You're definitely on the right track. As you said, just show that elements of are closed under addition and scalar multiplication. This will imply that zero is in there (as long as one element is there--then add it to itself multiplied by -1).

For the other one, just use the property of linearity: if f is linear, then f(a*x)=a*f(x) and f(x+y)=f(x)+f(y). That will show that the null set (usually called thekernel) is closed under addition and scalar multiplication. The kernel of a linear map is always a subspace.