Difference between revisions of "12240/Classnotes for Tuesday October 2"
(→Subspace) 

Line 12:  Line 12:  
Proof:  Proof:  
−  First direction:  +  First direction "=>": 
if a nonempty subset W ⊂ V is a subspace , then W is a vector space over the operations of V .  if a nonempty subset W ⊂ V is a subspace , then W is a vector space over the operations of V .  
Line 23:  Line 23:  
−  Second direction  +  Second direction "<=": 
if a nonempty subset W ⊂ V is closed under the operations of V and contains 0 of V  if a nonempty subset W ⊂ V is closed under the operations of V and contains 0 of V 
Latest revision as of 04:43, 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 nonempty 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 nonempty 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 nonempty 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