__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 $c\in\mathbb{R}$

__Theorem__ :

## Any two right cosets of $\frac{V}{W}$ are either disjoint or identical.

__Proof__ :

Let $(W+\alpha)$ and $(W+\beta)$ be any two right cosets of W in V where $\alpha, \beta \in V$.

__: $(W+\alpha)\cap (W+\beta)=\phi$ or $(W+\alpha)=(W+\beta)$__

**Claim**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.

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