Question

Which of the following subsets is a subspace of R³?

a. W = {(x1,x2, 1): x1 and x2 are real numbers}
b. W = {(X1,X1+ x3,x3): x1 and are real numbers}

Answer 1

To determine if a subset is a subspace of R³, we need to check if it satisfies the three conditions for a subspace: closure under addition, closure under scalar multiplication, and containment of the zero vector. Subset b, W = {(x1, x1 + x3, x3): x1 and x3 are real numbers}, satisfies all three conditions and is therefore a subspace of R³. Subset a, W = {(x1,x2, 1): x1 and x2 are real numbers}, does not satisfy the closure under addition condition and is not a subspace of R³.

Step-by-step explanation:

To determine if a subset is a subspace of ℝ³, we need to check if it satisfies the three conditions for a subspace:

1. Closure under addition: For any two vectors u and v in the subset, the sum u + v must also be in the subset.
2. Closure under scalar multiplication: For any vector u in the subset and any scalar c, the scalar multiple c⋅u must also be in the subset.
3. Containment of the zero vector: The zero vector must be in the subset.

Let's evaluate each subset to see if it satisfies these conditions:

a. W = {(x1,x2, 1): x1 and x2 are real numbers}

Closure under addition: Let u = (u1, u2, 1) and v = (v1, v2, 1), where u1, u2, v1, and v2 are real numbers. The sum u + v = (u1 + v1, u2 + v2, 2) is not in W because the third component is not equal to 1. Therefore, W is not closed under addition.

Closure under scalar multiplication: Let u = (u1, u2, 1) be in W and c be a scalar. The scalar multiple c⋅u = (c⋅u1, c⋅u2, c) is in W because the third component is always equal to 1. Therefore, W is closed under scalar multiplication.

Containment of the zero vector: The zero vector, (0, 0, 0), is not in W because the third component is not equal to 1. Therefore, W does not contain the zero vector.

Since W does not satisfy all three conditions, it is not a subspace of ℝ³.

b. W = {(x1, x1 + x3, x3): x1 and x3 are real numbers}

Closure under addition: Let u = (u1, u1 + u3, u3) and v = (v1, v1 + v3, v3), where u1, u3, v1, and v3 are real numbers. The sum u + v = (u1 + v1, (u1 + u3) + (v1 + v3), u3 + v3) is in W because the second component simplifies to u1 + v1 + u3 + v3, which is equal to (u1 + u3) + (v1 + v3). Therefore, W is closed under addition.

Closure under scalar multiplication: Let u = (u1, u1 + u3, u3) be in W and c be a scalar. The scalar multiple c⋅u = (c⋅u1, c⋅(u1 + u3), c⋅u3) is inW because the components satisfy x1 = c⋅u1, x1 + x3 = c⋅(u1 + u3), and x3 = c⋅u3. Therefore, W is closed under scalar multiplication.

Containment of the zero vector: The zero vector, (0, 0, 0), is in W because its components satisfy x1 = x3 = 0. Therefore, W contains the zero vector.

Since W satisfies all three conditions, it is a subspace of ℝ³.