12-240/Classnotes for Tuesday October 09: Difference between revisions
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(=>) every element of V can be written as a linear combination of elements of β in a unique way. |
(=>) every element of V can be written as a linear combination of elements of β in a unique way. |
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So, suppose <math>\sum \!\,</math> ai.ui = v = <math>\sum \!\,</math> bi.ui |
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=> <math>\sum \!\,</math> ai.ui - <math>\sum \!\,</math> bi.ui = 0 => <math>\sum \!\,</math> (ai-bi).ui = 0 |
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== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] == |
== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] == |
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Revision as of 16:09, 12 October 2012
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In this lecture, the professor concentrate on basics and related theorems.
Definition of basic
β [math]\displaystyle{ \subset \!\, }[/math] V is a basic if
1/ It generates ( span) V, span β = V
2/ It is linearly independent
theorems
1/ β is a basic of V iff every element of V can be written as a linear combination of elements of β in a unique way.
proof: ( in the case β is finite)
β = {u1, u2, ..., un}
(<=) need to show that β = span(V) and β is linearly independent.
The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given
Assume [math]\displaystyle{ \sum \!\, }[/math] ai.ui = 0 ai [math]\displaystyle{ \in\!\, }[/math] F, ui [math]\displaystyle{ \in\!\, }[/math] β
[math]\displaystyle{ \sum \!\, }[/math] ai.ui = 0 = [math]\displaystyle{ \sum \!\, }[/math] 0.ui
since 0 can be written as a linear combination of elements of β in a unique way, ai=0 [math]\displaystyle{ \forall\!\, }[/math] i
Hence β is linearly independent
(=>) every element of V can be written as a linear combination of elements of β in a unique way.
So, suppose [math]\displaystyle{ \sum \!\, }[/math] ai.ui = v = [math]\displaystyle{ \sum \!\, }[/math] bi.ui
=> [math]\displaystyle{ \sum \!\, }[/math] ai.ui - [math]\displaystyle{ \sum \!\, }[/math] bi.ui = 0 => [math]\displaystyle{ \sum \!\, }[/math] (ai-bi).ui = 0
Lecture notes scanned by Oguzhancan
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