User:Wongpak - Revision history

Wongpak at 18:06, 29 September 2006

2006-09-29T18:06:49Z

← Older revision		Revision as of 14:06, 29 September 2006
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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>.

Wongpak at 18:06, 29 September 2006

2006-09-29T18:06:39Z

← Older revision		Revision as of 14:06, 29 September 2006
Line 1:		Line 1:
	===Linear Combination===
	Definition: Let (''u''<sub>i</sub>) = (''u''<sub>1</sub>, ''u''<sub>2</sub>, ..., ''u''<sub>n</sub>) be a sequence of vectors in V. A sum of the form<br>
	::''a''<sub>i</sub> <math> \in </math> F, <math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub> = ''a''<sub>1</sub>''u''<sub>1</sub> + ''a''<sub>2</sub>''u''<sub>2</sub>+ ... +''a''<sub>n</sub>''u''<sub>n</sub>

	is called a "Linear Combination" of the ''u''<sub>i</sub>.

	===Span===		===Span===

Wongpak: /* Span */

2006-09-29T15:56:50Z

Span

← Older revision		Revision as of 11:56, 29 September 2006
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	'''Theorem''': For any <math>\mathcal{S} \subseteq</math> V, span <math>\mathcal{S}</math> is a subspace of V.		'''Theorem''': For any <math>\mathcal{S} \subseteq</math> V, span <math>\mathcal{S}</math> is a subspace of V.

	Proof:		Proof:<br>
⚫	# 0 <math> \in </math> span <math>\mathcal{S}</math>,
	# ~~Let ''x''~~ <math> \in </math> span <math>\mathcal{S}</math>~~, Let ''x''~~ <~~math> \in </math> span <math>\mathcal{S}</math~~>,		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'' = <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>).~~# ''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>.~~		<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>.

Wongpak at 15:53, 29 September 2006

2006-09-29T15:53:36Z

New page

===Linear Combination===
Definition: Let (''u''i) = (''u''1, ''u''2, ..., ''u''n) be a sequence of vectors in V. A sum of the form 
::''a''i <math> \in </math> F, <math>\sum_{i=1}^n</math> ''a''i''u''i = ''a''1''u''1 + ''a''2''u''2+ ... +''a''n''u''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>\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:
# 0 <math> \in </math> span <math>\mathcal{S}</math>,
# 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''i''u''i, ''u''i <math> \in \mathcal{S}</math>, ''y'' = <math>\sum_{i=1}^m</math> ''b''i''v''i, ''v''i <math> \in \mathcal{S}</math>.
<math>\Rightarrow</math> ''x''+''y'' = <math>\sum_{i=1}^n</math> ''a''i''u''i + <math>\sum_{i=1}^m</math> ''b''i''v''i = <math>\sum_{i=1}^{m+n}</math> ''c''i''w''i where ''c''i=(''a''1, ''a''2,...,''a''n, ''b''1, ''b''2,...,''b''m) and ''w''i=''c''i=(''u''1, ''u''2,...,''u''n, ''v''1, ''v''2,...,''v''m).# ''cx''= c<math>\sum_{i=1}^n</math> ''a''i''u''i=<math>\sum_{i=1}^n</math> (''ca''i)''u''i<math>\in </math> span <math>\mathcal{S}</math>.