PropositionOrthogonal complement: properties

In a subspace \(\VSpc{W}\) of \(\RNrSpc{n}\) the following hold:

  1. If \(S\subseteq T\subseteq \VSpc{W}\), then \(\OrthCmpl{T}\subseteq \OrthCmpl{S}\)

  2. \(\OrthCmpl{S}=\OrthCmpl{ \SpanOf{S} }\)

  3. If \(\VSpc{V}\) is a subspace of \(\VSpc{W}\), then \(\VSpc{V}\) and \(\OrthCmpl{V}\) have only the zero vector in common.

  4. For every subset \(S\) of \(\VSpc{W}\), \(\OrthCmpl{S} = S^{\bot\bot\bot}\).

Proof

i.   If \(S\subseteq T\) then \(\OrthCmpl{T}\subseteq \OrthCmpl{S}\)

If \(\Vect{x}\in \OrthCmpl{T}\), we need to show that \(\Vect{x}\in \OrthCmpl{S}\). So we need to show that \(\DotPr{ \Vect{x} }{ \Vect{s} } = 0\), for every \(s\in S\). – But if \(\Vect{s}\in S\), then also \(\Vect{s}\in T\), and so \(\DotPr{ \Vect{x} }{ \Vect{s} } = 0\) because \(\Vect{x} \in \OrthCmpl{T}\).

ii.

We need to verify the two properties below

  1. \(\OrthCmpl{S}\) contains \(\OrthCmpl{\SpanOf{S}}\)
  2. \(\OrthCmpl{\SpanOf{S}}\) contains \(\OrthCmpl{S}\)

The inclusion 1. holds by part (i), using that \(S\subseteq \SpanOf{S}\). To see that inclusion 2. holds, consider \(\Vect{x}\in \OrthCmpl{S}\). We need to show that \(\DotPr{ \Vect{x} }{ \Vect{v} } = 0\) for every \(\Vect{v} \in \span(S)\). Now \(\Vect{v} = t_1 \Vect{s}_1 + \cdots + t_k \Vect{s}_k\), for some \(\Vect{s}_1\), ... , \(\Vect{s}_k\) in \(S\), and \(t_1\), ... , \(t_k\) in \(\RNr\). Therefore

\(\DotPr{ \Vect{x} }{ \Vect{v} }\)\(=\)\(\DotPr{ \Vect{x} }{t_1 \Vect{s}_1 + \cdots + t_k \Vect{s}_k) }\)
\(\)\(=\)\(t_1(\DotPr{ \Vect{x} }{ \Vect{s}_1 }) + \cdots + t_k(\DotPr{ \Vect{x} }{ \Vect{s}_k })\)
\(\)\(=\)\(0\)

This proves part (ii) of the proposition.

iii.   \(\VSpc{V}\cap \OrthCmpl{\VSpc{V}} = \Vect{0}\).

Suppose \(\Vect{x}\) belongs to \(\VSpc{V}\) and to \(\OrthCmpl{\VSpc{V}}\), then \(\DotPr{ \Vect{x} }{ \Vect{x} } = 0\), and so \(\Vect{x} = \Vect{0}\).

iv.   \(S^{\bot} = S^{\bot\bot\bot}\)

We need to show that the two inclusions below hold

  1. \(\OrthCmpl{S}\subseteq S^{\bot\bot\bot}\)
  2. \(S^{\bot\bot\bot}\subseteq \OrthCmpl{S}\)

Consider \(\Vect{x}\in \OrthCmpl{S}\). We need to show that \(\DotPr{ \Vect{x} }{ \Vect{y} }=0\) for all \(\Vect{y}\in S^{\bot\bot}\). But \(\Vect{y}\) is in \(S^{\bot\bot}\) because \(\DotPr{ \Vect{x} }{ \Vect{y} }=0\) for each \(\Vect{x}\in \OrthCmpl{S}\), and this is exactly what we wanted to show.

The converse inclusion \(S^{\bot\bot\bot}\subseteq \OrthCmpl{S}\) holds because \(S\subseteq S^{\bot\bot}\).

This completes the proof of the proposition.