06-240/Classnotes For Thursday, September 28

Linear Combination

Definition: Let (u_i) = (u₁, u₂, ..., u_n) be a sequence of vectors in V. A sum of the form

a_i [math]\displaystyle{ \in }[/math] F, [math]\displaystyle{ \sum_{i=1}^n }[/math] a_iu_i = a₁u₁ + a₂u₂+ ... +a_nu_n

is called a "Linear Combination" of the u_i.

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].

Example 1. Let P₃([math]\displaystyle{ \Re }[/math])={ax³+bx²+cx+d}[math]\displaystyle{ \subseteq }[/math]P([math]\displaystyle{ \Re }[/math]), a, b, c, d, [math]\displaystyle{ \in \Re }[/math].
u₁=x³-2x²-5x-3
u₂=3x³-5x²-4x-9
v=2x³-2x²+12x-6
Let W=spab(u₁, u₂),
Does v [math]\displaystyle{ \in }[/math] W?
v is in W if v=a₁u₁+a₁u₂
for some a₁, a₂ [math]\displaystyle{ \in \Re }[/math].

If [math]\displaystyle{ \exists }[/math] a₁, a₂ [math]\displaystyle{ \in \Re }[/math],

2x3-2x2+12x-6	= a1(x-2x2-5x-3) + a2(3x3-5x2-4x-9)
	=(a1+3a2)x3 + (-2a1 -5a2)x2 + (-5a1-4a2)x + (-3a1-9a2)
[math]\displaystyle{ \Leftrightarrow }[/math]2	=a1+3a2
-2	=-2a1-5a2
12	=-5a1-4a2
-6	=-3a1-9a2

Solve the four equations above and we will get a₁=-4 and a₂=2.
Check if a₁=-4 and a₂=2 hold for all the 4 equations.
Since it's hold, [math]\displaystyle{ \Rightarrow }[/math] v [math]\displaystyle{ \in }[/math] W.