Question 3.30: (a) Find the reflection of R3 along the direction v3= (−1, 1...
(a) Find the reflection of R³ along the direction \overrightarrow{v_{3}} =\left(-1,1,-1\right) with respect to the plane \left(2,-2,3\right)+\ll \left(0,1,0\right),\left(0,-1,1\right)\gg.
(b) Show that
T\left(\overrightarrow{x}\right)=\overrightarrow{x_{0}}+\overrightarrow{x}A, where \overrightarrow{x_{0}}=\left(0,-2,-4\right) and A = \left[\begin{matrix} 1 & 0 & 0 \\ \\ 0 & \frac{5}{3} & \frac{4}{3} \\ \\ 0 & -\frac{4}{3} & -\frac{5}{3} \end{matrix} \right]
is a reflection. Determine its direction and plane of invariant points.
(a) In the affine basis B = \left\{\left(2,-2,3\right),\left(2,-1,3\right),\left(2,-3,4\right),\left(1,-1,2\right)\right\},the required T has the representation
\left[T\left(\overrightarrow{x}\right)\right]_{B}=\left[\overrightarrow{x}\right]_{B}\left[T\right]_{B}, where \left[T\right]_{B}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right].
While, in the natural affine basis N,
T\left(\overrightarrow{x}\right)=\left(2,-2,3\right)+\left(\overrightarrow{x}-\left(2,-2,3\right)\right)P^{-1}\left[T\right]_{B}P
where
P= P^{B}_{N}=\left[\begin{matrix} 0 & 1 & 0 \\ 0 & -1 & 1 \\ -1 & 1 & -1 \end{matrix} \right] \Rightarrow P^{-1}=\left[\begin{matrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{matrix} \right].
Therefore,
P^{-1}\left[T\right]_{B}P=\left[\begin{matrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right]\left[\begin{matrix} 0 & 1 & 0 \\ 0 & -1 & 1 \\ -1 & 1 & -1 \end{matrix} \right]
=\left[\begin{matrix} -1 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right], and
\left(2,-2,3\right)-\left(2,-2,3\right)P^{-1}\left[T\right]_{B}P=\left(2,-2,3\right)-\left(-2,2,-1\right)=\left(4,-4,4\right)
\Rightarrow T\left(\overrightarrow{x}\right)=\left(4,-4,4\right)+\overrightarrow{x}\left[\begin{matrix} -1 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] for \overrightarrow{x} \in R^{3}, or
y_{1}=4-x_{1} , y_{2}=-4+2x_{1}+x_{2} , y_{3}=4-2x_{1}+x_{3}.
(b) Since det A = −1, it is possible that T is a reflection. To make certainty, compute the characteristic polynomial det \left(A-tI_{3}\right)=-\left(t-1\right)^{2}\left(t+1\right).
So A has eigenvalues 1, 1 and −1. Moreover,
\left(A-I_{3}\right)\left(A+I_{3}\right)=\left[\begin{matrix} 0 & 0 & 0 \\ \\ 0 & \frac{2}{3} & \frac{4}{3} \\ \\ 0 & -\frac{4}{3} & -\frac{8}{3} \end{matrix} \right]\left[\begin{matrix} 2 & 0 & 0 \\ \\ 0 & \frac{8}{3} & \frac{4}{3} \\ \\ 0 & -\frac{4}{3} & -\frac{2}{3} \end{matrix} \right]=O_{3 \times 3}
indicates that A is diagonalizable and thus, the corresponding T is a reflection if \overrightarrow{x}\left(I_{3}-A\right) =\overrightarrow{x_{0}} has a solution. Now
\overrightarrow{x}\left(I_{3}-A\right) =\overrightarrow{x_{0}}
\Rightarrow x_{2}-2x_{3}-3=0 (the plane of invariant points).
So T is really a reflection.
Take eigenvectors \overrightarrow{v_{1}}=\left(2,0,0\right) and \overrightarrow{v_{2}}=\left(1,2,1\right) orresponding to 1 and \overrightarrow{v_{3}}=\left(0,1,2\right) corresponding to −1. Then
\frac{1}{2}\overrightarrow{x_{0}}+\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg =\left(0,-1,-2\right)+\left\{\left(2 \alpha_{1}+\alpha_{2},2\alpha_{2},\alpha_{2}\right)\mid \alpha_{1} , \alpha_{2} \in R\right\}
\Leftrightarrow \left\{\begin{matrix}x_{1}=2\alpha_{1}+\alpha_{2} \\x_{2}=-1+2\alpha_{2} \\ x_{3}=-2+\alpha_{2} , for \alpha_{1},\alpha_{2} \in R\end{matrix}\right.
\Leftrightarrow x_{2}-2x_{3}-3=0
indeed is the plane of invariant points. Also, \overrightarrow{v_{3}}=-\frac{1}{2}\overrightarrow{x_{0}} or just \overrightarrow{x_{0}} itself is the direction of T.
In the affine basis C = \left\{\overrightarrow{v_{3}},\overrightarrow{v_{3}}+\overrightarrow{v_{1}},\overrightarrow{v_{3}}+\overrightarrow{v_{2}},2\overrightarrow{v_{3}}\right\},
\left[T\left(\overrightarrow{x}\right)\right]_{C}=\left[\overrightarrow{x}\right]_{C}\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right].
See Fig. 3.58.
Remark The linear operator
T\left(\overrightarrow{x}\right)=\overrightarrow{x}A, where A =\left[\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right] (3.8.14)
is a standard orthogonal reflection of R³ in the direction \overrightarrow{v_{3} } =\left( \frac{1}{\sqrt{2} },-\frac{1}{\sqrt{2} },0 \right) with respect to the plane
\ll \overrightarrow{v_{1} },\overrightarrow{v_{2} } \gg=\left\{\left(x_{1},x_{2},x_{3} \right)\in R^{3}\mid x_{1}=x_{2}\right\}, where
\overrightarrow{v_{1} }=\left(\frac{1}{\sqrt{2} },\frac{1}{\sqrt{2} },0 \right) and \overrightarrow{v_{2} }=\overrightarrow{e_{3} }=\left(0,0,1\right).
See Fig. 3.59 and compare with Fig. 3.32 with a = b = c = 1. Notice that
A=P^{-1}\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right] P, where P = \left[\begin{matrix} \overrightarrow{v_{1} } \\ \overrightarrow{v_{2} } \\ \overrightarrow{v_{3} } \end{matrix} \right] =\left[\begin{matrix} \frac{1}{\sqrt{2} } & \frac{1}{\sqrt{2} } & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2} } & -\frac{1}{\sqrt{2} } & 0 \end{matrix} \right].
For details, refer to Case 7 below.
