NoteNextra-origin/pages/Math429/Math429_L9.md

Lecture 9

Chapter III Linear maps

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

Vector Space of Linear Maps 3A

Review

\mathscr{L}(V,W) = space of linear maps form V to W.

\mathscr{L}(V)=\mathscr{L}(V,V)

Key facts:

• \mathscr{L}(V,W) is a vector space
• given T\in\mathscr{L}(U,V),S\in \mathscr{L}(V,W), we have TS\in \mathscr{L}(V,W)
• not commutative

Null spaces and Range 3B

Definition 3.11

Null space and injectivity

For T\in \mathscr{L}(V,W), the null space of T, denoted as null(T) (sometime also noted as ker\ T), is a subset of V given by

ker\ T=null(T)=\{v\in V \vert Tv=0\}

Examples:

• 0\in \mathscr{L}(V,W), then null\ 0=V, null(I)=\{0\}
• T\in \mathscr{L}(\mathbb{R}^3,\mathbb{R}^2),T(x,y,z)=(x+y,y+z), to find the null space, we set T(x,y,z)=0, then x+y=0,y+z=0, x=-y,x=z. So null(T)=\{(x,-x,x)\in \mathbb{R}^2\vert x\in \mathbb{R}\}
• Let D\in \mathscr{L}(\mathscr{P}(\mathbb{R})),D(f)=f', null (D)= the set of constant functions. (because the derivatives of them are zero.)

Theorem 3.13

Given T\in \mathscr{L}(V,W), null(T) is a subspace of V.

Proof:

We check the conditions for the subspace.

• T0=0, so 0\in null(T)
• u,v\in null(T), then consider T(u+v)=Tu+Tv=0+0=0, so u+v\in null(T)
• Let v\in null (T),\lambda \in \mathbb{F}, then T(\lambda v)=\lambda (Tv)=\lambda 0=0, so \lambda v \in null (T)

So null(T) is a subspace.

Definition 3.14

A function f:V\to W is injective (also called one-to-one, 1-1) if for all u,v\in V, if Tv=Tu, then T=U.

Lemma 3.15

Let T\in \mathscr{L}(V,W) then T is injective if and only if null(T)=\{0\}

Proof:

\Rightarrow

Let T\in \mathscr{L}(V,W) be injective, and let v\in null (T). Then Tv=0=T0 so because T is injective v=0\implies null (T)=\{0\}

\Leftarrow

Suppose T\in \mathscr{L}(V,W) with null (T)=\{0\}. Let u,v\in V with Tu=Tv, Tu-Tv=0,T(u-v)=0,u-v=0,u=v, so T is injective

Definition 3.16

Range and surjectivity

For T\in \mathscr{L}(V,W) the range of T denoted range(T), is given by

range(T)=\{Tv\vert v\in V\}

Example:

• 0\in \mathscr{L}(V,W), range(0)=\{0\}
• I\in \mathscr{L}(V,W), range(I)=V
• Let T:\mathbb{R}\to \mathbb{R}^2 given by T(x)=(x,2x), range(T)=\{(x,2x)\vert x\in \mathbb{R}\}

Theorem 3.18

Given T\in \mathscr{L}(V,W), range(T) is a subspace of W

Proof:

Exercise, not interesting.

Definition 3.19

A function T:V\to W is surjective (also called onto) if range(T)=W

Theorem 3.21 (The Fundamental Theorem of Linear Maps, Rank-nullity Theorem)

Suppose V is finite dimensional, and T\in \mathscr{L}(V,W), then range(T) is finite dimensional (W don't need to be finite dimensional). and

dim(V)=dim(null (T))+dim(range(T))

Theorem 3.22

Let T\in \mathscr{L}(V,W) and suppose dim(V)>dim(W). Then T is not injective.

Proof:

By Theorem 3.21, dim(V)=dim(null (T))+dim(range(T)), dim(V)=dim(null (T))+dim(W), 0<dim (V)-dim(W)\leq dim(null(T))\implies dim(null(T))>0\implies null (T)\neq \{0\}

So T is not injective.