User:Wongpak: Difference between revisions

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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>.

Revision as of 10:56, 29 September 2006

Linear Combination

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

ai F, 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 V is any subset,

span := The set of all linear combination of vectors in
=

even if is empty.

Theorem: For any V, span is a subspace of V.

Proof:
1. 0 span .
2. Let x span , Let x span , x = aiui, ui , y = bivi, vi . x+y = aiui + bivi = ciwi where ci=(a1, a2,...,an, b1, b2,...,bm) and wi=ci=(u1, u2,...,un, v1, v2,...,vm).
3. cx= c aiui= (cai)ui span .