The shear transformation of \(\RNrSpc{n}\) parallel to the hyperspace perpendicular to a nonzero vector \(\Vect{a}\) is a linear transformation.
\[\FnctnDAAT{S}{\RNrSpc{n}}{\RNrSpc{n}},\quad \FnctnOf{S}{\Vect{x}} = \Vect{x} + (\DotPr{ \Vect{a} }{ \Vect{x} })\cdot \Vect{s}\]First we show that \(S\) commutes with vector addition; that is:
\[S(\Vect{x}+\Vect{y}) = S(\Vect{x}) + S(\Vect{y})\]So we compute:
| \(S(\Vect{x}+\Vect{y})\) | \(=\) | \(\Vect{x} + \Vect{y}\ +\ (\DotPr{\Vect{a}}{(\Vect{x}+\Vect{y})} ) \cdot \Vect{s}\) |
| \(\) | \(= \) | \(\Vect{x} + \Vect{y}\ +\ (\DotPr{\Vect{a}}{\Vect{x}}+ \DotPr{\Vect{a}}{\Vect{y}} ) \cdot \Vect{a}\) |
| \(\) | \(= \) | \(\left( \Vect{x} \ +\ (\DotPr{\Vect{a}}{\Vect{x}})\cdot \Vect{a}\right) \ +\ \left( \Vect{y}\ +\ (\DotPr{\Vect{a}}{\Vect{y}}) \cdot \Vect{a}\right)\) |
| \(\) | \(=\) | \(S(\Vect{x})\ +\ S(\Vect{y})\) |
Next we show that \(S\) commutes with scalar multiplication; that is
\[S(t\cdot \Vect{x}) = t\cdot S(\Vect{x})\]So we compute
| \(S(t\cdot \Vect{x})\) | \(=\) | \(t\cdot \Vect{x} + (\DotPr{\Vect{a}}{(t\cdot \Vect{x})})\cdot \Vect{a}\) |
| \(\) | \(= \) | \(t\cdot \Vect{x} + t\cdot \left( (\DotPr{\Vect{a}}{\Vect{x}})\cdot \Vect{a} \right)\) |
| \(\) | \(=\) | \(t\cdot \left( \Vect{x} + (\DotPr{\Vect{a}}{\Vect{x}})\cdot \Vect{a} \right)\) |
| \(\) | \(=\) | \(t\cdot S(\Vect{x})\) |
This proves that a shear transformation is linear.