12-240/Classnotes for Tuesday October 09: Difference between revisions
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In this lecture, the professor |
In this lecture, the professor concentrated on bases and related theorems. |
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== Definition of |
== Definition of basis == |
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β <math>\subset \!\,</math> V is a |
β <math>\subset \!\,</math> V is a basis if |
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1/ It generates ( |
1/ It generates (span) V, span β = V |
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2/ It is linearly independent |
2/ It is linearly independent |
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== |
== Theorems == |
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1/ β is a |
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. |
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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 |
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Assume <math>\sum \!\,</math> |
Assume <math>\sum \!\,</math> ai∙ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β |
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<math>\sum \!\,</math> |
<math>\sum \!\,</math> ai∙ui = 0 = <math>\sum \!\,</math> 0∙ui |
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since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\ |
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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Hence β is linearly independent |
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(=>) 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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Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0 |
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<math>\sum \!\,</math> (ai-bi)∙ui = 0 |
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β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i |
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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>. |
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== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] == |
== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] == |
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== Lecture notes uploaded by [[User:Grace.zhu|gracez]] == |
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Latest revision as of 06:06, 7 December 2012
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In this lecture, the professor concentrated on bases and related theorems.
Definition of basis
β 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 ai∙ui = 0 ai F, ui β
ai∙ui = 0 = 0∙ui
since 0 can be written as a linear combination of elements of β in a unique way, ai=0 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 ai∙ui = v = bi∙ui
Thus ai∙ui - bi∙ui = 0
(ai-bi)∙ui = 0
β is linear independent hence (ai - bi)= 0 i
i.e ai = bi, hence the combination is unique.
Clarification on lecture notes
On page 3, we find that then we say . The reason is, the Theorem 1.5 in the textbook.
Theorem 1.5: The span of any subset of a vector space is a subspace of . Moreover, any subspace of that contains must also contain
Since is a subset of , is a subspace of from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that . From the "Moreover" part of Theorem 1.5, since is a subspace of containing , must also contain .
Lecture notes scanned by Oguzhancan
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Lecture notes uploaded by gracez
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