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

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(Subspace)
 
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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 note ==
+
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:Mat 240.jpg|Page 1
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Image:12-240-Oct-2-Page-1.jpg|Page 1
Image:Mat 2401.jpg|Page 2
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Image:12-240-Oct-2-Page-2.jpg|Page 2
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Image:12-240-Oct-2-Page-3.jpg|Page 3
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Image:12-240-Oct-2-Page-4.jpg|Page 4
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Image:12-240-Oct-2-Page-5.jpg|Page 5
 
</gallery>
 
</gallery>

Latest revision as of 05:43, 7 December 2012

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: \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

Class Notes