proving vectors are not part of the subspace (1 Viewer)

[velox]

Do we have to prove these formally or can we just use a specific vector to be a counterexample?

TIA

 

[acmilan]

If you're asking what I'm thinking you are, formally. Do you have any examples?

 

[withoutaface]

Just show that the vector's projection onto the subspace isn't equal to the vector itself.

i.e. where the basis of the subspace is (a1, a2, a3) and we want to show v is not in that subspace, show that
v != (a1.v/a1.a1)a1+(a2.v/a2.a2)a2+(a3.v/a3.a3)a3
Though this only applies to inner product spaces methinks.

 

[Slidey]

Only two things are required to prove a subset H in V is a subpsace of V:
That H is closed under addition (for u and v in H, u+v is in H)
H is closed under scalar multplication: for c in R, u in H, cu is in H.

If you suspect a set is a not a subspace, test for the existence of the zero vector. The zero vector requires closure of addition and multiplication to exist, you see:
u+(-u)=0 in H (closed under addition)
-u exists because -u=-1*u (which is in any subspace).

 

[word.]

you also need to test for the zero vector, which also proves that the set is non-empty because
the zero vector can exist without closure under addition or scalar multiplication:

eg let S = {x in P₃(R) | x₁³ + x₂² + x₃ = 0}

If the zero vector (0,0,0) does not exist then clearly it's not a subspace but in this case it does exist... however it still isn't a subspace:
(0,1,-1) is in S but -1(0,1,-1) = (0,-1,1) is not thus it isn't closed under scalar multiplication and so S is not a subspace of P₃(R)

so to the original question, only a counter-example is required to prove that a subset S of a vector space V is not a subspace ...