In each of the following determine if the given set is a sub

In each of the following, determine if the given set is a subspace of R^2 or not. For each case in which the set is a subspace, provide a proof that it is subspace of R^2. For each case in which the set is not a subspace of R^2, state one of the properties of a subspace that does not hold and give a counter-example showing that the property fails. S_1 = {[x y] epsilon R^2| x greaterthanorequalto 0, y

Solution

a) S₁ = {(x,y)^T R²: x 0, y < 0} . If (x,y)^T is an arbitrary element of S₁ and a is a negative scalar, then a(x,y)^T = (ax,ay)^T. Since y < 0 and a < 0, hence ay > 0. Therefore, a(x,y)^T does not belong to S₁. Hence S₁ , not being closed under scalar multiplication,

b) S₂ = {(x,y)^T R² : y = 2x} .Let (x_1, y₁ )^T and (x₂, y₂)^T be 2 arbitrary elements of S₂ and let a be an arbitrary scalar. Then y₁ = 2x₁ and y₂ = 2x₂. Further, (x₁, y₁)^T + (x₂, y₂)^T = (x₁+ x₂ , y₁+ y₂ )^T. Since, y₁ + y₂ = 2x₁ +2x₂ = 2 (x₁ + x₂ ), hence (x₁, y₁)^T + (x₂, y₂)^T S₂ . Further a(x₁, y₁)^T = (ax₁ , ay₁ )^T. Since ay₁ = a(2x₁) = 2(ax₁), hence a(x₁, y₁)^T S_2. Then S₂ , being closed under vector addition scalar multiplication, is a vector space. Therefore S₂ is a subspace of R².

c) S₃ = { (x,y)^T R² : xy 0} . Since (2, 3) and ( -5,-1) both belong to S₃ and since (2,3) + ( -5,-1) = (-3, 2) does not belong to S₃ , hence S₃ is not closed under vector addition so that S₃ is not a vector space. Therefore S₃ is not a subspace of R².