OK, here goes another.
Prove that $ W_1 = ${$(a_1, a_2, \ldots, a_n) \in F^n : a_1 + a_2 + \cdots + a_n = 0$} is a subspace of $F^n$ but $ W_2 = ${$(a_1, a_2, \ldots, a_n) \in F^n : a_1 + a_2 + \cdots + a_n = 1$} is not.
OK. Any subspace has to contain the zero vector, be closed under addition and scalar multiplication by definition. So to prove this we first see whether the set $W_1$ meet those criteria. Plugging in 0 for $a_i$ obviously works, so the first condition is met.
Is it closed under addition? let $b_i$ be the components of an arbitrary vector in $W_1$. So ($b_1, b_2, \ldots, b_n) \in W_1$ and if we add it to $(a_1, a_2, \ldots, a_n)$ we get $(a_1 + b_1) + (a_2 + b_2) + \cdots +(a_n + b_n) = 0$. That's pretty clearly part of $W_1$ and thus closed under addition.
Next we see if it is closed under multiplication by a scalar. We pick an arbitrary scalar $c$ and multiply it by $(a_1, a_2, \ldots, a_n)$ to get $(ca_1, ca_2, \ldots, ca_n)$ and plugging that into the original condition we find that it doesn't matter what c is, because $ca_1 + ca_2 + \cdots + ca_n = 0$ and that's still in $W_1.$ Therefore $W_1$ is a subspace of $F^n$.
If we do the same procedure with $W_2$, though, we find that $0$ vector is not in the set. Because $a_1 + a_2 + \cdots + a_n = 1$ is a contradiction.
Further, we can see that it isn't closed under multiplication either. $ca_1 + ca_2 + \cdots + ca_n = c$ and that will only equal 1 if c=1, so the equation does not hold with an arbitrary $c$.
Therefore $W_2$ is NOT a subspace of $F^n$.
Any holes in this proof?
(Yeah, I have been bothering folks here a lot but I finally feel that I am getting the hang of this and I have an exam tomorrow night).
Best Answer
Your proof is fine. Also, to prove that a subset $S$ is not subspace, it suffices to show that at least one of the conditions ($0 \in S$, closure under addition and scalar multiplication) fails. So as soon as you show that $0 \notin S$, your proof is complete.