Let S1 and S2 be subspaces of a vector space V Let S1 S2 v

Let S_1 and S_2 be subspaces of a vector space V. Let S_1 S_2 = (v elementof V/c elementof S_1 or v elementof S_2), S_1 S_2 = {v elementof V/v elementof S_1 and v elementof S_2}, and let S_1 + S_2 = {v elementof V/v = x + y for x elementof S_1 and y elementof S_2}. a. Prove that S_1 S_2 is not in general a subspace of V. b. Prove that S_1 S_2 is a subspace of V. c. Prove that S_1 + S_2 is a subspace of V.

Solution

(a). We will prove the statement by a counter example. Let S₁ = {(x,0 ): x R}   and S₂ = {(0,y ): y R}   , both subspaces of R². Then S₁US₂ includes both (a,0)and (0,b)(where a and b are arbitrary real numbers). Further, (a,0)+ (0,b) S₁US₂ . Hence, S₁US_2, being not closed under addition is not a vector space.

(b) Since both S₁, S₂ are subspaces of V, therefore, the additive identity 0 both S₁, S₂ .This means that 0 S₁ S₂. Further, if u, w S₁ S₂ , then u, w S₁ and u, w S₂. Since S₁, S₂ are subspaces of V, these are closed under addition which means that u + w S₁ and u+ w S₂ , implying that u + w S₁ S₂. Further, any element u S₁ S₂ is in both S₁, S₂, so that cu both S₁, S₂ for any scalar c, implying that cu S₁ S₂. Thus, S₁ S₂ is a subspace of V.

(c) Let u₁ + w₁ and u₂ + w₂ with u_i S₁ and w_i S₂ be two arbitrary elements of S₁ + S₂. Then, their sum is

(u₁ + w₁) + (u₂ + w₂) = (u₁ + u₂) + (w₁ + w₂) S₁ + S₂ as u₁ + u₂ S₁ and w₁ + w₂ S_2. Further, c(u₁ + w₁) = (cu₁) + (cw₁) S₁ + S₂ as cu₁ S₁ and cw₁ S₂ ( where c is an arbitrary scalar).

Hence, S₁ + S₂ is a subspace.