Difference between revisions of "12-240/Classnotes for Tuesday October 09"

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(theorems)
(theorems)
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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.
 +
 +
So, suppose <math>\sum \!\,</math> ai.ui = v = <math>\sum \!\,</math> bi.ui
 +
 +
=> <math>\sum \!\,</math> ai.ui - <math>\sum \!\,</math> bi.ui = 0 => <math>\sum \!\,</math> (ai-bi).ui = 0
  
 
== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
 
== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==

Revision as of 17:09, 12 October 2012

In this lecture, the professor concentrate on basics and related theorems.

Definition of basic

β \subset \!\, 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 \sum \!\, ai.ui = 0 ai \in\!\, F, ui \in\!\, β

\sum \!\, ai.ui = 0 = \sum \!\, 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 \forall\!\, 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 \sum \!\, ai.ui = v = \sum \!\, bi.ui

=> \sum \!\, ai.ui - \sum \!\, bi.ui = 0 => \sum \!\, (ai-bi).ui = 0

Lecture notes scanned by Oguzhancan