06-240/Classnotes For Thursday, September 28: Difference between revisions
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<math>\Rightarrow</math> ''x''+''y'' = <math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub> + <math>\sum_{i=1}^m</math> ''b''<sub>i</sub>''v''<sub>i</sub> = <math>\sum_{i=1}^{m+n}</math> ''c''<sub>i</sub>''w''<sub>i</sub> where ''c''<sub>i</sub>=(''a''<sub>1</sub>, ''a''<sub>2</sub>,...,''a''<sub>n</sub>, ''b''<sub>1</sub>, ''b''<sub>2</sub>,...,''b''<sub>m</sub>) and ''w''<sub>i</sub>=''c''<sub>i</sub>=(''u''<sub>1</sub>, ''u''<sub>2</sub>,...,''u''<sub>n</sub>, ''v''<sub>1</sub>, ''v''<sub>2</sub>,...,''v''<sub>m</sub>).<br> |
<math>\Rightarrow</math> ''x''+''y'' = <math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub> + <math>\sum_{i=1}^m</math> ''b''<sub>i</sub>''v''<sub>i</sub> = <math>\sum_{i=1}^{m+n}</math> ''c''<sub>i</sub>''w''<sub>i</sub> where ''c''<sub>i</sub>=(''a''<sub>1</sub>, ''a''<sub>2</sub>,...,''a''<sub>n</sub>, ''b''<sub>1</sub>, ''b''<sub>2</sub>,...,''b''<sub>m</sub>) and ''w''<sub>i</sub>=''c''<sub>i</sub>=(''u''<sub>1</sub>, ''u''<sub>2</sub>,...,''u''<sub>n</sub>, ''v''<sub>1</sub>, ''v''<sub>2</sub>,...,''v''<sub>m</sub>).<br> |
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3. ''cx''= c<math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub>=<math>\sum_{i=1}^n</math> (''ca''<sub>i</sub>)''u''<sub>i</sub><math>\in </math> span <math>\mathcal{S}</math>. |
3. ''cx''= c<math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub>=<math>\sum_{i=1}^n</math> (''ca''<sub>i</sub>)''u''<sub>i</sub><math>\in </math> span <math>\mathcal{S}</math>. |
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To be continued ... |
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Revision as of 13:07, 29 September 2006
Linear Combination
Definition: Let (ui) = (u1, u2, ..., un) be a sequence of vectors in V. A sum of the form
- ai [math]\displaystyle{ \in }[/math] F, [math]\displaystyle{ \sum_{i=1}^n }[/math] 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 [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] aiui, ui [math]\displaystyle{ \in \mathcal{S} }[/math], y = [math]\displaystyle{ \sum_{i=1}^m }[/math] bivi, vi [math]\displaystyle{ \in \mathcal{S} }[/math].
[math]\displaystyle{ \Rightarrow }[/math] x+y = [math]\displaystyle{ \sum_{i=1}^n }[/math] aiui + [math]\displaystyle{ \sum_{i=1}^m }[/math] bivi = [math]\displaystyle{ \sum_{i=1}^{m+n} }[/math] ciwi where ci=(a1, a2,...,an, b1, b2,...,bm) and wi=ci=(u1, u2,...,un, v1, v2,...,vm).
3. cx= c[math]\displaystyle{ \sum_{i=1}^n }[/math] aiui=[math]\displaystyle{ \sum_{i=1}^n }[/math] (cai)ui[math]\displaystyle{ \in }[/math] span [math]\displaystyle{ \mathcal{S} }[/math].
To be continued ...