I am also given that if $S$ and $T$ are subspaces of a vector field, then the above are equivalent
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
Suppose $u \perp S$ and $v \perp T$; that is, $\langle u,s \rangle = 0$ and $\langle v,t \rangle = 0$ for all $s \in S$ and $t \in T$. You need to show that $u+v \perp S \cap T$; that is, $\langle u+v, x \rangle = 0$ for all $x \in S \cap T$. Can you do this?
Now suppose $S$ and $T$ are subspaces. Why does the converse to the above hold?
If you're still stuck, please post some of your working.
If $x\in S^\perp+T^\perp$, then we can write $x=u+v$ with $u\in S^\perp$ and $v\in T^\perp$. Now if $r\in S\cap T$, then $u\perp r$ and $v\perp r$, hence $x=u+v\perp r$, i.e. $x\in(S\cap T)^\perp$.