12-240/Classnotes for Tuesday October 09

DBN: Classes: 2012-13: 12-240 / Navigation Additions to this web site no longer count towards good deed points. # Week of... Notes and Links 1 Sep 10 About This Class, Tuesday, Thursday 2 Sep 17 HW1, Tuesday, Thursday, HW1 Solutions 3 Sep 24 HW2, Tuesday, Class Photo, Thursday 4 Oct 1 HW3, Tuesday, Thursday 5 Oct 8 HW4, Tuesday, Thursday 6 Oct 15 Tuesday, Thursday 7 Oct 22 HW5, Tuesday, Term Test was on Thursday. HW5 Solutions 8 Oct 29 Why LinAlg?, HW6, Tuesday, Thursday, Nov 4 is the last day to drop this class 9 Nov 5 Tuesday, Thursday 10 Nov 12 Monday-Tuesday is UofT November break, HW7, Thursday 11 Nov 19 HW8, Tuesday,Thursday 12 Nov 26 HW9, Tuesday , Thursday 13 Dec 3 Tuesday UofT Fall Semester ends Wednesday F Dec 10 The Final Exam (time, place, style, office hours times) Register of Good Deeds Add your name / see who's in!

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

Definition of basic

β [math]\displaystyle{ \subset \!\, }[/math] 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 [math]\displaystyle{ \sum \!\, }[/math] ai.ui = 0 ai [math]\displaystyle{ \in\!\, }[/math] F, ui [math]\displaystyle{ \in\!\, }[/math] β

[math]\displaystyle{ \sum \!\, }[/math] ai.ui = 0 = [math]\displaystyle{ \sum \!\, }[/math] 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 [math]\displaystyle{ \forall\!\, }[/math] 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 [math]\displaystyle{ \sum \!\, }[/math] ai∙ui = v = [math]\displaystyle{ \sum \!\, }[/math] bi∙ui

Thus [math]\displaystyle{ \sum \!\, }[/math] ai∙ui - [math]\displaystyle{ \sum \!\, }[/math] bi∙ui = 0

[math]\displaystyle{ \sum \!\, }[/math] (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 [math]\displaystyle{ \forall\!\, }[/math] i

i.e ai = bi, hence the combination is unique.

Clarification on lecture notes

On page 3, we find that [math]\displaystyle{ G \subseteq span(\beta) }[/math] then we say [math]\displaystyle{ span(G) \subseteq span(\beta) }[/math]. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset [math]\displaystyle{ S }[/math] of a vector space [math]\displaystyle{ V }[/math] is a subspace of [math]\displaystyle{ V }[/math]. Moreover, any subspace of [math]\displaystyle{ V }[/math] that contains [math]\displaystyle{ S }[/math] must also contain [math]\displaystyle{ span(S) }[/math]

Since [math]\displaystyle{ \beta }[/math] is a subset of [math]\displaystyle{ V }[/math], [math]\displaystyle{ span(\beta) }[/math] is a subspace of [math]\displaystyle{ V }[/math] from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that [math]\displaystyle{ G \subseteq span(\beta) }[/math]. From the "Moreover" part of Theorem 1.5, since [math]\displaystyle{ span(\beta) }[/math] is a subspace of [math]\displaystyle{ V }[/math] containing [math]\displaystyle{ G }[/math], [math]\displaystyle{ span(\beta) }[/math] must also contain [math]\displaystyle{ span(G) }[/math].

Lecture notes scanned by Oguzhancan

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Lecture notes uploaded by gracez

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