Vector Subspace Questions

Monday, July 30, 2018

9:15 AM

Prove that  is not a vector subspace

 

To prove something false, we need to find just one case that contradicts the proposal.

Use a real example, rather than a theoretical counter-proof.

 

We can prove that axiom three, subspaces are closed under scalar multiplication, does not hold true.

 

Hence, S is not a vector subspace as it is not closed under scalar multiplication

 

 

, A is a fixed matrix

Prove that it is a vector space

 

To prove that a subset is a subspace, we need to prove all three axioms

1. Existence of the Zero Vector
   
   Then
2. Closure under Addition
    

Then

Where

Therefore  is closed under addition

3. Closure under Scalar Multiplication

Then

So

 

 U is a subspace of  by the subspace thory

 

Show that S = {p in P2 : p(1) = 2} is not a subspace of P2 over F

 

Denote the zero polynomial of P2 by _0: F->F

 

0(1) = 0 neq 2 -> zero polynomial of P2 is not in S

.:. S is not a subspace of P2p

 

 

 

 

 

 

 

 

 

 

 

Show that S={x in r3: x1 – 2x2 + 3x3 = 0} is a subspace of R3

 

Zero vect of r3 _0 = (0,0,0) is in S because 0-2(0)+3(0) = 0

.:. _0 in S

--> we proved cond 1, now ne need to prove cond 2 and 3

 

 

USE A GENERAL CASE FOR PROVING EXISTENCE
(to prove against, use a specific case)

 

To prove that S is closed under addition

 

For any _u = (u1, u2, u3) and _v = (v1, v2, v3) IN s, so _u + _v = (u1+v1, u2+v2, u3+v3)

 

(u1+v1) - 2(u2+v2) + 3(u3+v3)

 

= u1-2u2+3u3 + v1-2v2+3v3

= 0 + 0 (_u, _v in S)

= 0

 

.:. u+v in s, and S is closed under addition

 

 

 

To prove that S s closed under scalar multiplication

For any lambda in R, and _v = (v1, v2, v3) in S

So lambda _v = (lambda v1, lambda v2, lambda v3)

Lambda v1 – 2 lambda v2 + 3 lambda v3 = lambda (v1 – 2v2 + 3v3) // properties of R

 

= lambda (0) // _v in s

= 0

 

.:. lambda v satisfies equation and is inside S

 

 

By the subspace theorem, S is a subspace of R3