12-240/Classnotes for Tuesday October 09

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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