The argument is entirely parallel to the proof that orthogonal projections are linear transformations:

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

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

So we compute:

\(M(\Vect{x}+\Vect{y})\)	\(=\)	\(\Vect{x} + \Vect{y}\ -\ 2\cdot \dfrac{ \DotPr{\Vect{a}}{(\Vect{x}+\Vect{y})} }{ \DotPr{\Vect{a}}{\Vect{a}} } \cdot \Vect{a}\)
\(\)	\(= \)	\(\Vect{x} + \Vect{y}\ -\ 2\cdot\dfrac{ \DotPr{\Vect{a}}{\Vect{x}}+ \DotPr{\Vect{a}}{\Vect{y}} }{ \DotPr{\Vect{a}}{\Vect{a}} } \cdot \Vect{a}\)
\(\)	\(= \)	\(\left( \Vect{x} \ -\ 2\cdot\dfrac{ \DotPr{\Vect{a}}{\Vect{x}} }{ \DotPr{\Vect{a}}{\Vect{a}} } \cdot \Vect{a}\right) \ +\ \left( \Vect{y}\ -\ 2\cdot \dfrac{ \DotPr{\Vect{a}}{\Vect{y}} }{ \DotPr{\Vect{a}}{\Vect{a}} } \cdot \Vect{a}\right)\)
\(\)	\(=\)	\(M(\Vect{x})\ +\ M(\Vect{y})\)

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

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

So we compute

\(M(t\cdot \Vect{x})\)	\(=\)	\(t\cdot \Vect{x} - 2\cdot \dfrac{ \DotPr{\Vect{a}}{(t\cdot \Vect{x})} }{ \DotPr{ \Vect{a} }{ \Vect{a} } }\cdot \Vect{a}\)
\(\)	\(= \)	\(t\cdot \Vect{x} - 2\cdot \dfrac{ t\cdot (\DotPr{\Vect{a}}{\Vect{x}}) }{ \DotPr{ \Vect{a} }{ \Vect{a} } }\cdot \Vect{a}\)
\(\)	\(=\)	\(t\cdot M(\Vect{x})\)

This proves that an orthogonal projection is a linear transformation.