Let \(S\) be a subset of an inner product space \(V\). Show that \((S^{\perp})^{\perp}\) contains the subspace spanned by \(S\). When \(V\) is finite-dimensional, show that \((S^{\perp})^{\perp}\) is the subspace spanned by \(S\).
Solution
Let \[ W=\operatorname{span}S. \] First observe that \[ S^\perp=W^\perp, \] because being orthogonal to every vector of \(S\) is equivalent, by linearity, to being orthogonal to every linear combination of vectors of \(S\).
1. One always has\[ \boxed{W\subseteq(S^\perp)^\perp.} \]
Take \(w\in W\) and \(u\in S^\perp=W^\perp\). By definition of \(W^\perp\), \[ (w\mid u)=0. \] Thus \(w\) is orthogonal to every element of \(S^\perp\), which means \[ w\in(S^\perp)^\perp. \]
2. If \(V\) is finite dimensional, equality holds.
In finite dimension the orthogonal-decomposition theorem gives \[ V=W\oplus W^\perp \] [@HK71]. Apply it again to the subspace \(W^\perp\): \[ V=W^\perp\oplus(W^\perp)^\perp. \] The first decomposition shows that \(W\) is a complementary subspace to \(W^\perp\) and is itself orthogonal to \(W^\perp\). The orthogonal complement is unique, hence \[ (W^\perp)^\perp=W. \] Since \(S^\perp=W^\perp\), \[ \boxed{(S^\perp)^\perp=\operatorname{span}S.} \]