(a) Find the reflection of R3 along the direction v3= (−1, 1,−1) with respect to the plane (2,−2, 3) + . (b) Show that T(x ^→) = x0^→ + xA^→, where x0^→ = (0,−2,−4) and A = [1 0 0 0 5/3 4/3 0 -4/3 -5/3] is a reflection. Determine its direction and plane of invariant points

Question

(a) Find the reflection of R³ along the direction [latex]\overrightarrow{v_{3}} =\left(-1,1,-1 ight)[/latex] with respect to the plane [latex]\left(2,-2,3 ight)+\ll \left(0,1,0 ight),\left(0,-1,1 ight)\gg.[/latex]

(b) Show that

[latex]T\left(\overrightarrow{x} ight)=\overrightarrow{x_{0}}+\overrightarrow{x}A, where \overrightarrow{x_{0}}=\left(0,-2,-4 ight) and A = \left[\begin{matrix} 1 & 0 & 0 \ \ 0 & \frac{5}{3} & \frac{4}{3} \ \ 0 & -\frac{4}{3} & -\frac{5}{3} \end{matrix} ight][/latex]

is a reflection. Determine its direction and plane of invariant points.

Accepted Answer

(a) In the affine basis [latex]B = \left\{\left(2,-2,3 ight),\left(2,-1,3 ight),\left(2,-3,4 ight),\left(1,-1,2 ight) ight\},[/latex]the required T has the representation

[latex]\left[T\left(\overrightarrow{x} ight) ight]_{B}=\left[\overrightarrow{x} ight]_{B}\left[T ight]_{B}, where \left[T ight]_{B}=\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{matrix} ight].[/latex]

While, in the natural affine basis N,

[latex]T\left(\overrightarrow{x} ight)=\left(2,-2,3 ight)+\left(\overrightarrow{x}-\left(2,-2,3 ight) ight)P^{-1}\left[T ight]_{B}P[/latex]

where

[latex]P= P^{B}_{N}=\left[\begin{matrix} 0 & 1 & 0 \ 0 & -1 & 1 \ -1 & 1 & -1 \end{matrix} ight] \Rightarrow P^{-1}=\left[\begin{matrix} 0 & -1 & -1 \ 1 & 0 & 0 \ 1 & 1 & 0 \end{matrix} ight].[/latex]

Therefore,

[latex]P^{-1}\left[T ight]_{B}P=\left[\begin{matrix} 0 & -1 & -1 \ 1 & 0 & 0 \ 1 & 1 & 0 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{matrix} ight]\left[\begin{matrix} 0 & 1 & 0 \ 0 & -1 & 1 \ -1 & 1 & -1 \end{matrix} ight][/latex]

[latex]=\left[\begin{matrix} -1 & 2 & -2 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight],[/latex] and

[latex]\left(2,-2,3 ight)-\left(2,-2,3 ight)P^{-1}\left[T ight]_{B}P=\left(2,-2,3 ight)-\left(-2,2,-1 ight)=\left(4,-4,4 ight)[/latex]

[latex]\Rightarrow T\left(\overrightarrow{x} ight)=\left(4,-4,4 ight)+\overrightarrow{x}\left[\begin{matrix} -1 & 2 & -2 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight] for \overrightarrow{x} \in R^{3},[/latex] or

[latex]y_{1}=4-x_{1} , y_{2}=-4+2x_{1}+x_{2} , y_{3}=4-2x_{1}+x_{3}.[/latex]

(b) Since det A = −1, it is possible that T is a reflection. To make certainty, compute the characteristic polynomial [latex]det \left(A-tI_{3} ight)=-\left(t-1 ight)^{2}\left(t+1 ight).[/latex]

So A has eigenvalues 1, 1 and −1. Moreover,

[latex]\left(A-I_{3} ight)\left(A+I_{3} ight)=\left[\begin{matrix} 0 & 0 & 0 \ \ 0 & \frac{2}{3} & \frac{4}{3} \ \ 0 & -\frac{4}{3} & -\frac{8}{3} \end{matrix} ight]\left[\begin{matrix} 2 & 0 & 0 \ \ 0 & \frac{8}{3} & \frac{4}{3} \ \ 0 & -\frac{4}{3} & -\frac{2}{3} \end{matrix} ight]=O_{3 imes 3}[/latex]

indicates that A is diagonalizable and thus, the corresponding T is a reflection if [latex]\overrightarrow{x}\left(I_{3}-A ight) =\overrightarrow{x_{0}}[/latex] has a solution. Now

[latex]\overrightarrow{x}\left(I_{3}-A ight) =\overrightarrow{x_{0}}[/latex]

[latex]\Rightarrow x_{2}-2x_{3}-3=0[/latex] (the plane of invariant points).

So T is really a reflection.

Take eigenvectors [latex]\overrightarrow{v_{1}}=\left(2,0,0 ight) and \overrightarrow{v_{2}}=\left(1,2,1 ight)[/latex] orresponding to 1 and [latex]\overrightarrow{v_{3}}=\left(0,1,2 ight)[/latex] corresponding to −1. Then

[latex]\frac{1}{2}\overrightarrow{x_{0}}+\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg =\left(0,-1,-2 ight)+\left\{\left(2 \alpha_{1}+\alpha_{2},2\alpha_{2},\alpha_{2} ight)\mid \alpha_{1} , \alpha_{2} \in R ight\}[/latex]

[latex]\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} ight.[/latex]

[latex]\Leftrightarrow x_{2}-2x_{3}-3=0[/latex]

indeed is the plane of invariant points. Also, [latex]\overrightarrow{v_{3}}=-\frac{1}{2}\overrightarrow{x_{0}}[/latex] or just [latex]\overrightarrow{x_{0}}[/latex] itself is the direction of T.

In the affine basis [latex]C = \left\{\overrightarrow{v_{3}},\overrightarrow{v_{3}}+\overrightarrow{v_{1}},\overrightarrow{v_{3}}+\overrightarrow{v_{2}},2\overrightarrow{v_{3}} ight\},[/latex]

[latex]\left[T\left(\overrightarrow{x} ight) ight]_{C}=\left[\overrightarrow{x} ight]_{C}\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{matrix} ight].[/latex]

See Fig. 3.58.

Remark The linear operator

[latex]T\left(\overrightarrow{x} ight)=\overrightarrow{x}A, where A =\left[\begin{matrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{matrix} ight] [/latex] (3.8.14)

is a standard orthogonal reflection of R³ in the direction [latex]\overrightarrow{v_{3} } =\left( \frac{1}{\sqrt{2} },-\frac{1}{\sqrt{2} },0 ight)[/latex] with respect to the plane

[latex]\ll \overrightarrow{v_{1} },\overrightarrow{v_{2} } \gg=\left\{\left(x_{1},x_{2},x_{3} ight)\in R^{3}\mid x_{1}=x_{2} ight\},[/latex] where

[latex]\overrightarrow{v_{1} }=\left(\frac{1}{\sqrt{2} },\frac{1}{\sqrt{2} },0 ight)[/latex] and [latex]\overrightarrow{v_{2} }=\overrightarrow{e_{3} }=\left(0,0,1 ight).[/latex]

See Fig. 3.59 and compare with Fig. 3.32 with a = b = c = 1. Notice that

[latex]A=P^{-1}\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{matrix} ight] P,[/latex] where [latex]P = \left[\begin{matrix} \overrightarrow{v_{1} } \ \overrightarrow{v_{2} } \ \overrightarrow{v_{3} } \end{matrix} ight] =\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} ight].[/latex]

For details, refer to Case 7 below.

Question 3.30: (a) Find the reflection of R3 along the direction v3= (−1, 1...

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