Let $U,W \subseteq V$, vector subspaces.

Question

Let $U,W \subseteq V$, vector subspaces then:

1) $U\subseteq W \implies W^{\perp}\subseteq U^{\perp}$

2) $(U + W)^{\perp}=U^{\perp}\cap W^{\perp}$

3) $(U\cap W)^{\perp}=U^{\perp}+W^{\perp}$

I am really begging and trying to learn this but I see not many people are intrested in linear algebra, anyway. I know $x\perp y\implies\langle x,y\rangle=0$, right?

so then to prove those should I pick some arbitrary $x,y\in\mathbb{R}^n$ one in $U$ and one in $W$ and try to prove them this way? Or is a more efficient way to do it?

Answer 1

1. Let $v\in W^\perp$, which means that $(\forall w\in W):\langle v,w\rangle=0$. You want to prove that $v\in U^\perp$. Take $u\in U$. But then $u\in W$ and therefore $\langle v,u\rangle=0$. So, $v\in U^\perp$.
2. Since $U\subset U+W$ and $V\subset U+W$, $(U+W)^\perp\subset U^\perp$ and $(U+W)^\perp\subset W^\perp.$ Therefore $(U+W)^\perp\subset U^\perp\cap W^\perp$. Now, take $v\in V\setminus(U+W)^\perp$. Then there is some $u\in U$ and there is some $w\in W$ such that $\langle v,u+w\rangle\neq0$. Therefore, the numbers $\langle v,u\rangle$ and $\langle v,w\rangle$ cannoy be both equal to $0$. In particular, $v\notin U^\perp\cap W^\perp$.
3. I will leave this one to you.