Difference between revisions of "12-240/Classnotes for Tuesday October 2"
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== Subspace == | == Subspace == | ||
− | Remind about the theorem of subspace: a subset W ⊂ V is a subspace iff is is closed under the operations of V | + | 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 "=>": | ||
− | == Class | + | 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: <math>\forall\!\,</math> a <math>\in\!\,</math> W: 0 + a = a | ||
+ | |||
+ | Moreover, a a <math>\in\!\,</math> 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 <math>\forall\!\,</math> x,y <math>\in\!\,</math> W => a,b <math>\in\!\,</math> V | ||
+ | |||
+ | While V is a vector space | ||
+ | |||
+ | thus x + y = y + x ( and the sum <math>\in\!\,</math> W since W is closed under addition) | ||
+ | |||
+ | VS2: (x + y) + z = x + (y + z) is proven similarly | ||
+ | |||
+ | VS3: As given, 0 of V <math>\in\!\,</math> W, pick any a in W ( possible since W is not empty) | ||
+ | |||
+ | So, a <math>\in\!\,</math> V hence a + 0 = a | ||
+ | |||
+ | Thus 0 is also additive identity element of W | ||
+ | |||
+ | == Class Notes == | ||
<gallery> | <gallery> | ||
− | Image: | + | Image:12-240-Oct-2-Page-1.jpg|Page 1 |
− | Image: | + | Image:12-240-Oct-2-Page-2.jpg|Page 2 |
+ | Image:12-240-Oct-2-Page-3.jpg|Page 3 | ||
+ | Image:12-240-Oct-2-Page-4.jpg|Page 4 | ||
+ | Image:12-240-Oct-2-Page-5.jpg|Page 5 | ||
</gallery> | </gallery> |
Latest revision as of 05:43, 7 December 2012
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The "vitamins" slide we viewed today is here.
Today, the professor introduces more about subspace, linear combination, and related subjects.
Subspace
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: a W: 0 + a = a
Moreover, a a 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 x,y W => a,b V
While V is a vector space
thus x + y = y + x ( and the sum W since W is closed under addition)
VS2: (x + y) + z = x + (y + z) is proven similarly
VS3: As given, 0 of V W, pick any a in W ( possible since W is not empty)
So, a V hence a + 0 = a
Thus 0 is also additive identity element of W