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


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.

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The answers below assume $V$ is finite dimensional; which is ok, since that's what your question seems to be assuming. – David Mitra Dec 16 '11 at 12:37
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 $W^{\perp\perp} = \overline{W}$. – kahen Dec 16 '11 at 13:34
up vote 3 down vote accepted

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.

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

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

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