12-240/Classnotes for Tuesday October 2: Difference between revisions

Revision as of 13:22, 4 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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The "vitamins" slide we viewed today is here.

Today, the professor introduces more about subspace, linear combination, and related subjects.

Remind about the theorem of subspace: a non-empty subset W ⊂ V is a subspace iff is is closed under the operations of V and contains 0 of V

Proof:

First direction:

if a non-empty subset W ⊂ V is a subspace , then W is a vector space over the operations of V .

=> + W is closed under the operations of V.

+ W has a unique identity of addition: $\forall \!\,$ a $\in \!\,$ W: 0 + a = a

Moreover, a a $\in \!\,$ V. Hence 0 is also identity of addtition of V

Second direction

if a non-empty subset W ⊂ V is closed under the operations of V and contains 0 of V

we need to prove that W is a vector space over operations of V, hence, and subspace of V.

Namely, we need to show that W satisfies all axioms of a vector space, but now we just consider some axioms and leave the rest to readers.

VS1: Consider $\forall \!\,$ x,y $\in \!\,$ W => a,b $\in \!\,$ V

While V is a vector space

thus x + y = y + x ( and the sum $\in \!\,$ W since W is closed under addition)

VS2: (x + y) + z = x + (y + z) is proven similarly

VS3: As given, 0 of V $\in \!\,$ W, pick any a in W ( possible since W is not empty)

So, a $\in \!\,$ V hence a + 0 = a

Thus 0 is also additive identity element of W

@@ Line 41: / Line 41: @@
 VS3: As given, 0 of V <math>\in\!\,</math> W, pick any a in W ( possible since W is not empty)
-=> a <math>\in\!\,</math> V  hence a + 0 = a
+So, a <math>\in\!\,</math> V  hence a + 0 = a
 Thus 0 is also additive identity element of W