Question 3.35: Give an affine transformation T(x^→ ) = x0^→ + x^→ A, where ...

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

det \left(A-tI_{3}\right)= \frac{1}{27}\left(-27t^{3}+27t^{2}+27t-27\right)=-\left(t-1\right)^{2}\left(t+1\right)

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

\overrightarrow{x} \left(A-I_{3}\right)=\overrightarrow{0}

\Rightarrow x_{1}-x_{2}-x_{3}=0.

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

\overrightarrow{x} \left(A+I_{3}\right)=\overrightarrow{0}

\Rightarrow 2x_{1}+x_{2}+x_{3}=0, x_{1}+2x_{2}-x_{3}=0.

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

\left[A\right]_{C}=QAQ^{-1}=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{matrix} \right] ,   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} \right] .

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

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

\overrightarrow{v_{1} }A=\overrightarrow{v_{1} },

\overrightarrow{v_{2} }A=\frac{1}{\sqrt{2}}\left(-\frac{1}{3},\frac{1}{3},\frac{4}{3}\right) = -\frac{1}{3}\overrightarrow{v_{2} }+\frac{4}{3\sqrt{2}}\overrightarrow{v_{3} },

\overrightarrow{v_{3} }A=\left(\frac{2}{3},-\frac{2}{3},\frac{1}{3}\right) =\frac{2\sqrt{2}}{3}\overrightarrow{v_{2} }+\frac{1}{3}\overrightarrow{v_{3} }

\Rightarrow \left[A\right]_{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} \right],  where 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} \right].

Also, the submatrix

\left[\begin{matrix}  -\frac{1}{3} & \frac{2\sqrt{2}}{3} \\ \\  \frac{2\sqrt{2}}{3} & \frac{1}{3} \end{matrix} \right]= \left[\begin{matrix} -1 & 0 \\  0 & 1 \end{matrix} \right]\left[\begin{matrix} \cos\theta & \sin\theta \\  -\sin\theta & \cos\theta \end{matrix} \right]

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

\overrightarrow{y}=\overrightarrow{x}P^{-1}\left(PAP^{-1}\right)P

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

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

\beta _{1}=\alpha _{1},

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

This means that, when the height \alpha _{1} of the point \overrightarrow{x} to the plane \ll \overrightarrow{v_{2} },\overrightarrow{v_{3} }\gg being fixed, the orthogonal projection \left(\alpha _{2},\alpha _{3}\right) of \overrightarrow{x} on \ll \overrightarrow{v_{2} },\overrightarrow{v_{3} }\gg is subject to a reflection with respect to the axis \ll \overrightarrow{v_{3} }\gg to the point \left(-\alpha _{2},\alpha _{3}\right) and then is rotated through the angle θ in the plane \ll \overrightarrow{v_{2} },\overrightarrow{v_{3} }\gg with \overrightarrow{0 } as center to the point \left(\beta _{2},\beta _{3}\right).Therefore, the point  \left(\beta _{1},\beta _{2},\beta _{3}\right), where \beta _{1}=\alpha _{1}, is the coordinate vector of \overrightarrow{x}A=T\left(\overrightarrow{x}\right) 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 T\left(\overrightarrow{x}\right)= \overrightarrow{x_{0} } +\overrightarrow{x}A 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 T\left(\overrightarrow{x}\right)= \overrightarrow{x_{0} } +\overrightarrow{x}A  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 N =\left\{\overrightarrow{0} ,\overrightarrow{e_{1} } ,\overrightarrow{e_{2} } ,\overrightarrow{e_{3} } \right\} ,

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

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

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

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

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

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

7. \overrightarrow{x} \rightarrow \overrightarrow{x_{0} }+\overrightarrow{x} : a translation along \overrightarrow{x_{0} }.

Readers shall practice more examples by themselves.