12-240/Classnotes for Tuesday October 2

From Drorbn
Revision as of 04:43, 7 December 2012 by Peterlue (talk | contribs) (→‎Subspace)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigationJump to search

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

Class Notes