06-240/Classnotes For Thursday, September 28

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Linear Combination

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

ai  \in F, \sum_{i=1}^n aiui = a1u1 + a2u2+ ... +anun

is called a "Linear Combination" of the ui.

Span

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


If \mathcal{S} \subseteq V is any subset,

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

even if \mathcal{S} is empty.

Theorem: For any \mathcal{S} \subseteq V, span \mathcal{S} is a subspace of V.

Proof:
1. 0  \in span \mathcal{S}.
2. Let x  \in span \mathcal{S}, Let x  \in span \mathcal{S}, \Rightarrow x = \sum_{i=1}^n aiui, ui  \in \mathcal{S}, y = \sum_{i=1}^m bivi, vi  \in \mathcal{S}. \Rightarrow x+y = \sum_{i=1}^n aiui + \sum_{i=1}^m bivi = \sum_{i=1}^{m+n} ciwi where ci=(a1, a2,...,an, b1, b2,...,bm) and wi=ci=(u1, u2,...,un, v1, v2,...,vm).
3. cx= c\sum_{i=1}^n aiui=\sum_{i=1}^n (cai)ui\in span \mathcal{S}.


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

If \exists a1, a2 \in \Re,

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)
 
\Leftrightarrow2
=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, \Rightarrow v \in W.