# Sum of Linear Subspaces

Let $$V$$ be a vector space over $$K$$ and $$S,U,W \subset V$$ linear subspaces of $$V$$. $$U+W$$ is defined as $$\{ u+w | u \in U, w \in W \}$$. Is it true, that

a) $$U+W = \{ u-w | u \in U, w \in W \}$$

b) $$S \cap (U + W) = (S \cap U) + (S \cap W)$$?

My thoughts for a) so far had been that due to $$-1 \in K$$, $$-w$$ also has to be in $$W$$ if $$w \in W$$.

• If $w\in W$ is $-w\in W$? – Doug M Jan 7 '19 at 19:27
• @DougM OP already mentioned that – Shubham Johri Jan 7 '19 at 19:28
• Well, that takes care of part (a): $\{u - w: u \in U, w \in W\} = \{u + (-w): u \in U, -w \in W\} = U + W$. As for part (b), what have you tried there, OP? – bounceback Jan 7 '19 at 19:31
• @bounceback I was thinking about rewriting it like $S \cap \{ u + w | u \in U, w \in W \} = \{ u + w | u \in (U \cap S), w \in (W \cap S) \} = (S \cap U) + (S \cap W)$ but I am not sure if this is a valid operation. – Tim Jan 7 '19 at 19:35
• No, your first equality is false - see Bernard's answer below for the counterexample – bounceback Jan 7 '19 at 19:38

For $$a)$$ it's fine, as $$w\in W\iff -w\in W$$.
For $$b)$$, a counter-example: take $$V=K^2$$ ($$K$$ is the base field)$$,$$U={(x,0),\;x\in K}, $$\;W=\{(0,x),\;x\in K\}$$ and $$\;S=\{(x,x),\;x\in K\}$$.
Then $$U+W=V$$, so $$S\cap(U+W)=S$$, but $$S\cap U=\{0\}=S\cap W$$, so $$\;S\cap U+S\cap W=\{0\}$$.