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Math429
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Math429 L8

Lecture 8

Chapter III Linear maps

Assumption: U,V,WU,V,W are vector spaces (over F\mathbb{F})

Vector Space of Linear Maps 3A

Definition 3.1

A linear map from VV to WW is a function from T:VWT:V\to W with the following properties:

  1. Additivity: T(u+v)=T(u)+T(v),u,vVT(u+v)=T(u)+T(v),\forall u,v\in V
  2. Homogeneity: T(λv)=λT(v),λF,vVT(\lambda v)=\lambda T(v),\forall \lambda \in \mathbb{F},v\in V

Notation

  • Tv=T(v)Tv=T(v)
  • L(V,W)\mathscr{L}(V,W) denotes the set of linear maps from VV to WW. (homomorphism, Hom(V,W)Hom(V,W))
  • L(V)\mathscr{L}(V) denotes the set of linear maps from VV to VV. (endomorphism, End(V)End(V))

Example

  • zero map 0(v)L(V,W)0(v)\in \mathscr{L}(V,W) 0(v)=00(v)=0
  • identity map IL(V,W)I\in \mathscr{L}(V,W), I(v)=vI(v)=v
  • scaling map TL(V,W)T\in \mathscr{L}(V,W), T(v)=av,aFT(v)=av,a\in \mathbb{F}
  • differentiation map DL(Pm(F),Pm1(F))D\in \mathscr{L}(\mathscr{P}_m(\mathbb{F}),\mathscr{P}_{m-1}(\mathbb{F})), D(f)=fD(f)=f'

Lemma 3.10

T0=0T0=0 for TL(V,W)T\in \mathscr{L}(V,W)

Proof:

T(0+0)=T(0)+T(0)T(0+0)=T(0)+T(0)

Theorem 3.4 Linear map lemma

Suppose v1,...,vnv_1,...,v_n is a basis for VV, and suppose w1,...,wnWw_1,...,w_n\in W are arbitrary vector. Then, there exists a unique linear map. T:VWT:V\to W such that Tvi=wiT_{v_i}=w_i for i=1,...,ni=1,...,n

Proof:

First we show existence.

by constrains,

T(c1v1,...+cnvn)=c1w1+...+cnwnT(c_1 v_1,...+c_n v_n)=c_1w_1+...+c_n w_n

T is well defined because v1,....vnv_1,....v_n are a basis.

Need to show that TT is a linear map.

  • Additivity: let u,vVu,v\in V and suppose a1,...,an,b1,...,bnFa_1,...,a_n,b_1,...,b_n\in \mathbb{F} with u=a1v1+....+anvn,v=b1v1+...+b2vnu=a_1v_1+....+a_n v_n ,v=b_1v_1+...+b_2v_n, then T(u+v)=T((a1+b1)v1+...+(an+bn)vn)=Tu+TvT(u+v)=T((a_1+b_1)v_1+...+(a_n+b_n)v_n)=Tu+Tv

Proof for homogeneity used for exercise.

Need to show TT is unique. Let SL(V,W)S\in\mathscr{L}(V,W) such that Svi=wi,i=1,...,nSv_i=w_i,i=1,...,n

S(c1v1+...+cnvn)=S(c1v1)+S(...)+S(cnvn)=c1S(v1)+...+cnS(vn)+c1w1+...+cnwnS(c_1 v_1+...+c_n v_n)=S(c_1v_1)+S(...)+S(c_n v_n)=c_1S(v_1)+...+c_nS(v_n) +c_1w_1+...+c_nw_n

Then S=TS=T

Definition 3.5

Let S,TL(V,W)S,T\in \mathscr{L}(V,W), then define

  • (S+T)L(V,W)(S+T)\in\mathscr{L}(V,W) by (S+T)(v)=Sv+Tv(S+T)(v)=Sv+Tv
  • for λF\lambda \in \mathbb{F}, (λT)L(V,W)(\lambda T)\in \mathscr{L}(V,W), (λT)(v)=λT(v)(\lambda T)(v)=\lambda T(v)

Exercises: Show that S+TS+T and λT\lambda T are linear maps.

Theorem 3.6

L(V,W)\mathscr{L}(V,W) is a vector space.

Sketch of proof:

  • additive identity: 0(v)=00(v)=0
  • associativity:
  • commutativity:
  • additive inverse: T(1)T=TT\to (-1)T=-T
  • scalar multiplication 1T=T1T=T
  • distributive

Definition 3.7

Multiplication for linear map: (ST)v=S(T(v))=(ST)(v)(ST)v=S(T(v))=(S\circ T)(v) Not commutative but associative.