}$and$Y=\mathbb {R}$then the assignment$(1,0)\to -1$and$(0,1)\to 2$can be linearly extended from the linearly independent set of vectors$S:=\{(1,0),(0,1)\}$to a linear map on$\operatorname {span} \{(1,0),(0,1)\}=\mathbb {R} ^{2}.$The unique linear extension$F:\mathbb {R} ^{2}\to \mathbb {R}$is the map that sends$(x,y)=x(1,0)+y(0,1)\in \mathbb {R} ^{2}$to $$F(x,y)=x(-1)+y(2)=-x+2y.$$

Every (scalar-valued)linear functional$f$defined on avector subspaceof a real or complex vector space$X$has a linear extension to all of$X.$ Indeed, theHahn–Banach dominated extension theoremeven guarantees that when this linear functional$f$is dominated by some givenseminorm$p:X\to \mathbb {R}$(meaning that$|f(m)|\leq p(m)$holds for all$m$in the domain of$f$) then there exists a linear extension to$X$that is also dominated by$p.$

Matrices

Main article:Transformation matrix

If$V$and$W$arefinite-dimensionalvector spaces and abasisis defined for each vector space, then every linear map from$V$to$W$can be represented by amatrix.This is useful because it allows concrete calculations. Matrices yield examples of linear maps: if$A$is a real$m\times n$matrix, then$f(\mathbf {x} )=A\mathbf {x}$describes a linear map$\mathbb {R} ^{n}\to \mathbb {R} ^{m}$(seeEuclidean space).

Let$\{\mathbf {v} _{1},\ldots,\mathbf {v} _{n}\}$be a basis for$V$.Then every vector$\mathbf {v} \in V$is uniquely determined by the coefficients$c_{1},\ldots,c_{n}$in the field$\mathbb {R}$: $$\mathbf {v} =c_{1}\mathbf {v} _{1}+\cdots +c_{n}\mathbf {v} _{n}.$$

If${\textstyle f:V\to W}$is a linear map, $$f(\mathbf {v} )=f(c_{1}\mathbf {v} _{1}+\cdots +c_{n}\mathbf {v} _{n})=c_{1}f(\mathbf {v} _{1})+\cdots +c_{n}f\left(\mathbf {v} _{n}\right),$$

which implies that the functionfis entirely determined by the vectors$f(\mathbf {v} _{1}),\ldots,f(\mathbf {v} _{n})$.Now let$\{\mathbf {w} _{1},\ldots,\mathbf {w} _{m}\}$be a basis for$W$.Then we can represent each vector$f(\mathbf {v} _{j})$as $$f\left(\mathbf {v} _{j}\right)=a_{1j}\mathbf {w} _{1}+\cdots +a_{mj}\mathbf {w} _{m}.$$

Thus, the function$f$is entirely determined by the values of$a_{ij}$.If we put these values into an$m\times n$matrix$M$,then we can conveniently use it to compute the vector output of$f$for any vector in$V$.To get$M$,every column$j$of$M$is a vector $${\begin{pmatrix}a_{1j}\\\vdots \\a_{mj}\end{pmatrix}}$$ corresponding to$f(\mathbf {v} _{j})$as defined above. To define it more clearly, for some column$j$that corresponds to the mapping$f(\mathbf {v} _{j})$, $$\mathbf {M} ={\begin{pmatrix}\ \cdots &a_{1j}&\cdots \ \\&\vdots &\\&a_{mj}&\end{pmatrix}}$$ where$M$is the matrix of$f$.In other words, every column$j=1,\ldots,n$has a corresponding vector$f(\mathbf {v} _{j})$whose coordinates$a_{1j},\cdots,a_{mj}$are the elements of column$j$.A single linear map may be represented by many matrices. This is because the values of the elements of a matrix depend on the bases chosen.

The matrices of a linear transformation can be represented visually:

1. Matrix for${\textstyle T}$relative to${\textstyle B}$:${\textstyle A}$
2. Matrix for${\textstyle T}$relative to${\textstyle B'}$:${\textstyle A'}$
3. Transition matrix from${\textstyle B'}$to${\textstyle B}$:${\textstyle P}$
4. Transition matrix from${\textstyle B}$to${\textstyle B'}$:${\textstyle P^{-1}}$

The relationship between matrices in a linear transformation

Such that starting in the bottom left corner${\textstyle \left[\mathbf {v} \right]_{B'}}$and looking for the bottom right corner${\textstyle \left[T\left(\mathbf {v} \right)\right]_{B'}}$,one would left-multiply—that is,${\textstyle A'\left[\mathbf {v} \right]_{B'}=\left[T\left(\mathbf {v} \right)\right]_{B'}}$.The equivalent method would be the "longer" method going clockwise from the same point such that${\textstyle \left[\mathbf {v} \right]_{B'}}$is left-multiplied with${\textstyle P^{-1}AP}$,or${\textstyle P^{-1}AP\left[\mathbf {v} \right]_{B'}=\left[T\left(\mathbf {v} \right)\right]_{B'}}$.

Examples in two dimensions

In two-dimensionalspaceR²linear maps are described by 2 × 2matrices.These are some examples:

• rotation
  • by 90 degrees counterclockwise:$$\mathbf {A} ={\begin{pmatrix}0&-1\\1&0\end{pmatrix}}$$
  • by an angleθcounterclockwise:$$\mathbf {A} ={\begin{pmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{pmatrix}}$$
• reflection
  • through thexaxis:$$\mathbf {A} ={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}$$
  • through theyaxis:$$\mathbf {A} ={\begin{pmatrix}-1&0\\0&1\end{pmatrix}}$$
  • through a line making an angleθwith the origin:$$\mathbf {A} ={\begin{pmatrix}\cos 2\theta &\sin 2\theta \\\sin 2\theta &-\cos 2\theta \end{pmatrix}}$$
• scalingby 2 in all directions:$$\mathbf {A} ={\begin{pmatrix}2&0\\0&2\end{pmatrix}}=2\mathbf {I}$$
• horizontal shear mapping:$$\mathbf {A} ={\begin{pmatrix}1&m\\0&1\end{pmatrix}}$$
• skew of theyaxis by an angleθ:$$\mathbf {A} ={\begin{pmatrix}1&-\sin \theta \\0&\cos \theta \end{pmatrix}}$$
• squeeze mapping:$$\mathbf {A} ={\begin{pmatrix}k&0\\0&{\frac {1}{k}}\end{pmatrix}}$$
• projectiononto theyaxis:$$\mathbf {A} ={\begin{pmatrix}0&0\\0&1\end{pmatrix}}.$$

If a linear map is only composed of rotation, reflection, and/or uniform scaling, then the linear map is aconformal linear transformation.

Vector space of linear maps

The composition of linear maps is linear: if$f:V\to W$and${\textstyle g:W\to Z}$are linear, then so is theircomposition${\textstyle g\circ f:V\to Z}$.It follows from this that theclassof all vector spaces over a given fieldK,together withK-linear maps asmorphisms,forms acategory.

Theinverseof a linear map, when defined, is again a linear map.

If${\textstyle f_{1}:V\to W}$and${\textstyle f_{2}:V\to W}$are linear, then so is theirpointwisesum$f_{1}+f_{2}$,which is defined by$(f_{1}+f_{2})(\mathbf {x} )=f_{1}(\mathbf {x} )+f_{2}(\mathbf {x} )$.

If${\textstyle f:V\to W}$is linear and${\textstyle \ Alpha }$is an element of the ground field${\textstyle K}$,then the map${\textstyle \ Alpha f}$,defined by${\textstyle (\ Alpha f)(\mathbf {x} )=\ Alpha (f(\mathbf {x} ))}$,is also linear.

Thus the set${\textstyle {\mathcal {L}}(V,W)}$of linear maps from${\textstyle V}$to${\textstyle W}$itself forms a vector space over${\textstyle K}$,sometimes denoted${\textstyle \operatorname {Hom} (V,W)}$.Furthermore, in the case that${\textstyle V=W}$,this vector space, denoted${\textstyle \operatorname {End} (V)}$,is anassociative algebraundercomposition of maps,since the composition of two linear maps is again a linear map, and the composition of maps is always associative. This case is discussed in more detail below.

Given again the finite-dimensional case, if bases have been chosen, then the composition of linear maps corresponds to thematrix multiplication,the addition of linear maps corresponds to thematrix addition,and the multiplication of linear maps with scalars corresponds to the multiplication of matrices with scalars.

Endomorphisms and automorphisms

Main articles:EndomorphismandAutomorphism

A linear transformation${\textstyle f:V\to V}$is anendomorphismof${\textstyle V}$;the set of all such endomorphisms${\textstyle \operatorname {End} (V)}$together with addition, composition and scalar multiplication as defined above forms anassociative algebrawith identity element over the field${\textstyle K}$(and in particular aring). The multiplicative identity element of this algebra is theidentity map${\textstyle \operatorname {id}:V\to V}$.

An endomorphism of${\textstyle V}$that is also anisomorphismis called anautomorphismof${\textstyle V}$.The composition of two automorphisms is again an automorphism, and the set of all automorphisms of${\textstyle V}$forms agroup,theautomorphism groupof${\textstyle V}$which is denoted by${\textstyle \operatorname {Aut} (V)}$or${\textstyle \operatorname {GL} (V)}$.Since the automorphisms are precisely thoseendomorphismswhich possess inverses under composition,${\textstyle \operatorname {Aut} (V)}$is the group ofunitsin the ring${\textstyle \operatorname {End} (V)}$.

If${\textstyle V}$has finite dimension${\textstyle n}$,then${\textstyle \operatorname {End} (V)}$isisomorphicto theassociative algebraof all${\textstyle n\times n}$matrices with entries in${\textstyle K}$.The automorphism group of${\textstyle V}$isisomorphicto thegeneral linear group${\textstyle \operatorname {GL} (n,K)}$of all${\textstyle n\times n}$invertible matrices with entries in${\textstyle K}$.

Kernel, image and the rank–nullity theorem

Main articles:Kernel (linear algebra),Image (mathematics),andRank of a matrix

If${\textstyle f:V\to W}$is linear, we define thekerneland theimageorrangeof${\textstyle f}$by $${\begin{aligned}\ker(f)&=\{\,\mathbf {x} \in V:f(\mathbf {x} )=\mathbf {0} \,\}\\\operatorname {im} (f)&=\{\,\mathbf {w} \in W:\mathbf {w} =f(\mathbf {x} ),\mathbf {x} \in V\,\}\end{aligned}}$$

${\textstyle \ker(f)}$is asubspaceof${\textstyle V}$and${\textstyle \operatorname {im} (f)}$is a subspace of${\textstyle W}$.The followingdimensionformula is known as therank–nullity theorem: $$\dim(\ker(f))+\dim(\operatorname {im} (f))=\dim(V).$$

The number${\textstyle \dim(\operatorname {im} (f))}$is also called therankof${\textstyle f}$and written as${\textstyle \operatorname {rank} (f)}$,or sometimes,${\textstyle \rho (f)}$;the number${\textstyle \dim(\ker(f))}$is called thenullityof${\textstyle f}$and written as${\textstyle \operatorname {null} (f)}$or${\textstyle \nu (f)}$.If${\textstyle V}$and${\textstyle W}$are finite-dimensional, bases have been chosen and${\textstyle f}$is represented by the matrix${\textstyle A}$,then the rank and nullity of${\textstyle f}$are equal to the rank and nullity of the matrix${\textstyle A}$,respectively.

Cokernel

Main article:Cokernel

A subtler invariant of a linear transformation${\textstyle f:V\to W}$is thecokernel,which is defined as $$\operatorname {coker} (f):=W/f(V)=W/\operatorname {im} (f).$$

This is thedualnotion to the kernel: just as the kernel is asubspace of thedomain,the co-kernel is aquotientspaceof thetarget.Formally, one has theexact sequence $$0\to \ker(f)\to V\to W\to \operatorname {coker} (f)\to 0.$$

These can be interpreted thus: given a linear equationf(v) =wto solve,

• the kernel is the space ofsolutionsto thehomogeneousequationf(v) = 0, and its dimension is the number ofdegrees of freedomin the space of solutions, if it is not empty;
• the co-kernel is the space ofconstraintsthat the solutions must satisfy, and its dimension is the maximal number of independent constraints.

The dimension of the co-kernel and the dimension of the image (the rank) add up to the dimension of the target space. For finite dimensions, this means that the dimension of the quotient spaceW/f(V) is the dimension of the target space minus the dimension of the image.

As a simple example, consider the mapf:R²→R²,given byf(x,y) = (0,y). Then for an equationf(x,y) = (a,b) to have a solution, we must havea= 0 (one constraint), and in that case the solution space is (x,b) or equivalently stated, (0,b) + (x,0), (one degree of freedom). The kernel may be expressed as the subspace (x,0) <V:the value ofxis the freedom in a solution – while the cokernel may be expressed via the mapW→R,${\textstyle (a,b)\mapsto (a)}$:given a vector (a,b), the value ofais theobstructionto there being a solution.

An example illustrating the infinite-dimensional case is afforded by the mapf:R^∞→R^∞,${\textstyle \left\{a_{n}\right\}\mapsto \left\{b_{n}\right\}}$withb₁= 0 andb_{n+ 1}=a_nforn> 0. Its image consists of all sequences with first element 0, and thus its cokernel consists of the classes of sequences with identical first element. Thus, whereas its kernel has dimension 0 (it maps only the zero sequence to the zero sequence), its co-kernel has dimension 1. Since the domain and the target space are the same, the rank and the dimension of the kernel add up to the samesumas the rank and the dimension of the co-kernel (${\textstyle \aleph _{0}+0=\aleph _{0}+1}$), but in the infinite-dimensional case it cannot be inferred that the kernel and the co-kernel of anendomorphismhave the same dimension (0 ≠ 1). The reverse situation obtains for the maph:R^∞→R^∞,${\textstyle \left\{a_{n}\right\}\mapsto \left\{c_{n}\right\}}$withc_n=a_{n+ 1}.Its image is the entire target space, and hence its co-kernel has dimension 0, but since it maps all sequences in which only the first element is non-zero to the zero sequence, its kernel has dimension 1.

Index

For a linear operator with finite-dimensional kernel and co-kernel, one may defineindexas: $$\operatorname {ind} (f):=\dim(\ker(f))-\dim(\operatorname {coker} (f)),$$ namely the degrees of freedom minus the number of constraints.

For a transformation between finite-dimensional vector spaces, this is just the difference dim(V) − dim(W), by rank–nullity. This gives an indication of how many solutions or how many constraints one has: if mapping from a larger space to a smaller one, the map may be onto, and thus will have degrees of freedom even without constraints. Conversely, if mapping from a smaller space to a larger one, the map cannot be onto, and thus one will have constraints even without degrees of freedom.

The index of an operator is precisely theEuler characteristicof the 2-term complex 0 →V→W→ 0. Inoperator theory,the index ofFredholm operatorsis an object of study, with a major result being theAtiyah–Singer index theorem.

Algebraic classifications of linear transformations

No classification of linear maps could be exhaustive. The following incomplete list enumerates some important classifications that do not require any additional structure on the vector space.

LetVandWdenote vector spaces over a fieldFand letT:V→Wbe a linear map.

Monomorphism

Tis said to beinjectiveor amonomorphismif any of the following equivalent conditions are true:

1. Tisone-to-oneas a map ofsets.
2. kerT= {0_V}
3. dim(kerT) = 0
4. Tismonicor left-cancellable, which is to say, for any vector spaceUand any pair of linear mapsR:U→VandS:U→V,the equationTR=TSimpliesR=S.
5. Tisleft-invertible,which is to say there exists a linear mapS:W→Vsuch thatSTis theidentity maponV.

Epimorphism

Tis said to besurjectiveor anepimorphismif any of the following equivalent conditions are true:

1. Tisontoas a map of sets.
2. cokerT= {0_W}
3. Tisepicor right-cancellable, which is to say, for any vector spaceUand any pair of linear mapsR:W→UandS:W→U,the equationRT=STimpliesR=S.
4. Tisright-invertible,which is to say there exists a linear mapS:W→Vsuch thatTSis theidentity maponW.

Isomorphism

Tis said to be anisomorphismif it is both left- and right-invertible. This is equivalent toTbeing both one-to-one and onto (abijectionof sets) or also toTbeing both epic and monic, and so being abimorphism.

IfT:V→Vis an endomorphism, then:

• If, for some positive integern,then-th iterate ofT,Tⁿ,is identically zero, thenTis said to benilpotent.
• IfT²=T,thenTis said to beidempotent
• IfT=kI,wherekis some scalar, thenTis said to be a scaling transformation or scalar multiplication map; seescalar matrix.

Change of basis

Main articles:Basis (linear algebra)andChange of basis

Given a linear map which is anendomorphismwhose matrix isA,in the basisBof the space it transforms vector coordinates [u] as [v] =A[u]. As vectors change with the inverse ofB(vectors coordinates arecontravariant) its inverse transformation is [v] =B[v'].

Substituting this in the first expression $$B\left[v'\right]=AB\left[u'\right]$$ hence $$\left[v'\right]=B^{-1}AB\left[u'\right]=A'\left[u'\right].$$

Therefore, the matrix in the new basis isA′=B⁻¹AB,beingBthe matrix of the given basis.

Therefore, linear maps are said to be 1-co- 1-contra-variantobjects, or type (1, 1)tensors.

Continuity

Main articles:Continuous linear operatorandDiscontinuous linear map

Alinear transformationbetweentopological vector spaces,for examplenormed spaces,may becontinuous.If its domain and codomain are the same, it will then be acontinuous linear operator.A linear operator on a normed linear space is continuous if and only if it isbounded,for example, when the domain is finite-dimensional.An infinite-dimensional domain may havediscontinuous linear operators.

An example of an unbounded, hence discontinuous, linear transformation is differentiation on the space of smooth functions equipped with the supremum norm (a function with small values can have a derivative with large values, while the derivative of 0 is 0). For a specific example,sin(nx)/nconverges to 0, but its derivativecos(nx)does not, so differentiation is not continuous at 0 (and by a variation of this argument, it is not continuous anywhere).

Applications

A specific application of linear maps is forgeometric transformations,such as those performed incomputer graphics,where the translation, rotation and scaling of 2D or 3D objects is performed by the use of atransformation matrix.Linear mappings also are used as a mechanism for describing change: for example in calculus correspond to derivatives; or in relativity, used as a device to keep track of the local transformations of reference frames.

Another application of these transformations is incompiler optimizationsof nested-loop code, and inparallelizing compilertechniques.

See also

• Additive map– Z-module homomorphism
• Antilinear map– Conjugate homogeneous additive map
• Bent function– Special type of Boolean function
• Bounded operator– Linear transformation between topological vector spaces
• Cauchy's functional equation– Functional equation
• Continuous linear operator
• Linear functional– Linear map from a vector space to its field of scalarsPages displaying short descriptions of redirect targets
• Linear isometry– Distance-preserving mathematical transformationPages displaying short descriptions of redirect targets
• Category of matrices
• Quasilinearization

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