Give an affine transformation T(x^→ ) = x0^→ + x^→ A, where A =[1/3 2/3 2/3 2/3 1/3 -2/3 2/3 -2/3 1/3]. Determine these x0^→ so that each such T is an orthogonal reflection, and the direction and the plane of invariant points.

Question

Give an affine transformation

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

Determine these [latex]\overrightarrow{x_{0} }[/latex] so that each such T is an orthogonal reflection, and the direction and the plane of invariant points.

Accepted Answer

It is obvious that A is orthogonal and det A = −1. If we pick [latex]\overrightarrow{x_{0} } \in Im\left(A-I_{3} ight) ,[/latex]then the associated T is an orthogonal reflection.
The characteristic polynomial of A is

[latex]det \left(A-tI_{3} ight)= \frac{1}{27}\left(-27t^{3}+27t^{2}+27t-27 ight)=-\left(t-1 ight)^{2}\left(t+1 ight)[/latex]

and hence A has eigenvalue 1 of multiplicity 2 and another eigenvalue −1.
Solve

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

[latex]\Rightarrow x_{1}-x_{2}-x_{3}=0.[/latex]

Take unit eigenvectors [latex]\overrightarrow{u_{1} }=\frac{1}{\sqrt{2}}\left(1,1,0 ight)[/latex] and [latex]\overrightarrow{u_{2} }=\frac{1}{\sqrt{6}}\left(1,-1,2 ight)[/latex] so that [latex]\overrightarrow{u_{1} }\bot\overrightarrow{u_{2} }.[/latex] Solve

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

[latex]\Rightarrow 2x_{1}+x_{2}+x_{3}=0, x_{1}+2x_{2}-x_{3}=0.[/latex]

Take a unit vector [latex]\overrightarrow{u_{3} }=\frac{1}{\sqrt{3}}\left(1,-1,-1 ight).[/latex] Then, [latex]C = \left\{\overrightarrow{u_{1} },\overrightarrow{u_{2} },\overrightarrow{u_{3} } ight\}[/latex] is an orthonormal basis for R³. In C,

[latex]\left[A ight]_{C}=QAQ^{-1}=\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1\end{matrix} ight] , where Q =\left[\begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ \ \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} \ \ \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \end{matrix} ight] .[/latex]

Note that [latex]Im \left(A-I_{3} ight)=Ker \left(A+I_{3} ight)=Ker \left(A-I_{3} ight)^{ \bot}.[/latex] So there are three different ways to compute [latex]Im \left(A-I_{3} ight)= \ll \left(1,-1,-1 ight)\gg.[/latex] Pick up any point [latex]\overrightarrow{x_{0} }[/latex] on [latex]Im \left(A-I_{3} ight),[/latex] say [latex]\overrightarrow{x_{0} }=\overrightarrow{0 }[/latex] for simplicity, then the direction of the reflection [latex]T =\overrightarrow{x}A[/latex] is [latex]\overrightarrow{u_{3} }[/latex] and the plane of invariant points is [latex] \ll \overrightarrow{u_{1} },\overrightarrow{u_{2} }\gg=\left\{\overrightarrow{x} \in R^{3} \mid x_{1}-x_{2}-x_{3}=0 ight\}= \ll \overrightarrow{u_{3} } \gg^{ \bot}.[/latex] See Fig. 3.72(a).

On the other hand, take [latex]\overrightarrow{v_{1} }= \overrightarrow{u_{1} }= \frac{1}{\sqrt{2}}\left(1,1,0 ight).[/latex] Then [latex]\ll \overrightarrow{v_{1} } \gg^{ \bot}=\left\{\overrightarrow{x} \mid x_{1}+x_{2}=0 ight\}.[/latex] Choose [latex]\overrightarrow{v_{2} }= \frac{1}{\sqrt{2}}\left(1,-1,0 ight), \overrightarrow{v_{3} }= \overrightarrow{e_{3} }=\left(0,0,1 ight)[/latex] so that [latex]B = \left\{ \overrightarrow{v_{1} }, \overrightarrow{v_{2} }, \overrightarrow{v_{3} } ight\}[/latex] forms an orthonormal basis for R³. Since

[latex]\overrightarrow{v_{1} }A=\overrightarrow{v_{1} },[/latex]

[latex]\overrightarrow{v_{2} }A=\frac{1}{\sqrt{2}}\left(-\frac{1}{3},\frac{1}{3},\frac{4}{3} ight) = -\frac{1}{3}\overrightarrow{v_{2} }+\frac{4}{3\sqrt{2}}\overrightarrow{v_{3} },[/latex]

[latex]\overrightarrow{v_{3} }A=\left(\frac{2}{3},-\frac{2}{3},\frac{1}{3} ight) =\frac{2\sqrt{2}}{3}\overrightarrow{v_{2} }+\frac{1}{3}\overrightarrow{v_{3} }[/latex]

[latex]\Rightarrow \left[A ight]_{B}=PAP^{-1}=\left[\begin{matrix} & 0 & 0 \ \ 0 & -\frac{1}{3} & \frac{2\sqrt{2}}{3} \ \ 0 & \frac{2\sqrt{2}}{3} & \frac{1}{3} \end{matrix} ight],[/latex] where [latex]P = \left[\begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \ \ 0 & 0 & 1 \end{matrix} ight].[/latex]

Also, the submatrix

[latex]\left[\begin{matrix} -\frac{1}{3} & \frac{2\sqrt{2}}{3} \ \ \frac{2\sqrt{2}}{3} & \frac{1}{3} \end{matrix} ight]= \left[\begin{matrix} -1 & 0 \ 0 & 1 \end{matrix} ight]\left[\begin{matrix} \cos heta & \sin heta \ -\sin heta & \cos heta \end{matrix} ight][/latex]

where [latex]\cos heta =\frac{1}{3}, \sin heta=-\frac{2\sqrt{2}}{3} .[/latex] To interpret [latex]T\left(\overrightarrow{x} ight)= \overrightarrow{x}A[/latex] in B, take any point [latex]\overrightarrow{x} \in R^{3}.[/latex] Then, letting [latex]\overrightarrow{y}=\overrightarrow{x}A, [/latex]

[latex]\overrightarrow{y}=\overrightarrow{x}P^{-1}\left(PAP^{-1} ight)P[/latex]

[latex]\Rightarrow \left[\overrightarrow{y} ight] _{B}=\left[\overrightarrow{x} ight] _{B}\left(PAP^{-1} ight),[/latex] i.e. letting [latex]\left[\overrightarrow{x} ight] _{B}=\left(\alpha _{1},\alpha _{2},\alpha _{3} ight)[/latex] and

[latex]\left[\overrightarrow{y} ight] _{B}= \left(\beta _{1},\beta _{2},\beta _{3} ight),[/latex]

[latex]\beta _{1}=\alpha _{1},[/latex]

[latex]\left(\beta _{2},\beta _{3} ight)=\left(\alpha _{2},\alpha _{3} ight)\left[\begin{matrix} -1 & 0 \ 0 & 1 \end{matrix} ight]\left[\begin{matrix} \cos heta & \sin heta \ -\sin heta & \cos heta \end{matrix} ight].[/latex]

This means that, when the height [latex]\alpha _{1}[/latex] of the point [latex]\overrightarrow{x}[/latex] to the plane [latex]\ll \overrightarrow{v_{2} },\overrightarrow{v_{3} }\gg[/latex] being fixed, the orthogonal projection [latex]\left(\alpha _{2},\alpha _{3} ight)[/latex] of [latex]\overrightarrow{x}[/latex] on [latex]\ll \overrightarrow{v_{2} },\overrightarrow{v_{3} }\gg[/latex] is subject to a reflection with respect to the axis [latex]\ll \overrightarrow{v_{3} }\gg[/latex] to the point [latex]\left(-\alpha _{2},\alpha _{3} ight)[/latex] and then is rotated through the angle θ in the plane [latex]\ll \overrightarrow{v_{2} },\overrightarrow{v_{3} }\gg[/latex] with [latex]\overrightarrow{0 }[/latex] as center to the point [latex]\left(\beta _{2},\beta _{3} ight).[/latex]Therefore, the point [latex]\left(\beta _{1},\beta _{2},\beta _{3} ight),[/latex] where [latex]\beta _{1}=\alpha _{1},[/latex] is the coordinate vector of [latex]\overrightarrow{x}A=T\left(\overrightarrow{x} ight)[/latex] in B. See Fig. 3.72(b). Compare with Figs. 2.109 and 2.110.

Just like (2.8.38) and (2.8.39), an affine transformation [latex]T\left(\overrightarrow{x} ight)= \overrightarrow{x_{0} } +\overrightarrow{x}A[/latex] on R³ can be expressed as a composite of a finite number of reflections, stretches and shearings and then followed by a translation.

For example, let [latex]T\left(\overrightarrow{x} ight)= \overrightarrow{x_{0} } +\overrightarrow{x}A[/latex] where A is the one in Example 3.6 of Sec. 3.7.5. Then, A is the composite of the following affine transformations in [latex]N =\left\{\overrightarrow{0} ,\overrightarrow{e_{1} } ,\overrightarrow{e_{2} } ,\overrightarrow{e_{3} } ight\} ,[/latex]

1.[latex]\left[\begin{matrix} 1 & 0 & 0 \ 4 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight][/latex] : a shearing in the direction [latex]\overrightarrow{e_{1} }[/latex] with [latex]\ll \overrightarrow{e_{1} },\overrightarrow{e_{3} }\gg[/latex] as the plane of invariant points and coefficient 4.

2.[latex]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{matrix} ight][/latex] : a shearing in the direction [latex]\overrightarrow{e_{1} }[/latex] with [latex]\ll \overrightarrow{e_{1} },\overrightarrow{e_{2} }\gg[/latex] as the plane of invariant points and coefficient 1.

3. [latex]\left[\begin{matrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{matrix} ight]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 5 \end{matrix} ight]=\left[\begin{matrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 5 \end{matrix} ight]:[/latex]a two-way stretch having scale factor −1 along [latex]\ll \overrightarrow{e_{2} }\gg[/latex] and 5 along [latex]\ll \overrightarrow{e_{3} }\gg[/latex] and [latex]\ll \overrightarrow{e_{1} }\gg[/latex] the line of invariant points. Note that the former factor represents an orthogonal reflection in the direction [latex]\overrightarrow{e_{2} }.[/latex]

4. [latex]\left[\begin{matrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight][/latex] : a shearing in the direction [latex]\overrightarrow{e_{2} }[/latex] with [latex]\ll \overrightarrow{e_{2} },\overrightarrow{e_{3} }\gg[/latex] as the plane of invariant points and coefficient 1.

5. [latex]\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 5 \ 0 & 0 & 1 \end{matrix} ight][/latex] : a shearing in the direction [latex]\overrightarrow{e_{3} }[/latex] with [latex]\ll \overrightarrow{e_{1} },\overrightarrow{e_{3} }\gg[/latex] as the plane of invariant points and coefficient 5.

6.[latex]\left[\begin{matrix} 1 & 0 & -5 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight][/latex] : a shearing in the direction [latex]\overrightarrow{e_{3} }[/latex] with [latex]\ll \overrightarrow{e_{2} },\overrightarrow{e_{3} }\gg[/latex] as the plane of invariant points and coefficient -5.

7. [latex]\overrightarrow{x} ightarrow \overrightarrow{x_{0} }+\overrightarrow{x}[/latex] : a translation along [latex]\overrightarrow{x_{0} }.[/latex]

Readers shall practice more examples by themselves.

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