How to show that $(U^{\bot})^{\bot}=U$, if $U$ is a linear subspace of $V$ and $V$ is finite-dimensional?

Question

Let $V$ be an Euclidean vector space with scalar product $(.|.)$. If $S ⊂ V$ is any subset of $V$ , define the orthogonal complement of $S$ by $$S^{\bot}=\left\{v\in V| \forall s\in S:\left(s|v\right)=0\right\}$$

I need to show that $(U^{\bot})^{\bot}=U$, if $U$ is a linear subspace of $V$ and $V$ is finite-dimensional.

Answer 1

$$x\in U\implies \langle x, u\rangle =0\;\;\;\forall\;u'\in U^\perp\implies U\subset\left(U^\perp\right)^\perp$$

Now:

$$\dim V=n\;,\;\;\dim U=k\implies \dim U^\perp = n-k\implies \dim\left(U^\perp\right)^\perp=n-(n-k)=k\implies$$

$$\begin{cases}U\subset\left(U^\perp\right)^\perp\\{}\\\dim U=\dim\left(U^\perp\right)^\perp\end{cases}\;\;\implies u=\left(U^\perp\right)^\perp$$

Answer 2

It is immediately clear from the definition that every vector of $U$ is orthogonal to every vector of $U^\perp$, so that $U\subseteq U^{\perp\perp}$; the hard part is showing the opposite inclusion. A dimension argument is an easy and perfectly standard way to do this, but just for the fun of it I'd like to try a different approach.

I must use finite dimensionality, since the result is not true without. I'll use it in the following form: every linear form on $V$ is of the form $x\mapsto(x\mid v)$, that is taking the scalar product with a fixed vector$~v$. This is easy to prove using an orthonormal basis $e_1,\ldots,e_n$ of $V$: given a linear form $\alpha:V\to\Bbb R$ set $v=\alpha(e_1)e_1+\cdots+\alpha(e_n)e_n$ (without finite dimension, one cannot form this linear combination at all). Now $\alpha(e_i)=(e_i\mid v)$, so the linear forms $\alpha$ and $x\mapsto(x\mid v)$ coincide on a basis, hence on all vectors.

Now let $w\notin U$; we need to find a $v\in U^\perp$ with $w\not\perp v$ to show that $w\notin U^{\perp\perp}$. Take a basis of $U$, add the linearly independent vector $w$, and complete to a basis of $V$ (using the incomplete basis theorem). Let $\alpha$ be the function taking the coordinate of $w$ for this basis, and let $v$ be the vector associated to$~\alpha$ as above. By construction $\alpha$ vanishes on (the chosen basis of) $U$, so $v\in U^\perp$, but not at$~w$, so $w\not\perp v$.