Determine Whether the Following Sets Form Subspaces of R^3

(a) To show that this set is a subspace of R^3, we need to show that it is closed under vector addition and scalar multiplication. Let's take two vectors in this set, call them u and v: u = (u1, u2, u3) such that u1 + u3 = 1 v = (v1, v2, v3) such that v1 + v3 = 1 Now, let's add these two vectors together: u + v = (u1 + v1, u2 + v2, u3 + v3) We want to show that u + v is also in the set, i.e. (u1 + v1) + (u3 + v3) = 1. But we know that u1 + u3 = 1 and v1 + v3 = 1, so (u1 + v1) + (u3 + v3) = (u1 + u3) + (v1 + v3) = 1 + 1 = 2 So u + v is not in the set, and therefore this set is not closed under vector addition. Thus, it is not a subspace of R^3.

(b) Let's take two vectors in this set, call them u and v: u = (u1, u2, u3) such that u1 = u2 = u3 v = (v1, v2, v3) such that v1 = v2 = v3 Now, let's add these two vectors together: u + v = (u1 + v1, u2 + v2, u3 + v3) We want to show that u + v is also in the set, i.e. (u1 + v1) = (u2 + v2) = (u3 + v3). But we know that u1 = u2 = u3 and v1 = v2 = v3, so (u1 + v1) = (u2 + v2) = (u3 + v3) Thus, u + v is in the set. Now, let's take a scalar c and a vector u in the set: u = (u1, u2, u3) such that u1 = u2 = u3 cu = (cu1, cu2, cu3) We want to show that cu is also in the set, i.e. cu1 = cu2 = cu3. But we know that u1 = u2 = u3, so cu1 = cu2 = cu3 Thus, cu is in the set. Therefore, this set is closed under vector addition and scalar multiplication, and is a subspace of R^3.

(c) Let's take two vectors in this set, call them u and v: u = (u1, u2, u3) such that u3 = u1 + u2 v = (v1, v2, v3) such that v3 = v1 + v2 Now, let's add these two vectors together: u + v = (u1 + v1, u2 + v2, u3 + v3) We want to show that u + v is also in the set, i.e. (u1 + v1) + (u2 + v2) = (u1 + v1) + (u2 + v2). But we know that u3 = u1 + u2 and v3 = v1 + v2, so u3 + v3 = (u1 + u2) + (v1 + v2) = (u1 + v1) + (u2 + v2) Thus, u + v is in the set. Now, let's take a scalar c and a vector u in the set: u = (u1, u2, u3) such that u3 = u1 + u2 cu = (cu1, cu2, cu3) We want to show that cu is also in the set, i.e. cu3 = cu1 + cu2. But we know that u3 = u1 + u2, so cu3 = c(u1 + u2) = cu1 + cu2 Thus, cu is in the set. Therefore, this set is closed under vector addition and scalar multiplication, and is a subspace of R^3.

(d) Let's take two vectors in this set, call them u and v: u = (u1, u2, u3) such that u3 = u1 or u3 = u2 v = (v1, v2, v3) such that v3 = v1 or v3 = v2 Now, let's add these two vectors together: u + v = (u1 + v1, u2 + v2, u3 + v3) We want to show that u + v is also in the set, i.e. (u1 + v1) = (u3 + v3) or (u2 + v2) = (u3 + v3). There are two cases:

• If u3 = u1, then v3 = v1 and (u1 + v1) = (u3 + v3)
• If u3 = u2, then v3 = v2 and (u2 + v2) = (u3 + v3) In both cases, u + v is in the set. Now, let's take a scalar c and a vector u in the set: u = (u1, u2, u3) such that u3 = u1 or u3 = u2 cu = (cu1, cu2, cu3) We want to show that cu is also in the set, i.e. cu3 = cu1 or cu3 = cu2. There are two cases:
• If u3 = u1, then cu3 = cu1
• If u3 = u2, then cu3 = cu2 In both cases, cu is in the set. Therefore, this set is closed under vector addition and scalar multiplication, and is a subspace of R^3.