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Problem

Let $V$ be a real inner product space and $U \subset V$. Show that $(U^\perp)^\perp=U$.

Progress

Clearly for $x\in U$ we have that $\langle x,v \rangle=0$ for all $v \in U^\perp$. This immediately yields that $x \in (U^\perp)^\perp$ and so $U \subset (U^\perp)^\perp$.

Taking $x \in (U^\perp)^\perp$, we have that $\langle x,v \rangle=0$ for all $v \in U^\perp$. Not sure how to move it on from here though; any assistance would be much appreciated. Regards.

  • 1
    Indeed it is false for e.g. an infinite dimensional Hilbert space $V$ with a proper dense subspace $W$. This is because the orthogonal complement of a subset in a Hilbert space is always closed, so we have that $W^{\perp\perp} = \overline{W}$.2011-12-16

4 Answers 4

4

Since you already know that $U\subseteq U^{\perp\perp}$, it will suffice that these two spaces have the same dimension. Since $\dim(U^{\perp\perp})=\dim(V)-\dim(U^\perp)=\dim(V)-\dim(V)+\dim(U)=\dim(U),$ you are done.

1

If $\dim V = n$ and $\dim U = k$, then $\dim U^\perp = n-k$. Then argue using orthonormal basis.

0

Hint: use the orthogonal decomposition $V=U\oplus U^\perp$.

0

Using the orthogonal decomposition $V=U\oplus U^\perp$ mentioned in another answer here, every vector $v\in V$ may be decomposed as $v=p+n$ with $p\in U$ and $n\in U^\perp$. Then $\begin{align} v\in(U^\perp)^\perp&\iff\forall n'\in U^\perp\ \langle v,n'\rangle=0\\ &\iff\forall n'\in U^\perp\ \langle n+p,n'\rangle=0\\ &\iff\forall n'\in U^\perp\ \langle n,n'\rangle+\langle p,n'\rangle=0\\ &\iff\forall n'\in U^\perp\ \langle n,n'\rangle=0\\ &\iff n=\mathbf 0\\ &\iff v\in U \end{align}$ and so $(U^\perp)^\perp=U$.