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HomeMathematicsAlgebraAny two right cosets of V/W are either disjoint or identical

Any two right cosets of V/W are either disjoint or identical

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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 ​\( \alpha, \beta \in V \)​.
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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