Difference between revisions of "12240/Classnotes for Tuesday October 09"
(→Clarification on lecture notes) 

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2/ It is linearly independent  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.  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.  
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 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
(aibi)∙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 .