Dimension theorem of a quotient space

Dimension theorem of a quotient space
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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