So I'm trying to show that if $U \subseteq V$, then $V^\perp \subseteq U^\perp$, where ${}^\perp$ denotes orthogonal complement.
I started by assuming a vector $v \in V^\perp$, so $\langle v,w \rangle = 0$ for all $w \in V$.
So if $w$ happens to be inside $U$, it means that $\langle v,w \rangle = 0$ for all $w \in U$ and thus $v$ is inside $U^\perp$.
But what about the other case? i.e. when $w$ is inside $V$ but outside $U$? How can I proceed then?
Thank you
Best Answer
You have already shown all that needs to be shown. If $w$ is not in $U$, then whether $\langle v,w \rangle = 0$ or not is irrelevant for determining whether $v \in U^\perp$.
In other words, $v \in U^\perp$ means $\forall w. w \in U \Rightarrow \langle v,w \rangle = 0$, so it does not matter what happens when $w \notin U$.
Let's see if I can write out a proof as explanatory as I can.
We want to show that $V^\perp \subseteq U^\perp$. So, given any $w \in V^\perp$, we need to show that $w \in U^\perp$. We know that $\langle w,v \rangle = 0$ for all $v \in V$, by definition of $V^\perp$.
Now $w \in U^\perp$ means that for all $u \in U$, $\langle w,u \rangle = 0$. To show this, suppose we are given $u \in U$. Since $U \subseteq V$, we have $u \in V$. Then $\langle w,u \rangle = 0$, because $w \in V^\perp$ and $u \in V$.
In general, if $U \subseteq V$, then any statement that is true for all elements of $V$ is automatically true for all elements of $U$. So if $\langle w,v \rangle = 0$ for all $v \in V$, then $\langle w,u \rangle = 0$ for all $u \in U$, simply because $V$ contains $U$.