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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 $w \in W$.

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  • $\begingroup$ If $w\in W$ is $-w\in W$? $\endgroup$ – Doug M Jan 7 '19 at 19:27
  • $\begingroup$ @DougM OP already mentioned that $\endgroup$ – Shubham Johri Jan 7 '19 at 19:28
  • $\begingroup$ 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? $\endgroup$ – bounceback Jan 7 '19 at 19:31
  • $\begingroup$ @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. $\endgroup$ – Tim Jan 7 '19 at 19:35
  • $\begingroup$ No, your first equality is false - see Bernard's answer below for the counterexample $\endgroup$ – bounceback Jan 7 '19 at 19:38
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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 $\;S\cap U+S\cap W=\{0\}$.

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