The solution set S of the equation y=ax+b, S={(x, y) ∈R^2: y=ax+b}, is not a vector space using regular vector addition and regular scalar multiplication.
I'm having a hard time understanding why that is and which axiom it violates. Let me know where I went wrong with my proofs.
Let a = 6 and b = 8, and y = 6(x) + 8
Closed Under Addition:
(u, 6u + 8) + (v, 6v + 8) = u + v + 6(u) + 6(v) + 16 = (u + v, 6u + 6v + 16) is in S
Commutative:
(u, 6u + 8) + (v, 6v + 8) = u + v + (6(u) + 6(v) + 16) = (v + 6v + 8) + (u + 6u + 8) = (v, 6u + 8)+ (u, 6u + 8)
Associative Property: (u,6u+8) + [(v, 6v + 8) + w] = u + (6u + 8) + v + (6v + 8) + w = [u + + (6u + 8) + v + (6v + 8)]
Additive Identity: (u, 6u+8) + (0,0) = u + 6u + 8 + 0 + 0 = (u, 6u + 8)
Additive Inverse: (u,6u+8) + (-u, -(6u+8)) = u + 6u + 8 – u – 6u – 8 = 0
Scalar multiplication: Let c be a scalar, c(u, 6u + 8) = cu + 6cu + 8c = (6cu, 6cu + 8c) is in S
Distributive Property 1: c[(u, 6u + 8) + (v, 6v + 8)] = cu + cv + 6cu + 6cv + 16c = (cu, 6cu + 8c) + (cv, 6cv + 8c)
Distributive Property 2: let d be a scalar, (c + d)(u, 6u + 8) = (c+d)(u + 6u + 8) = uc + 6cu + 8c + du + 6du + 8d = (cu, 6cu + 8d) + (du, 6du + 8d)
Associative Property: c(du) = c(du + 6du + 8d) = cdu + 6cdu + 8cd = cd(u, 6u + 8)
Scalar Identity: 1(u, 6u + 8) = 1(u + 6u + 8) = u + 6u + 8 = (u, 6u + 8)
Best Answer
You mentioned a vector space (over $\mathbb{R}$), so this is an abelian group, i.e., you need an (additive) identity. Since you are considering a subspace of $\mathbb{R}^2$, the identity is clearly $(0,0)$, i.e., the zero vector. The rest follows from what has already been mentioned.