# Translates of a subspace, Quotient spaces

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• null$(A)=$ null $T=\{v\in V:Tv=0\}$, colspace$(A)$ = range $T=\{Tv:v\in V\}$
• dim null $T$ + dim range $T$ = dim $V$
• Linear equation: $Ax=b$
• Solved using elementary row operations
• Solution (if it exists): $u+$ null$(A)$

## Translates of a subspace

$U\subseteq V$, subspace

For $v\in V$, the translate of $U$ by $v$ is defined as $v+U$.

• If $v\in U$, $v+U=U$

• If $v\notin U$, $(v+U) \cap U=\emptyset$

For $v,w\in V$, the translates $v+U$ and $w+U$ are either equal or disjoint. Partial overlap of two translates is not possible.

• if $v+U$ and $w+U$ are disjoint, we are done.

• if not, let $x\in v+U$ and $x\in w+U$.

• $x=v+u_1=w+u_2$, where $u_1,u_2\in U$.

• $w=v+u_3$, where $u_3=u_1-u_2\in U$.

• so, $w+U=v+(u_3+U)=v+U$ and the two translates are equal.

## Quotient space

Quotient space, denoted $V/U$, is the set of all translates of $U$.

• How to find all translates?

Let $W$ be a subspace s.t. $V=U\oplus W$.

• For any translate $v+U$, there exists $w\in W$ such that $v+U=w+U$.

• For $w_1,w_2\in W$, $w_1\ne w_2$, the translates $w_1+U$ and $w_2+U$ are disjoint.

$V/U=\{w+U:w\in W\}$, where $W$ is a subspace satisfying $V=U\oplus W$.

• Translates of $U$ partition $V$

## Quotient space of null of a linear map

$T:V\to W$, linear map. Let us consider $V/$null $T$.

• $W$: subspace such that $V=\text{null }T\oplus W$

• $V/\text{null }T=\{w+\text{null }T:w\in W\}$

$T(w+\text{null }T)=Tw$ for any $w\in W$

• All vectors in $w+\text{null }T$ mapped to same vector $Tw$ by $T$

$Tw_1\ne Tw_2$ for $w_1,w_2\in W$, $w_1\ne w_2$

• Two different vectors in $W$ mapped to distinct vectors by $T$

Linear map $T$ maps the quotient space $V/\text{null }T$ to range $T$ in a one-to-one manner.

## Example

$A=\begin{bmatrix} 1&3\\ 2&6 \end{bmatrix}$

range $T=$ span$\{(1,2),(3,6)\}=$ span$\{(1,2)\}$, dim null $T=1$

Elementary row operations $Ax\to \begin{bmatrix} 1&3\\ 0&0 \end{bmatrix}$

null $T=$ span$\{(-3,1)\}$

$W=$ span$\{(1,0)\}$

$\begin{bmatrix} 1&3\\ 2&6 \end{bmatrix}\left(a\begin{bmatrix}1\\0\end{bmatrix}+b\begin{bmatrix}-3\\1\end{bmatrix}\right)=a\begin{bmatrix}1\\2\end{bmatrix}$

## Solving linear equations

Linear map $T$ maps the quotient space $V/\text{null }T$ to range $T$ in a one-to-one manner.

$Ax=b$

• if $b$ is not in range $A$

• no solution
• if $b$ is in range $A$

• find one $u$ such that $Au=b$

• solution: $u+$ null $A$