The following content is from "Linear Algebra Done Right" by Sheldon Axler
Corollary: Suppose $U$ is a finite-dimensional subspace of $V$. Then $$U = (U ^\perp)^\perp.$$
We need to prove the following: i) $U \subseteq (U^\perp)^\perp$ and
ii) $(U^\perp)^\perp \subseteq U$
Proof i): Supposer $u \in U$. Then $\langle u,v \rangle = 0$ for $v \in U^\perp$. Since all vector $u$ is orthogonal to $v$. Then $u \in U$. Hence $u \in (U ^\perp)^\perp$
I do understand the proof for i) as stated in the above.
But I'm confused about the proof in ii)
ii) Let $u \in (U ^\perp)^\perp$ From Proposition 2, we have that $V = U \bigoplus U^\perp$. and so $u$ can be written as $u=v+w$ where $v \in U$ and $w \in U^\perp$. But $u – v = w$, so $u – v \in U^\perp$.
Now we already have that $u \in (U ^\perp)^\perp$ and $v \in (U ^\perp)^\perp$ and $u – v \in (U ^\perp)^\perp$. Therfore, $u – v \in U ^\perp \cap (U^\perp)^\perp$.
***Here is where I don't understand! Why we can let $u = v + w$, but since $u \in (U ^\perp)^\perp$? and Why does $u – v \in U ^\perp \cap (U^\perp)^\perp$?
Best Answer
Since $V=U\oplus U^\perp$, every vector in $V$ can be written as a sum of a vector in $U$ and a vector in $U^\perp$.
This holds, in particular, for $u\in(U^\perp)^\perp$. So, according to Axler, write $$ u=v+w,\qquad v\in U, w\in U^\perp $$ Now apply the assumption that $u\in(U^\perp)^\perp$ and the fact that $v\in U\subseteq(U^\perp)^\perp$ to conclude that $$ w=u-v\in(U^\perp)^\perp $$ as well. On the other hand, $w\in U^\perp$ by hypothesis. Therefore $$ w=u-v\in U^\perp\cap(U^\perp)^\perp $$ A general result about orthogonal complements is that, for every subspace $X$, $X\cap X^\perp=\{0\}$. This holds in particular for $X=U^\perp$.
Hence $w=u-v=0$, so $u=v\in U$.