- Advertisement -

__Quotient Space__:

Â Â If V is a vector space over â€‹\( \mathbb{R} \)â€‹ and W is a subspace of V then â€‹\( \frac{V}{W}=\{w+\alpha | \alpha\in V\} \)â€‹ is quotient space and two operations addition and scalar multiplication on â€‹\( \frac{V}{W} \)â€‹Â is defined as follows:

Â Â Let â€‹\( \alpha, \beta \)â€‹ be any two arbitrary element of V then â€‹\( W+\alpha, W+\beta\in \frac{V}{W} \)â€‹

â€‹\( (W+\alpha)+(W+\beta)=W+\alpha+\beta \)â€‹Â and

â€‹\( c(W+\alpha)=W+c\alpha \)â€‹, for any â€‹\( c\in\mathbb{R} \)â€‹

__Theorem__:

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

__Proof__:

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

**Claim** : â€‹\( (W+\alpha)\cap (W+\beta)=\phi \)â€‹ or â€‹\( (W+\alpha)=(W+\beta) \)â€‹

Suppose, if possible, â€‹\( (W+\alpha)\cap (W+\beta)\ne\phi \)â€‹

Then, we have to prove that â€‹\( (W+\alpha)=(W+\beta) \)â€‹.

As â€‹\( (W+\alpha)\cap (W+\beta)\ne\phi \)â€‹, there exist a vector â€‹\( v\in V \)â€‹ such that â€‹\( v\in(W+\alpha)\cap (W+\beta) \)â€‹.

â€‹\( \implies v\in W+\alpha and v\in W+\beta \)â€‹

As â€‹\( v\in W+\alpha\implies \exists w_1\in W \)â€‹ such that â€‹\( v=w_1+\alpha \)â€‹

Similarly, â€‹\( v\in W+\beta\implies \exists w_2\in W \)â€‹ such that â€‹\( v=w_2+\beta \)â€‹

â€‹\( \therefore w_1+\alpha=w_2+\beta \)â€‹

â€‹\( \implies \alpha-\beta=w_2-w_1 \)â€‹

Since,\( w_1, w_2\in W \)and W is a subspace of V, â€‹\( \therefore w_2-w_1\in W \)â€‹.

â€‹\( \therefore (\alpha-\beta) \)â€‹ is also a vector in W. Let â€‹\( u=\alpha-\beta \)â€‹ be a vector in W.

Now, to prove that (i) â€‹\( W+\alpha\subseteq W+\beta \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii) â€‹\( W+\beta\subseteq W+\alpha \)â€‹

Let x be any vector in â€‹\( W+\alpha \)â€‹.

We will prove that â€‹\( x\in W+\beta \)â€‹.

Now, as â€‹\( x\in W+\alpha\implies \exists w\in W \)â€‹ such that,

â€‹\( x=W+\alpha \)â€‹

Â Â â€‹\( =w+(\alpha-\beta)+\beta \)â€‹

Â Â â€‹\( =w+u+\beta \)â€‹

Â Â â€‹\( =w’+\beta \)â€‹

â€‹\( \implies x\in W+\beta \)â€‹

â€‹\( \therefore W+\alpha\subseteq W+\beta \)â€‹

Let â€‹\( y\in W+\beta \)

\( \implies y=w’_1+\beta, w’_1\in W \)â€‹

â€‹Â Â Â Â \( =w’_1+w”_1+\alpha\in W+\alpha \)â€‹

â€‹\( \therefore W+\beta\subseteq W+\alpha \)â€‹

â€‹\( W+\alpha=W+\beta \)â€‹

Hence proved.

- Advertisement -