# 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