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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).

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  • $\begingroup$ This looks good to me! You might just want to be careful with your wording; the fact that the $0$ vector is not in $W_{2}$ is not really a contradiction, but your reasoning is all correct here. $\endgroup$ – Alex Wertheim Jun 17 '13 at 4:40
  • $\begingroup$ thanks a lot. This is a help to know I did this right. $\endgroup$ – Jesse Jun 17 '13 at 4:42
  • $\begingroup$ (+1) for remembering to check whether the set contains $0$. $\endgroup$ – user70962 Jun 17 '13 at 6:07
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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.

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  • $\begingroup$ Again thanks. And as a subsidiary question: Let S be a nonempty set and K a field. For any s_0 in S, prove that if f is in F(S,K), with f(s_0) = 0, then that too is a subspace. I would say it's because given the f(s_0) condition, no matter what function I put in for f it is zero, so zero is in the set. And no matter what function I put in to add to it it's still the same function because f(s_0) = 0 and is still in the set. And the same thing happens if I multiply by an arbitrary scalar. Yes? $\endgroup$ – Jesse Jun 17 '13 at 4:51
  • $\begingroup$ @Jesse I think you understand what's going on. If you could give a formal proof, I'll critique it. $\endgroup$ – Ink Jun 17 '13 at 5:00
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You definitely got the right idea and the only thing I'd criticize is you're not very formal and too much verbiage. Here is a formal proof of you question which is more concise:

Prove that $W =\{(a_1, \ldots, a_n) \in \mathbb{F}^n \mid a_1 + \cdots + a_n =0\}$ is a subspace of $\mathbb{F}^n$.

Proof: Obviously, $0 = (0_1, \ldots, 0_n) \in W$ as $0_1 + \ldots + 0_n = 0$. Let $x = (x_1, \ldots, x_n) \in W$ and $y = (y_1, \ldots, y_n) \in W$. We have $x + y = (x_1 + y_1, \ldots, x_n + y_n) \in W$ as $$(x_1 + y_1) + \cdots + (x_n + y_n) = (x_1 + \cdots + x_n) + (y_1 + \cdots + y_n) = 0 + 0 = 0.$$ Furthermore, $cx = (cx_1, \ldots, cx_n) \in W$ as

$$(cx_1) + \cdots + (cx_n) = c(x_1 + \cdots + x_n) = c \cdot 0 = 0.$$

Therefore, $W$ is a subspace of $\mathbb{F}^n$.

Hope this helps!

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