Lecture 8

Chapter III Linear maps

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

Vector Space of Linear Maps 3A

Definition 3.1

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

Additivity: T(u+v)=T(u)+T(v),\forall u,v\in V
Homogeneity: T(\lambda v)=\lambda T(v),\forall \lambda \in \mathbb{F},v\in V

Notation

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

Example

zero map 0(v)\in \mathscr{L}(V,W) 0(v)=0
identity map I\in \mathscr{L}(V,W), I(v)=v
scaling map T\in \mathscr{L}(V,W), T(v)=av,a\in \mathbb{F}
differentiation map D\in \mathscr{L}(\mathscr{P}_m(\mathbb{F}),\mathscr{P}_{m-1}(\mathbb{F})), D(f)=f'

Lemma 3.10

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

Proof:

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

Theorem 3.4 Linear map lemma

Suppose v_1,...,v_n is a basis for V, and suppose w_1,...,w_n\in W are arbitrary vector. Then, there exists a unique linear map. T:V\to W such that T_{v_i}=w_i for i=1,...,n

Proof:

First we show existence.

by constrains,

T(c_1 v_1,...+c_n v_n)=c_1w_1+...+c_n w_n

T is well defined because v_1,....v_n are a basis.

Need to show that T is a linear map.

Additivity: let u,v\in V and suppose a_1,...,a_n,b_1,...,b_n\in \mathbb{F} with u=a_1v_1+....+a_n v_n ,v=b_1v_1+...+b_2v_n, then T(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 T is unique. Let S\in\mathscr{L}(V,W) such that Sv_i=w_i,i=1,...,n


S(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=T

Definition 3.5

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

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

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

Theorem 3.6

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

Sketch of proof:

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

Definition 3.7

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

2.5 KiB Raw Blame History