# Let U,W sube V, vector subspaces then: 1) U sube W => W^(_|_) sube U^(_|_) 2) (U+W)^(_|_)=U^(_|_) nn W^(_|_)

Let $U,W\subseteq V$, vector subspaces then:
1) $U\subseteq W\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{W}^{\perp }\subseteq {U}^{\perp }$
2) $\left(U+W{\right)}^{\perp }={U}^{\perp }\cap {W}^{\perp }$
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Julianne Mccoy
1)Let $v\in {W}^{\perp }$, which means that $\left(\mathrm{\forall }w\in W\right):⟨v,w⟩=0$. You want to prove that $v\in {U}^{\perp }$. Take $u\in U$. But then $u\in W$ and therefore $⟨v,u⟩=0$. So, $v\in {U}^{\perp }$
2)Since $U\subset U+W$ and $V\subset U+W$, $\left(U+W{\right)}^{\perp }\subset {U}^{\perp }$ and $\left(U+W{\right)}^{\perp }\subset {W}^{\perp }.$. Therefore $\left(U+W{\right)}^{\perp }\subset {U}^{\perp }\cap {W}^{\perp }$. Now, take $v\in V\setminus \left(U+W{\right)}^{\perp }$. Then there is some $u\in U$ and there is some $w\in W$ such that $⟨v,u+w⟩\ne 0$. Therefore, the numbers $⟨v,u⟩$ and $⟨v,w⟩$ cannoy be both equal to 0. In particular, $v\notin {U}^{\perp }\cap {W}^{\perp }$