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Dimension theorem of a quotient space: If W be a subspace of a finite dimensional vector space V over ​\( \mathbb{R} \)​ then ​\( dim\frac{V}{W}=\dim V – \dim W \)​

Theorem:

If W be a subspace of a finite dimensional vector space V over ​\( \mathbb{R} \)​ then ​\( dim\frac{V}{W}=\dim V – \dim W \)​.

Proof:

Let m be the dim W. 
\( S=\{\alpha_1, \alpha_2, \alpha_3, … , \alpha_m\} \)​ be the basis of subspace W.
Since, S is a basis of W and W is a subspace of real vector space V which is finite dimension ,
\( \therefore \)​ S can be extended in the form of basis of V.
\( S’=\{\alpha_1, \alpha_2,\alpha_3, … , \alpha_m,\beta_1,\beta_2,\beta_3, … , \beta_n\} \)​ be the extension of set S which forms the basis of vector space V.
\( \therefore \dim V=m+n \)
\( \therefore \dim V-\dim W=m+n-m=n \)
To prove that ​\( \dim \frac{V}{W}=n \)

i.e. to prove that ​

\[ S_1=\{W+\beta_1,W+\beta_2,W+\beta_3, … ,W+\beta_n\} \]

​ forms set of basis of ​\( \frac{V}{W} \)

First, we prove that ​\( S_1 \)​ is linearly independent.

Consider, ​

\[ a_1(W+\beta_1)+a_2(W+\beta_2)+ … +a_n(W+\beta_n)=W \]

\[ \implies W+a_1\beta_1+W+a_2\beta_2+ … +W+a_n\beta_n=W \]

\( \implies W+a_1\beta_1+a_2\beta_2+ … +a_n\beta_n=W \)
\( \implies a_1\beta_1+a_2\beta_2+ … +a_n\beta_n\in W \)
Since the set ​\( S=\{\alpha_1, \alpha_2, \alpha_3, … , \alpha_m\} \)​ is basis of W,
\( \exists b_1, b_2, … ,b_m\in \mathbb{R} \)​, such that,

\[ a_1\beta_1+a_2\beta_2+ … +a_n\beta_n=b_1\alpha_1+b_2\alpha_2+ … +b_m\alpha_m \]

\[ a_1\beta_1+a_2\beta_2+ … +a_n\beta_n+(-b_1)\alpha_1+(-b_2)\alpha_2+ … +(-b_m)\alpha_m=0 \]

\( \therefore a_1=0, a_2=0, … , a_n=0 \)
\( \therefore \{W+\beta_1, W+\beta_2, … , W+\beta_n\} \)​ is linearly independent. ….. (1)
To prove that ​\( S_1=\{W+\beta_1,W+\beta_2, … ,W+\beta_n\} \)​ spans ​\( \frac{V}{W} \)​.
i.e. ​\( L(S_1)=\frac{V}{W} \)
Let ​\( W+\alpha \)​ be any element in ​\( \frac{V}{W} \)​.
As ​\( W+\alpha\in\frac{V}{W}\implies\alpha\in V \)
Since, ​\( S’=\{\alpha_1, \alpha_2,\alpha_3, … , \alpha_m,\beta_1,\beta_2,\beta_3, … , \beta_n\} \)​ basis of V, ​\( \exists \)​ scalars ​\( c_1, c_2, …, c_m, d_1, d_2, …, d_n \)​ such that,

\[ \alpha=c_1\alpha_1+c_2\alpha_2+ … +c_m\alpha_m+d_1\beta_1+d_2\beta_2+ … +d_n\beta_n \]

\[ W+\alpha=W+c_1\alpha_1+c_2\alpha_2+ … +c_m\alpha_m+d_1\beta_1+d_2\beta_2+ … +d_n\beta_n \]

Since, ​\( \{\alpha_1, \alpha_2, \alpha_3, … , \alpha_m\} \)​ is a basis of W and W is subspace of V,
\( \therefore c_1\alpha_1+c_2\alpha_2+ … +c_m\alpha_m\in W \)
\( c_1\alpha_1+c_2\alpha_2+ … +c_m\alpha_m=\gamma\in W \)

\[ \therefore W+\alpha=W+\gamma+d_1\beta_1+d_2\beta_2+ … +d_n\beta_n \]\[ \therefore W+\alpha=W+d_1\beta_1+d_2\beta_2+ … +d_n\beta_n …… (\because \gamma\in W) \]

\[ \therefore W+\alpha=d_1(W+\beta_1)+d_2(W+\beta_2)+ … +d_n(W+\beta_n) \]

Hence, ​\( W+\alpha\in\frac{V}{W} \)​ can be expressed as linear combination of elements of ​\( S_1 \)​ i.e. ​\( L(S_1)=\frac{V}{W} \)​ ….. (2)
From (1) and (2), ​\( S_1 \)​ forms basis of ​\( \frac{V}{W} \)​.
\( \therefore \dim \frac{V}{W}=n \)
                    ​\( =(m+n)-m \)
                    ​\( =\dim V-\dim W \)
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