Final answer:
To show that the set W is a subspace of R3, it needs to be proven that it satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector.
Step-by-step explanation:
To show that the set W is a subspace of R3, we need to prove that it satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector.
Closure under addition:
Let (a1, a1+4b1, b1) and (a2, a2+4b2, b2) be two vectors in W. We need to show that their sum is also in W. The sum of these vectors is (a1+a2, a1+a2+4(b1+b2), b1+b2). Since a1+a2, a1+a2+4(b1+b2), and b1+b2 are all real numbers, the sum of the two vectors is in W. Therefore, W is closed under addition.
Closure under scalar multiplication:
Let (a, a+4b, b) be a vector in W and c be a real number. We need to show that c(a, a+4b, b) is also in W. The product of the vector and scalar is (ca, ca+4(cb), cb). Since ca, ca+4(cb), and cb are all real numbers, the product is in W. Therefore, W is closed under scalar multiplication.
Contains the zero vector:
The zero vector is (0, 0+4(0), 0) = (0, 0, 0). Since all the real numbers in the vector are 0, the zero vector is in W. Therefore, W contains the zero vector.
Since the set W satisfies all three properties of a subspace, it is a subspace of R3 with the standard operations.