09 - Quotient Space

Quotient Space lemma

Let $V$ be a vector space over field $\mathbf{F}$ and let $U$ be a subspace of $V$

The set of left cosets

$$V/U=\{v+U\mid v\in V\}$$

with the operations

$$\begin{array}{rl}(v+U)+(w+U)& :=v+w+U\\ a(v+U)& :=av+U\end{array}$$

for $v,w\in V$ and $a\in \mathbf{F}$

This makes it a vector space and is called the quotient space

Proof - Well-defined

Assume $v+U=v^{\prime }+U$ and $w+U=w^{\prime }+U$ then

$$v=v^{\prime }+u,w=w^{\prime }+\tilde{u}\text{ for some }u,\tilde{u}\in U$$ $$\begin{array}{rl}(v+U)+(w+U)& =v+w+U\\ & =v^{\prime }+u+w^{\prime }+\tilde{u}+U\\ & =v^{\prime }+w^{\prime }+U\\ & =(v^{\prime }+U)+(w^{\prime }+U)\end{array}$$

The operations satisfy the vector space axioms which follow from the fact that operations in $V$ do

Basis for the Quotient Space

Let $V$ be a vector space over field $\mathbf{F}$ and let $U$ be a subspace of $V$
Let $\mathcal{E}$ be a basis of $U$ extended to basis $\mathcal{B}$ of $V$

Let $V/U$ be the quotient space then define

$$\overline{\mathcal{B}}:=\{e+U\mid e\in \mathcal{B}\setminus \mathcal{E}\}\subseteq V/U$$

Note that $\mathcal{B}\setminus \mathcal{E}$ means elements in $\mathcal{B}$ not in $\mathcal{E}$
Then

$$\overline{\mathcal{B}}\text{ is a basis for }V/U$$

Proof - Spanning

Let $v+U\in V/U$ then
There exists $e_1,\cdots ,e_k\in \mathcal{E}$ and $e_{k+1},\cdots ,e_n\in \mathcal{B}\setminus \mathcal{E}$ and $a_1,\cdots ,a_n\in \mathbf{F}$ such that

$$v=a_1{e}_1+\cdots +a_k{e}_k+a_{k+1}{e}_{k+1}+\cdots +a_n{e}_n$$

as $\mathcal{B}$ is spanning

Hence

$$\begin{array}{rl}v+U& =a_{k+1}{e}_{k+1}+\cdots +a_n{e}_n+U\\ & =a_{k+1}(e_{k+1}+U)+\cdots +a_n(e_n+U)\end{array}$$

Thus $\overline{\mathcal{B}}$ is spanning

Proof - Independence

Assume for $a_1,\cdots ,a_r\in \mathbf{F}$ and $e_1,\cdots ,e_r\in \mathcal{B}\setminus \mathcal{E}$ that

$$a_1(e_1+U)+\cdots +a_r(e_r+U)=0+U=U$$

So

$$a_1{e}_1+\cdots +a_r{e}_r\in U$$

Hence

$$a_1{e}_1+\cdots +a_r{e}_r=b_1{e}_1^{\prime }+\cdots +b_s{e}_s^{\prime }$$

where $e_1^{\prime },\cdots ,e_s^{\prime }\in \mathcal{E}$ and for some $b_1,\cdots ,b_s\in \mathbf{F}$ as $\mathcal{E}$ spans $U$
So $a_1=\cdots =a_r=-b_1=\cdots =-b_s=0$ as $\mathcal{B}$ is linearly independent
Hence $\overline{\mathcal{B}}$ is linearly independent

Basis Extension through Quotient Space

Let $U\subset V$ be vector spaces, with $\mathcal{E}$ a basis for $U$

Let $\mathcal{F}\subset V$ be a set of vectors such that

$$\{v+U:v\in \mathcal{F}\}\text{ is a basis for }V/U$$

Then

$$\mathcal{E}\cup \mathcal{F}f\text{ is a basis for }V$$

Dimension Formula for Quotient Spaces corollary

Let $U\subset V$ be a finite dimensional vector space then

$$\dim (V)=\dim (U)+\dim (V/U)$$

Induced Linear Map on Quotients lemma

Let $T:V\to W$ be a linear map and let $A\subseteq V,V\subseteq W$ be subspaces

Formula $\overline{T}(v+A):=T(v)+B$ is a well-defined linear map of quotients

$$\overline{T}:V/A\to W/B⟺T(A)\subseteq B$$

Proof - $⟸$

Assume $T(A)\subseteq B$ then $\overline{T}$ is linear if well-defined
Assume $v+A=v^{\prime }+A$ then $v=v^{\prime }+a$ for some $a\in A$ so

$$\begin{array}{rlrl}\overline{T}(v+A)& =T(v)+B& & \text{ by defintion}\\ & =T(v^{\prime }+a)+B& & \\ & =T(v^{\prime })+T(a)+B& & \text{ by linearity of }T\\ & =T(v^{\prime })+B& & \text{ as }T(A)\subseteq B\\ & =\overline{T}(v^{\prime }+A)& & \end{array}$$

Hence $\overline{T}$ is well-defined

Proof - $⟹$

Assume that $\overline{T}$ is well-defined and let $a\in A$ then

$$B=0_{W/B}=\overline{T}(0_{V/A})=\overline{T}(A)=\overline{T}(a+A)=T(a)+B$$

Hence $T(a)\in B$ so $T(A)\subseteq B$