The orthogonal complement of a set \(S\) in a subvector space \(V\) of \(\RNrSpc{n}\) is a subspace of \(V\).
The orthogonal complement of a set \(S\) in a subvector space \(V\) of \(\RNrSpc{n}\) is a subspace of \(V\).
We need to show that \(\VSpc{W} \DefEq \OrthCmpl{S}\) satisfies the three defining properties of a subvector space.
The \(\Vect{0}\)-vector belongs to \(W\) because \(\DotPr{\Vect{0}}{ \Vect{s} }=0\), for every vector \(\Vect{s}\) in \(S\).
To see that \(W\) is closed under vector addition let \(\Vect{x}\) and \(\Vect{y}\) be in \(W\). This means that, for every \(\Vect{s}\) in \(S\),
\[\DotPr{ \Vect{x} }{ \Vect{s} } = 0 = \DotPr{ \Vect{y} }{ \Vect{s} }\]But then
\[\DotPr{ (\Vect{x} + \Vect{y}) }{ \Vect{s} } = \DotPr{ \Vect{x} }{ \Vect{s} } + \DotPr{ \Vect{y} }{ \Vect{s} } = 0+0 = 0\]So \(\Vect{x} + \Vect{y}\) belongs to \(\VSpc{W}\) as required.
To see that \(\VSpc{W}\) is closed under scalar multiplication, let \(\Vect{x}\) in \(W\). This means that \(\DotPr{ \Vect{x} }{ \Vect{s} }=0\) for every \(\Vect{s}\) in \(S\). Now if \(t\) in \(\RNr\) is arbitrary, then
\[\DotPr{ (t \Vect{x}) }{ \Vect{s} } = t\cdot (\DotPr{ \Vect{x} }{ \Vect{s} }) = 0 \]So \(t\Vect{x}\) belongs to \(\VSpc{W}\) as required.
We have seen that \(\OrthCmpl{S}\) contains \(\Vect{0}\) and is closed under addition and scalar multiplication. Therefore \(\OrthCmpl{S}\) is a subvector space of \(\RNrSpc{n}\).