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

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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]] ==

Revision as of 01:07, 21 October 2012

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

Contents

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

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