Question

Let V be the vector space of real-valued functions, y = f(x).(With the operations that vector addition is the addition of functions, and scalar multiplication is multiplying a function by a constant.) Let H be the set of real-valued functions that go through the origin, (0,0). (With the same operations as V) Show that H is a subspace of V.

Answer 1

The set H is a subspace of V because it is closed under vector addition and scalar multiplication, and contains the zero vector, all real-valued functions in H pass through the origin, satisfying substitution requirements.

To demonstrate that the set H is a subspace of the vector space V, we must verify that H satisfies three conditions: closure under vector addition, closure under scalar multiplication, and the presence of the zero vector. Firstly, if we take any two real-valued functions f(x) and g(x) that pass through the origin, their sum f(x) + g(x) also passes through the origin since (f+g)(0) = f(0) + g(0) = 0 + 0 = 0, ensuring that H is closed under addition. Secondly, for any function f(x) in H and a scalar α, the function αf(x) will also pass through the origin because (αf)(0) = αf(0) = α·0 = 0, confirming closure under scalar multiplication.

Finally, H contains the zero vector (the function 0(x) = 0 for all x), as it meets the criteria of passing through the origin. Since all three conditions are met, H is indeed a subspace of V.

The set H, consisting of real-valued functions passing through the origin, is a subspace of the vector space V of real-valued functions as it is closed under vector addition, scalar multiplication, and contains the zero vector.