User:Wongpak: Difference between revisions
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===Span=== |
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span(''u''<sub>i</sub>):= The set of all possible linear combinations of the ''u''<sub>i</sub>'s. |
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If <math>\mathcal{S} \subseteq</math> V is any subset, |
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: |
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{| border="0" cellpadding="0" cellspacing="0" |
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|- |
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|span <math>\mathcal{S}</math> |
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|:= The set of all linear combination of vectors in <math>\mathcal{S}</math> |
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|- |
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| |
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|=<math>\left \{ \sum_{i=0}^n a_i u_i, a_i \in \mbox{F}, u_i \in \mathcal{S} \right \} \ni 0</math> |
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even if <math>\mathcal{S}</math> is empty. |
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'''Theorem''': For any <math>\mathcal{S} \subseteq</math> V, span <math>\mathcal{S}</math> is a subspace of V. |
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Proof:<br> |
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1. 0 <math> \in </math> span <math>\mathcal{S}</math>.<br> |
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2. Let ''x'' <math> \in </math> span <math>\mathcal{S}</math>, Let ''x'' <math> \in </math> span <math>\mathcal{S}</math>, |
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<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>. |
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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> |
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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>. |
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