User:Wongpak

Span

span(u_i):= The set of all possible linear combinations of the u_i's.

If [math]\displaystyle{ \mathcal{S} \subseteq }[/math] V is any subset,

span [math]\displaystyle{ \mathcal{S} }[/math]	:= The set of all linear combination of vectors in [math]\displaystyle{ \mathcal{S} }[/math]
	=[math]\displaystyle{ \left \{ \sum_{i=0}^n a_i u_i, a_i \in \mbox{F}, u_i \in \mathcal{S} \right \} \ni 0 }[/math]

even if [math]\displaystyle{ \mathcal{S} }[/math] is empty.

Theorem: For any [math]\displaystyle{ \mathcal{S} \subseteq }[/math] V, span [math]\displaystyle{ \mathcal{S} }[/math] is a subspace of V.

Proof:
1. 0 [math]\displaystyle{ \in }[/math] span [math]\displaystyle{ \mathcal{S} }[/math].
2. Let x [math]\displaystyle{ \in }[/math] span [math]\displaystyle{ \mathcal{S} }[/math], Let x [math]\displaystyle{ \in }[/math] span [math]\displaystyle{ \mathcal{S} }[/math], [math]\displaystyle{ \Rightarrow }[/math] x = [math]\displaystyle{ \sum_{i=1}^n }[/math] a_iu_i, u_i [math]\displaystyle{ \in \mathcal{S} }[/math], y = [math]\displaystyle{ \sum_{i=1}^m }[/math] b_iv_i, v_i [math]\displaystyle{ \in \mathcal{S} }[/math]. [math]\displaystyle{ \Rightarrow }[/math] x+y = [math]\displaystyle{ \sum_{i=1}^n }[/math] a_iu_i + [math]\displaystyle{ \sum_{i=1}^m }[/math] b_iv_i = [math]\displaystyle{ \sum_{i=1}^{m+n} }[/math] c_iw_i where c_i=(a₁, a₂,...,a_n, b₁, b₂,...,b_m) and w_i=c_i=(u₁, u₂,...,u_n, v₁, v₂,...,v_m).
3. cx= c[math]\displaystyle{ \sum_{i=1}^n }[/math] a_iu_i=[math]\displaystyle{ \sum_{i=1}^n }[/math] (ca_i)u_i[math]\displaystyle{ \in }[/math] span [math]\displaystyle{ \mathcal{S} }[/math].