Q: Determine whether the set,

is a subspace of

A: I said yes, because...

Firstly, the set contains the zero vector, since the set consists of all functions with zero area on the interval ; thus, is non-empty.

Now, let , then...

is closed under addition.

Suppose again that and let , then....

is closed under scalar multiplication.

In general, is it true that a set made some function equal to zero will automatically contain the zero vector and be nonempty?