03. Quotient Spaces

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

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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

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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

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

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Dimension Formula for Quotient Spaces corollary

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

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

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08 - First Isomorphism Theorem - V2

First Isomorphism Theorem - V2

Let $T:V\to W$ be a linear map of vector spaces over $\mathbf{F}$ then

$$\begin{array}{rl}\overline{T}:V/\mathrm{Ker}(T)& \to \mathrm{Im}(T)\\ v+\mathrm{Ker}(T)& \mapsto T(v)\end{array}$$

is an isomorphism of vector spaces

Proof - Short

It follows from the first isomorphism theorem for groups that $T$ is an isomorphism of (abelian) groups.
As $T$ is also compatible with scalar multiplication then

$$T\text{ is a linear isomorphism}$$

Proof - Detailed

Well-Defined

$$\begin{array}{rl}& v+\mathrm{Ker}(T)=v^{\prime }+\mathrm{Ker}(T)\\ & ⟹v=v^{\prime }+u\text{ for some }u\in \mathrm{Ker}(T)\\ & ⟹\overline{T}(v+\mathrm{Ker}(T)):=T(v)\\ & =T(v^{\prime }+u)=T(v^{\prime })=:\overline{T}(v^{\prime }+\mathrm{Ker}(T))\end{array}$$

Linear

$$\begin{array}{rl}& \overline{T}(a(v+\mathrm{Ker}(T))+(v^{\prime }+\mathrm{Ker}(T)))\\ & =\overline{T}(av+v^{\prime }+\mathrm{Ker}(T))\\ & :=T(av+v^{\prime })\\ & =aT(v)+T(v^{\prime })\\ & =:a\overline{T}(v+\mathrm{Ker}(T))+\overline{T}(v^{\prime }+\mathrm{Ker}(T))\\ & \end{array}$$

Surjective

$$\begin{array}{rl}& w\in \mathrm{Im}T\\ & ⟹\exists v\in V:T(v)=w\\ & ⟹\overline{T}(v+\mathrm{Ker}(T))=T(v)=w\\ & ⟹w\in \mathrm{Im}\overline{T}\end{array}$$

Injective

$$\begin{array}{rl}& v+\mathrm{Ker}(T)\in \mathrm{Ker}(\overline{T})\\ & ⟹\overline{T}(v+\mathrm{Ker}(T))=T(v)=0\\ & ⟹v\in \mathrm{Ker}(T)\\ & ⟹v+\mathrm{Ker}(T)=0+\mathrm{Ker}(T)\end{array}$$

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09 - Rank-Nullity Theorem

Rank-Nullity Theorem

If $T:V\to W$ is linear transformation and $V$ is finite dimensional then

$$\dim (V)=\dim (\mathrm{Ker}(T))+\dim (\mathrm{Im}T)$$

Proof

Apply dimension formula for quotient spaces to $U=\ker T$ then

$$\dim (V)=\dim (\mathrm{Ker}(T))+\dim (V/\mathrm{Ker}(T))$$

But by First Isomorphism Theorem - V2 then

$$\dim (V/\mathrm{Ker}(T))=\dim (\mathrm{Im}(T)$$

So the result follows

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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 $T(A)\subseteq B$

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10 - Block Matrix Decomposition of a Linear Map

Block Matrix Decomposition

Let $V,W$ be finite dimensional vector spaces with $A\subseteq V$ and $B\subseteq W$
Let $\mathcal{B}=\{e_1,\cdots ,e_n\}$ be a basis for $V$ and $\mathcal{E}=\{e_1,\cdots ,e_k\}$ a basis for $A$ (with $k\le n$)
Let ${\mathcal{B}}^{\prime }=\{e_1^{\prime },\cdots ,e_m^{\prime }\}$ be a basis for $W$ and ${\mathcal{E}}^{\prime }=\{e_1^{\prime },\cdots ,e_l^{\prime }\}$ a basis for $B$

So we have induced bases for $V/A$ and $W/B$ given by

$$\begin{array}{rl}\overline{\mathcal{B}}& =e_{k+1}+A,\cdots ,e_n+A\\ \overline{{\mathcal{B}}^{\prime }}& =e_{l+1}^{\prime }+B,\cdots ,e_m^{\prime }+B\end{array}$$

Let $T:V\to W$ be a linear map such that $T(A)\subseteq B$ then $T$ induces a map $\overline{T}$ on quotients and restricts to a linear map

$$T{\mid }_A:A\to B\text{ with }T{\mid }_A(v)=T(v)\text{ for }v\in A$$

Using notation

$${}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}}=(a_{ij}{)}_{1\le i\le m,1\le j\le n}$$

so

$$T(e_j)=a_{1j}{e}_1^{\prime }+\cdots +a_{mj}{e}_m^{\prime }$$

Then there is block matrix decomposition

$${}_{{\mathcal{B}}^{\prime }}[T{]}_{\mathcal{B}}=\left(\begin{array}{cc}{}_{{\mathcal{E}}^{\prime }}[T{\mid }_A{]}_{\mathcal{E}}& \ast \\ 0& {}_{\overline{{\mathcal{B}}^{\prime }}}[\overline{T}{]}_{\overline{\mathcal{B}}}\end{array}\right)$$

where ${}_{\overline{{\mathcal{B}}^{\prime }}}[\overline{T}{]}_{\overline{\mathcal{B}}}=(a_{ioj}{)}_{l+1\le i\le m,k+1\le j\le n}$

Proof

For $j\le k$ then $T(e_j)\in B$ so $a_{ij}=0$ for $i>l$
$a_{ij}$ is equal to $(i,j)-$th entry of ${}_{{\mathcal{E}}^{\prime }}[T{\mid }_A{]}_{\mathcal{E}}$ for $i\le l$

For the bottom right corner of the matrix

$$\begin{array}{rl}\overline{T}(e_j+A)& =T(e_j)+B\\ & =a_{1j}{e}_1^{\prime }+\cdots +a_{mj}{e}_m^{\prime }+B\\ & =a_{l+1,j}(e_{l+1}^{\prime }+B)+\cdots +a_{mj}(e_m^{\prime }+B)\end{array}$$

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