Given subspaces \(\VSpc{V}\subseteq \VSpc{W}\) of \(\RNrSpc{n}\), every \(\Vect{x}\) in \(\VSpc{W}\) has a unique sum expression
| \(\Vect{x}\) | \(=\) | \(\Vect{x}_V + \Vect{x}_{\bot}\) |
where \(\Vect{x}_V=\OrthoPrjctn{V}{\Vect{x}}\) is in \(\VSpc{V}\) and \(\Vect{x}_{\bot}\DefEq \Vect{x}- \OrthoPrjctn{V}{\Vect{x}}\) is in \(\OrthCmpl{V}\).
If \(\VSpc{V}\) consists only of the zero vector, we have no choice but to set
\[\Vect{x}_{V}\DefEq \Vect{0} \quad\text{and}\quad \Vect{x}_{\bot}\DefEq \Vect{x}\]These vectors are the only ones satisfying the requirements of the proposition.
Turning to the case where \(V\) has nonzero vectors, we know that
| \(\Vect{x}_V\) | \(=\) | \((\DotPr{ \Vect{x} }{ \Vect{b}_1 }) \Vect{b}_1 + \cdots + (\DotPr{ \Vect{x} }{ \Vect{b}_r })\Vect{b}_r\) |
belongs to \(V\), as it is a linear combination of the vectors \(\Vect{b}_1\), ... , \(\Vect{b}_r\) forming an ONB \(\EuScript{B}\) of \(V\). So we show next that \(\Vect{x}_{\bot}\) is in \(V^{\bot}\). It suffices to show that \(\Vect{x}_{\bot}\) is in \(\EuScript{B}^{\bot}\). So we compute
| \(\DotPr{ \Vect{x}_{\bot} }{ \Vect{b}_j }\) | \(=\) | \(\DotPr{ \left[ \Vect{x} - (\DotPr{ \Vect{x} }{ \Vect{b}_1 }) \Vect{b}_1 - \cdots - (\DotPr{ \Vect{x} }{ \Vect{b}_r })\Vect{b}_r\right] }{ \Vect{b}_j }\) |
| \(\) | \(=\) | \(\DotPr{ \Vect{x} }{ \Vect{b}_j } - (\DotPr{ \Vect{x} }{\Vect{b}_1})(\DotPr{ \Vect{b}_1 }{ \Vect{b}_j }) - \cdots - (\DotPr{ \Vect{x} }{\Vect{b}_j})(\DotPr{ \Vect{b}_j }{ \Vect{b}_j }) - \cdots - (\DotPr{ \Vect{x} }{\Vect{b}_r})(\DotPr{ \Vect{b}_r }{ \Vect{b}_j })\) |
| \(\) | \(=\) | \(\DotPr{ \Vect{x} }{ \Vect{b}_j }\ -\ \DotPr{ \Vect{x} }{ \Vect{b}_j }\) |
| \(\) | \(=\) | \(0\) |
as was to be shown. So the decomposition of \(\Vect{x}\) as the sum \(\Vect{x}_V + \Vect{x}_{\bot}\) has the stated properties. It remains to show that it is the only decomposition with these properties. So consider the situation
| \(\Vect{x}\) | \(=\) | \(\Vect{y} + \Vect{z}\) |
with \(\Vect{y}\in V\) and \(\Vect{z}\in V^{\bot}\). It follows that
| \(\Vect{x}_V - \Vect{y}\) | \(=\) | \(\Vect{z} - \Vect{x}_{\bot}\) |
The vector \(\Vect{x}_V - \Vect{y}\) belongs to \(V\), while \(\Vect{z} - \Vect{x}_{\bot}\) belongs to \(V^{\bot}\). Now the only vector common to \(V\) and \(V^{\bot}\) is the zero vector. Therefore
\[\Vect{x}_V - \Vect{y} = \Vect{0} \quad\text{and}\quad \Vect{z} - \Vect{x}_{\bot} = \Vect{0}\]This gives \(\Vect{x}_V=\Vect{y}\) and \(\Vect{z}=\Vect{x}_{\bot}\), implying that the decomposition of \(\Vect{x}\) as a sum of a vector in \(V\) and another in \(V^{\bot}\) is unique.