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

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{{12-240/Navigation}}
 
{{12-240/Navigation}}
In this lecture, the professor concentrate on basics and related theorems.
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In this lecture, the professor concentrated on bases and related theorems.
== Definition of basic ==
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== Definition of basis ==
β <math>\subset \!\,</math> V is a basic if
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β <math>\subset \!\,</math> V is a basis if
  
1/ It generates ( span) V, span β = V
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1/ It generates (span) V, span β = V
  
 
2/ It is linearly independent
 
2/ It is linearly independent
  
== theorems ==
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== 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.
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1/ β is a basis 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)
 
proof: ( in the case β is finite)
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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
 
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>\sum \!\,</math> ai.ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β
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Assume <math>\sum \!\,</math> ai∙ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β
  
<math>\sum \!\,</math> ai.ui = 0 = <math>\sum \!\,</math> 0.ui
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<math>\sum \!\,</math> ai∙ui = 0 = <math>\sum \!\,</math> 0∙ui
  
 
since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i
 
since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i
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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
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So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui
  
Thus <math>\sum \!\,</math> ai.ui - <math>\sum \!\,</math> bi.ui = 0  
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Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0  
  
<math>\sum \!\,</math> (ai-bi).ui = 0
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<math>\sum \!\,</math> (ai-bi)∙ui = 0
  
 
β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i
 
β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i
  
 
i.e ai = bi, hence the combination is unique.
 
i.e ai = bi, hence the combination is unique.
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== Clarification on lecture notes ==
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On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.
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<b>Theorem 1.5:</b> The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>
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Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math> span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math> span(\beta)</math> must also contain <math> span(G)</math>.
  
 
== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
 
== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
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Image:12-240-1009-6.jpg|Page 6
 
Image:12-240-1009-6.jpg|Page 6
 
Image:12-240-1009-7.jpg|Page 7
 
Image:12-240-1009-7.jpg|Page 7
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</gallery>
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== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
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<gallery>
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Image:12-240-O9-1.jpg|Page 1
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Image:12-240-O9-2.jpg|Page 2
 
</gallery>
 
</gallery>

Latest revision as of 06:06, 7 December 2012

In this lecture, the professor concentrated on bases and related theorems.

Contents

Definition of basis

β \subset \!\, V is a basis if

1/ It generates (span) V, span β = V

2/ It is linearly independent

Theorems

1/ β is a basis 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

Thus \sum \!\, ai∙ui - \sum \!\, bi∙ui = 0

\sum \!\, (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 \forall\!\, i

i.e ai = bi, hence the combination is unique.

Clarification on lecture notes

On page 3, we find that G \subseteq span(\beta) then we say span(G) \subseteq span(\beta). The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain span(S)

Since \beta is a subset of V, span(\beta) is a subspace of V from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that G \subseteq span(\beta). From the "Moreover" part of Theorem 1.5, since  span(\beta) is a subspace of V containing G,  span(\beta) must also contain  span(G).

Lecture notes scanned by Oguzhancan

Lecture notes uploaded by gracez