# Proof of a Vector Space

Let $F$ be a field and let $(V, +, F)$ be a vector space over $F$. If $W_1$ and $W_2$ are subspaces of $F$, prove that $W_1 - W_2 = \{v \in V | v = w_1 - w_2 \text{ for some } w_1 \in W_1, w_2 \in W_2 \}$ is a subspace of $V$.

To be clear, I know that $0$ is an element of $W_1,W_2$ because both are subspaces, so $0$ is also an element of $W_1-W_2$. I know both are closed under scalar multiplication and addition because of the same reason. So how would I show $W_1-W_2$ is closed under scalar multiplication and addition?

• You are almost there, just write everything out. For scalar multiplication say, note that $\lambda\left(w_1-w_2\right)=\lambda\,w_1-\lambda\,w_2$. For addition, suppose you took two elements in your "formal difference": $w_1-w_2,\;\omega_1-\omega_2$. Add these to get... – lulu Aug 31 '15 at 23:14
• (w1 + omega1) - (w2 + omega2)? – Jay3 Aug 31 '15 at 23:20
• Perfect. To be complete, you should check additive inverses. – lulu Aug 31 '15 at 23:22
• Would that be in this case (w2-w1) where w2 is still an element of W2 and w1 is still an element of W1? – Jay3 Aug 31 '15 at 23:26
• You got it. If you want to dig deeper, suppose you had an explicit basis for each of $W_1$ and $W_2$. Find one for the difference. – lulu Aug 31 '15 at 23:44

If I were you, I would try demonstrating the $W=W_1-W_2 \neq \emptyset$ by showing an element which is in it. Here is a hint:
We know that since $W_1, W_2$ are subspace that $0 \in W$. Since $0 = 0 - 0 \in V$ because $0 \in W_1, W_2$.
Let $x \in W$ and $y \in W$, then $x = w_1 - w_2$, $y=w_3 -w_4$ for some $w_1, w_2 \in W_1, w_3, w_4 \in W_2$. And further, $x+y = (w_1-w_2)-(w_4-w_3) \in W$, since $W_i$ is a subspace.
Let $a \in \mathbb{F}$, then $av \in W$; since $av =a(w_1 -w_2)=aw_1-aw_2$ for some $w_1, w_2 \in W_1, W_2$. This is because $W_i$ is a subspace, i.e. a vector space.