Final answer:
To verify that W is not a subspace of R^2, we can find a vector that does not satisfy the requirement of being the square of the first component.
Step-by-step explanation:
To verify that W is not a subspace of R^2, we need to show that it fails one of the three conditions for a subspace: closed under addition, closed under scalar multiplication, and contains the zero vector.
Consider the vector (1,1) in W. The second component is 1^2 = 1, which does not satisfy the requirement of being the square of the first component. Therefore, (1,1) is not in W and W is not closed under addition.
Since W fails the closed under addition condition, it cannot be a subspace of R^2.