First we need to show that \(P\) commutes with vector addition; that is:

\[P(\Vect{x}+\Vect{y}) = P(\Vect{x}) + P(\Vect{y})\]

So we compute:

\(P(\Vect{x}+\Vect{y})\)	\(=\)	\(\Vect{x} + \Vect{y}\ -\ \dfrac{ \DotPr{\Vect{a}}{(\Vect{x}+\Vect{y})} }{ \DotPr{\Vect{a}}{\Vect{c}} } \cdot \Vect{a}\)
\(\)	\(= \)	\(\Vect{x} + \Vect{y}\ -\ \dfrac{ \DotPr{\Vect{a}}{\Vect{x}}+ \DotPr{\Vect{a}}{\Vect{y}} }{ \DotPr{\Vect{a}}{\Vect{a}} } \cdot \Vect{a}\)
\(\)	\(= \)	\(\left( \Vect{x} \ -\ \dfrac{ \DotPr{\Vect{a}}{\Vect{x}} }{ \DotPr{\Vect{a}}{\Vect{a}} } \cdot \Vect{a}\right) \ +\ \left( \Vect{y}\ -\ \dfrac{ \DotPr{\Vect{a}}{\Vect{y}} }{ \DotPr{\Vect{a}}{\Vect{a}} } \cdot \Vect{a}\right)\)
\(\)	\(=\)	\(P(\Vect{x})\ +\ P(\Vect{y})\)

Next we need to show that \(P\) commutes with scalar multiplication; that is

\[P(t\cdot \Vect{x}) = t\cdot P(\Vect{x})\]

So we compute

\(P(t\cdot \Vect{x})\)	\(=\)	\(t\cdot \Vect{x} - \dfrac{ \DotPr{\Vect{a}}{(t\cdot \Vect{x})} }{ \DotPr{ \Vect{a} }{ \Vect{a} } }\cdot \Vect{a}\)
\(\)	\(= \)	\(t\cdot \Vect{x} - \dfrac{ t\cdot (\DotPr{\Vect{a}}{\Vect{x}}) }{ \DotPr{ \Vect{a} }{ \Vect{a} } }\cdot \Vect{a}\)
\(\)	\(=\)	\(t\cdot P(\Vect{x})\)

This proves that an orthogonal projection is a linear transformation.