33 C
Maharashtra
Friday, June 9, 2023
HomeMathematicsAlgebraDimension theorem of a quotient space: If W be a subspace of...

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

## 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 ​$$\frac{V}{W}$$​.
$$\therefore \dim \frac{V}{W}=n$$
​$$=(m+n)-m$$
​$$=\dim V-\dim W$$
RELATED ARTICLES

### Any two closed subset of metric space are connected iff they are disjoint

$${}$$