12-240/Classnotes for Tuesday October 09

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

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

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Lecture notes uploaded by gracez

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