NoteNextra-origin/pages/Math429/Math429_L15.md

Lecture 15

Chapter III Linear maps

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

Products and Quotients of Vector Spaces 3E

Quotient Space

Idea: For a vector space V and a subspace U. Construct a new vector space V/U which is elements of V up to equivalence by U.

Definition 3.97

For v\in V and U a subspace of V. Then v+U=\{v+u\vert u\in U\} is the translate of U by v. (also called a coset of U)

Example

Let U\subseteq \mathbb{R}^2 be U=\{(x,2x)\vert x\in \mathbb{R}\}, v=(5,3)\in\mathbb{R}^2, v+U=\{(x+3.5, 2x)\vert x\in \R\}

Describe the solutions to (p(x))'=x^2, p(x)=\frac{1}{3}x^3+c. Let u\in \mathscr{P}(\mathbb{R}) be the constant functions then the set of solutions to (p(x))'=x^2 is \frac{1}{3}x^3+U

Definition 3.99

Suppose U is a subspace of V, then the quotient space V/U is given by

V/U=\{v+U\vert v\in V\}

This is not subset of V.

Example:

Let U\subseteq \mathbb{R}^2 be U=\{(x,2x)\vert x\in \mathbb{R}\}, then \mathbb{R}^2/U is the set of all lines of slope 2 in \mathbb{R}^2

Lemma 3.101

Let U be a subspace of V and v,w\in V then the following are equivalent

a) v-w\in U
b) v+U=w+U
c) (v+U)\cap(w+U)\neq \phi

Proof:

• a\implies b

Suppose v-w\in U, we wish to show that v+U=w+U.

Let u\in U then v+u=w+((v-w)+u)\in w+U

So v+U\in w+U and by symmetry, w+U\subseteq v+U so v+U=w+U

• b\implies c

u\neq \phi \implies v+U=w+U\neq \phi

• c\implies a

Suppose (v+U)\cap (w+U)\neq\phi So let u_1,u_2\in U be such that v+u_1=w+u_2 but then v-w=u_2-u_1\in U

Definition 3.102

Let U\subseteq V be a subspace, define the following:

• (v+U)+(w+U)=(v+w)+U
• \lambda (v+U)=(\lambda v)+U

Theorem 3.103

Let U\in V be a subspace, then V/U is a vector space.

Proof:

Assume for now that Definition 3.102 is well defined.

• commutativity: by commutativity on V.
• associativity: by associativity on V.
• distributive: law by V.
• additive identity: 0+U.
• additive inverse: -v+U.
• multiplicative identity: 1(v+U)=v+U

Why is 3.102 well defined.

Let v_1,v_2,w_1,w_2\in V such that v_1+U=v_2+U and w_1+U=w_2+U

Note by lemma 3.101

v_1-v_2\in U and w_1-w_2\in U \implies

(v_1+w_1)-(v_2+w_2)\in U \implies

(v_1+w_1)+U=(v_2+w_2)+U=(v_1+U)+(w_1+U)=(v_2+U)+(w_2+U)

same idea for scalar multiplication.

Definition 3.104

Let U\subseteq V. The quotient map is

\pi:V\to V/U, \pi (v)=v+U

Lemma 3.104.1

\pi is a linear map

Theorem 3.105

Let V be finite dimensional U\subseteq V then dim(V/U)=dim\ V-dim\ U

Proof:

Note null\ pi=U, since if \pi(v)=0=0+u\iff v\in U

By the Fundamental Theorem of Linear Maps says

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

but \pi is surjective, so we are done.

Theorem 3.106

Suppose T\in \mathscr{L}(V,W) then,

Define \tilde{T}:V/null\ T\to \tilde{W} by \tilde{T}(v+null\ T) Then we have the following.

1. \tilde{T}\circ\pi =T
2. \tilde{T} is injective
3. range \tilde{T}=range\ T
4. V/null\ T and range\ T are isomorphic