12-240/Classnotes for Tuesday October 09: Difference between revisions

Revision as of 16:13, 12 October 2012

#	Week of...	Notes and Links
Additions to this web site no longer count towards good deed points.
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)
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In this lecture, the professor concentrate on basics and related theorems.

β $\subset \!\,$ V is a basic if

1/ It generates ( span) V, span β = V

2/ It is linearly independent

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 $\sum \!\,$ 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.

@@ Line 35: / Line 35: @@
 <math>\sum \!\,</math> (ai-bi).ui = 0
-β is linear independent hence (ai - bi)= 0 <math>\for all\!\</math> i
+β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i
 i.e ai = bi, hence the combination is unique.