# Quotient Space :

If V is a vector space over $\mathbb{R}$ and W is a subspace of V then $\frac{V}{W}=\{w+\alpha | \alpha\in V\}$ is quotient space and two operations addition and scalar multiplication on $\frac{V}{W}$ is defined as follows:
Let $\alpha, \beta$ be any two arbitrary element of V then $W+\alpha, W+\beta\in \frac{V}{W}$
$(W+\alpha)+(W+\beta)=W+\alpha+\beta$ and
$c(W+\alpha)=W+c\alpha$, for any $c\in\mathbb{R}$

# Proof :

Let $(W+\alpha)$ and $(W+\beta)$ be any two right cosets of W in V where $\alpha, \beta \in V$.
Claim : $(W+\alpha)\cap (W+\beta)=\phi$ or $(W+\alpha)=(W+\beta)$
Suppose, if possible, $(W+\alpha)\cap (W+\beta)\ne\phi$
Then, we have to prove that $(W+\alpha)=(W+\beta)$.
As $(W+\alpha)\cap (W+\beta)\ne\phi$, there exist a vector $v\in V$ such that $v\in(W+\alpha)\cap (W+\beta)$.
$\implies v\in W+\alpha$ and $v\in W+\beta$
As $v\in W+\alpha\implies \exists w_1\in W$ such that $v=w_1+\alpha$
Similarly, $v\in W+\beta\implies \exists w_2\in W$ such that $v=w_2+\beta$
$\therefore w_1+\alpha=w_2+\beta$
$\implies \alpha-\beta=w_2-w_1$
Since, $w_1, w_2\in W$ and W is a subspace of V, $\therefore w_2-w_1\in W$.
$\therefore (\alpha-\beta)$ is also a vector in W. Let $u=\alpha-\beta$ be a vector in W.
Now, to prove that (i) $W+\alpha\subseteq W+\beta$
(ii) $W+\beta\subseteq W+\alpha$
Let x be any vector in $W+\alpha$.
We will prove that $x\in W+\beta$.
Now, as $x\in W+\alpha\implies \exists w\in W$ such that,
$x=W+\alpha$
$=w+(\alpha-\beta)+\beta$
$=w+u+\beta$
$=w'+\beta$
$\implies x\in W+\beta$
$\therefore W+\alpha\subseteq W+\beta$
Let $y\in W+\beta\implies y=w'_1+\beta, w'_1\in W$
$=w'_1+w''_1+\alpha\in W+\alpha$
$\therefore W+\beta\subseteq W+\alpha$
$W+\alpha=W+\beta$
Hence proved.