Question

Question: Given \({\bf{u}} \ne {\bf{0}}\) in \({\mathbb{R}^n}\), let \(L = {\bf{Span}}\left\{ {\bf{u}} \right\}\). Show that the mapping \({\bf{x}} \mapsto {\bf{pro}}{{\bf{j}}_L}{\bf{x}}\) is a linear transformation.

Short answer

The mapping \({\bf{x}} \mapsto {\rm{pro}}{{\rm{j}}_L}{\bf{x}}\) is a linear transformation.

Step-by-step solution

Write orthogonal projection of y onto line L

The orthogonal projection of y onto the line L can be expressed as:

\(\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\)

The transformation can be represented as shown below:

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

Use the inner product for the transformation

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

\(\begin{array}{c}T\left( {c{\bf{x}} + d{\bf{y}}} \right) = \frac{{\left( {c{\bf{x}} + d{\bf{y}}} \right) \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = \frac{{c{\bf{x}} \cdot {\bf{u}} + d{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = \frac{{c{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}} + \frac{{d{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = cT\left( {\bf{x}} \right) + dT\left( {\bf{y}} \right)\end{array}\)

Therefore, the mapping \({\bf{x}} \mapsto {\rm{pro}}{{\rm{j}}_L}{\bf{x}}\) is a linear transformation.