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I would like if someone could look over my proof. It feels odd to me.

Let $W$ be a subspace of a vector space $V$ over a field $F$. Prove that $v + W = \{v + w \mid w \in W\}$ is a subspace of $V$ if and only if $v \in W$.

Proof: ($\Rightarrow$) Suppose $v + W$ is a subspace of $V$. Note that $v = v + 0_W \in v + W$ and as a result $(-1)v = -v \in v + W$ as $v + W$ is a subspace. Therefore, $-v = v + w$ for some $w \in W$. Solving for $w$, we get $w = -2v$ and since $W$ is a subspace $(\frac{-1}{2})w = (\frac{-1}{2})(-2v) = v \in W$ as desired.

($\Leftarrow$) Suppose $v \in W$. As $W$ is a subspace, $-v \in W$ and therefore $0 = v + (-v) \in v + W$. If $x, y \in v + W$ then $x = v + w$ and $y = v + w'$ for some $w,w' \in W$. Then $x + y = (v + w) + (v + w') = v + (v + w + w') \in v + W$ as $v,w,w'\in W$ and $W$ is a subspace. Furthermore, note that $cx = c(v + w) = cv + cw = v + (cv + cw - v) \in v + W$ as $v,w \in W$ and $W$ is a subspace. Therefore, $v + W$ is a subspace of $V$ as it contains the $0$ element and it is closed under addition and scalar multiplication. $_\Box$

I'd appreciate any feedback. Thank-you.

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    Argumentation using coset properties might be simpler. For $\Rightarrow$ I'D simply write that if $v+W$ ia a subspace, it contains $0$, as does $0+W$, hence $v\in v+W=0+W=W$. For $\Leftarrow$ similary: $v\in W$ implies $-v\in W$ implies $0\in v+W$ imlies $v+W=0+W=W$. - Otherwise wour proof is fine, except that it uses the unjustified assumption that $2$ is invertible in the ground field. – Hagen von Eitzen May 21 '13 at 14:16
  • Cosets were introduced in a problem so all I know about them is that $v + W = {v + w \mid w \in W}$ so I can't really follow your proof. Why don't I have the existence of the additive/multiplicative inverse of 2? $F$ is a field so I would think that it is alright. –  May 21 '13 at 14:55

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Argumentation using coset properties might be simpler. For $\Longrightarrow$, I'd simply write that if $v+W$ is a subspace, it contains $0$, as does $0+W$, hence $v \in v+W=0+W=W$. For $\Longleftarrow$ similary: $v \in W$ implies $−v \in W$ implies $0 \in v+W$ implies $v+W=0+W=W$. Otherwise, your proof is fine, except that it uses the unjustified assumption that $2$ is invertible in the ground field.

(Originally posted as a comment by Hagen von Eitzen.)

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