Friday, February 3, 2023
HomeMathematicsAlgebraAny two right cosets of V/W are either disjoint or identical

# 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.
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