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.
\(S_1=\{W+\beta_1,W+\beta_2,W+\beta_3, … ,W+\beta_n\} \)
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 \( \frac{V}{W} \).
\( \therefore \dim \frac{V}{W}=n \)
\( =(m+n)-m \)
\( =\dim V-\dim W \)
$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 \( \frac{V}{W} \).
\( \therefore \dim \frac{V}{W}=n \)
\( =(m+n)-m \)
\( =\dim V-\dim W \)
