Difference between revisions of "User:Wongpak"

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===Span===
 
span(''u''<sub>i</sub>):= The set of all possible linear combinations of the ''u''<sub>i</sub>'s.
 
 
 
If <math>\mathcal{S} \subseteq</math> V is any subset,
 
:
 
{| border="0" cellpadding="0" cellspacing="0"
 
|-
 
|span <math>\mathcal{S}</math>
 
|:= The set of all linear combination of vectors in <math>\mathcal{S}</math>
 
|-
 
|
 
|=<math>\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>\mathcal{S}</math> is empty.
 
 
'''Theorem''': For any <math>\mathcal{S} \subseteq</math> V, span <math>\mathcal{S}</math> is a subspace of V.
 
 
Proof:<br>
 
1. 0 <math> \in </math> span <math>\mathcal{S}</math>.<br>
 
2. Let ''x'' <math> \in </math> span <math>\mathcal{S}</math>, Let ''x'' <math> \in </math> span <math>\mathcal{S}</math>,
 
<math>\Rightarrow</math> ''x'' = <math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub>, ''u''<sub>i</sub> <math> \in \mathcal{S}</math>, ''y'' = <math>\sum_{i=1}^m</math> ''b''<sub>i</sub>''v''<sub>i</sub>, ''v''<sub>i</sub> <math> \in \mathcal{S}</math>.
 
<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>
 
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>.
 

Latest revision as of 14:06, 29 September 2006