Question: Given \({\bf{u}} \ne {\bf{0}}\) in \({\mathbb{R}^n}\), let \(L = {\bf{Span}}\left\{ {\bf{u}} \right\}\). For y in \({\mathbb{R}^n}\), the reflection of y in L is the point \({\bf{ref}}{{\bf{l}}_L}y\) defined by

\({\bf{ref}}{{\bf{l}}_L}{\bf{y}} = {\bf{2}} \cdot {\bf{pro}}{{\bf{j}}_L}{\bf{y}} - {\bf{y}}\)

See the figure, which shows that \({\bf{ref}}{{\bf{l}}_L}{\bf{y}}\) is the sum of \({\bf{\hat y}} = {\bf{pro}}{{\bf{j}}_L}{\bf{y}}\) and \(\widehat {\bf{y}} - {\bf{y}}\). Show that the mapping \({\bf{y}} \mapsto {\bf{ref}}{{\bf{l}}_L}{\bf{y}}\) is a linear transformation.

The projection for x is given by;

\({\rm{pro}}{{\rm{j}}_L}{\bf{x}} = \frac{{{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\)

The transformation becomes:

\(\begin{array}{c}T\left( {\bf{x}} \right) = {\rm{pro}}{{\rm{j}}_L}{\bf{x}}\\ = \frac{{{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\end{array}\)

Apply an inner product for the transformation \(T\left( {\bf{x}} \right)\) by using the transformation \(T\left( {\bf{x}} \right) = \frac{{{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\).

\(\begin{array}{c}T\left( {a{\bf{x}} + b{\bf{y}}} \right) = \frac{{\left( {a{\bf{x}} + b{\bf{y}}} \right) \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = \frac{{b{\bf{x}} \cdot {\bf{u}} + b{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = \frac{{a{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}} + \frac{{b{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = aT\left( {\bf{x}} \right) + bT\left( {\bf{y}} \right)\end{array}\)

Therefore, \(T\left( {\bf{x}} \right)\) is a linear transformation.

Let, \({T_1}\left( {\bf{y}} \right) = {\rm{ref}}{{\rm{l}}_L}{\bf{y}} = 2 \cdot {\rm{pro}}{{\rm{j}}_L}{\bf{y}} - {\bf{y}}\).

Check inner product for the transformation \({T_1}\) as shown below:

\(\begin{array}{c}{T_1}\left( {c{\bf{y}} + d{\bf{z}}} \right) = {\rm{re}}{{\rm{f}}_L}\left( {c{\bf{y}} + d{\bf{z}}} \right)\\ = 2 \cdot {\rm{pro}}{{\rm{j}}_L}\left( {c{\bf{y}} + d{\bf{z}}} \right) - \left( {c{\bf{y}} + d{\bf{z}}} \right)\\ = 2 \cdot \left( {\left( c \right){\rm{pro}}{{\rm{j}}_L}{\bf{y}} + \left( d \right){\rm{pro}}{{\rm{j}}_L}{\bf{z}}} \right) - c{\bf{y}} - d{\bf{z}}\\ = 2c \cdot \left( {{\rm{pro}}{{\rm{j}}_L}{\bf{y}}} \right) + 2d \cdot \left( {{\rm{pro}}{{\rm{j}}_L}{\bf{z}}} \right) - c{\bf{y}} - d{\bf{z}}\\ = c\left( {2 \cdot {\rm{pro}}{{\rm{j}}_L}{\bf{y}} - {\bf{y}}} \right) + d\left( {2 \cdot {\rm{pro}}{{\rm{j}}_L}{\bf{z}} - {\bf{z}}} \right)\\ = c\left( {{\rm{ref}}{{\rm{l}}_L}{\bf{y}}} \right) + d\left( {{\rm{ref}}{{\rm{l}}_L}{\bf{z}} - {\bf{z}}} \right)\\ = c{T_1}\left( {\bf{y}} \right) + d{T_1}\left( {\bf{z}} \right)\end{array}\)

The above equation shows that \({T_1}\) is a linear transformation.

Thus, it is proved that \({\bf{y}} \mapsto {\rm{ref}}{{\rm{l}}_L}{\bf{y}}\) is a linear transformation.