Show that mapping {y}mapsto{r}{e}{f}{l}_{{L}}{y} is a linear transformation.

Question
Transformation properties
asked 2021-01-19
Show that mapping \({y}\mapsto{r}{e}{f}{l}_{{L}}{y}\) is a linear transformation.

Answers (1)

2021-01-20
Given information:
\({u}\ne{0}\in{R}^{n}{\quad\text{and}\quad}{L}={S}{p}{a}{n}{\left\lbrace{u}\right\rbrace}.\)
For y in \(R^{n},\ \text{the reflection of y in L is the point}\ refl_{L} y\ \text{defined by}\ refl_{L} y = 2\ \cdot\ proj_{L} y\ -\ y.\)
\(refl_{L} y\ \text{is the sum of haty}\ = proj_{L} y\) and hath - y.
Calculation:
Consider \(T (y) = refl_{L} y.\)
Substitute \(2\ \cdot\ proj_{L} y\ -\ y\ \text{for}\ refl_{L} y.\)
\({T}{\left({y}\right)}={2}\cdot{p}{r}{o}{j}_{{L}}{y}-{y}{\left({1}\right)}\)
Apply Theorem 1 (b) as shown below.
Let u, v, and w be vectors in \(R^{n}\), and let c be a scalar. Then
\({\left({u}+{v}\right)}\cdot{w}={u}\cdot{w}+{v}\cdot{w}\)
For any vectors y and z in \(R^{n}\), and any scalars c and d, the properties of inner product as shown below.
Substitute \(cy\ +\ dz\) for y in Equation (1).
\({T}{\left({c}{y}+{\left.{d}{z}\right.}\right)}={2}\cdot{p}{r}{o}{j}_{{L}}{\left({c}{y}+{\left.{d}{z}\right.}\right)}-{\left({c}{y}+{\left.{d}{z}\right.}\right)}\)
\(={2}{\left({c}{p}{r}{o}{j}_{{L}}{y}+{d}{p}{r}{o}{j}_{{L}}{z}\right)}-{\left({c}{y}+{\left.{d}{z}\right.}\right)}\)
\(={2}{c}{p}{r}{o}{j}_{{L}}{y}-{c}{y}+{2}{d}{p}{r}{o}{j}_{{L}}{z}-{\left.{d}{z}\right.}\)
\(={c}{\left({2}{p}{r}{o}{j}_{{L}}{y}-{y}\right)}+{d}{\left({2}{p}{r}{o}{j}_{{L}}{z}-{z}\right)}\)
\(= cT (y)\ +\ dT (z)\)
Therefore, the mapping {y}\mapsto{r}{e}{f}{l}_{{L}} y is a linear transformation.
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