Question 3.34: Give an affine transformation T(x^→ ) = x0^→ + x^→A, where A...
Give an affine transformation
T\left(\overrightarrow{x} \right)=\overrightarrow{x_{0}}+\overrightarrow{x}A, where A = \left[\begin{matrix} -\frac{1}{3} & \frac{2}{3} & -\frac{2}{3} \\ \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\ \\ \frac{2}{3} & -\frac{1}{3} & -\frac{2}{3} \end{matrix} \right] .
Try to determine these \overrightarrow{x_{0}} so that T is a rotation. In this case, determine the axis and the angle of the rotation, also the rotational plane.
The three row (or column) vectors of A are of unit length and are perpendicular to each other. So A^{*}=A^{-1} and A is orthogonal. Direct computation shows that
det A = \frac{1}{27}\left(4+4+4+8+8-1\right)=\frac{1}{27}\cdot27=1.
Therefore the associated T is a rotation of R³ if \overrightarrow{x_{0}} is chosen from Im\left(I_{3}-A\right).
The characteristic polynomial of A is
det \left(A-tI_{3}\right)=-\frac{1}{3}\left(t-1\right)\left(3t^{2}+4t+3\right)
and thus A has a real eigenvalue 1 of multiplicity one. Solve
\overrightarrow{x} \left(A-I_{3}\right)=\left(x_{1},x_{2},x_{3}\right)\left[\begin{matrix} -\frac{4}{3} & \frac{2}{3} & -\frac{2}{3} \\ \\ \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ \\ \frac{2}{3} & -\frac{1}{3} & -\frac{5}{3} \end{matrix} \right] =\overrightarrow{0}
\Rightarrow 2x_{1}-x_{2}-x_{3}=0, 2x_{1}-x_{2}+5x_{3}=0
\Rightarrow \overrightarrow{x}=x_{1}\left(1,2,0\right) for x_{1} \in R.
Hence, take an eigenvector \overrightarrow{v_{1}}=\frac{1}{\sqrt{5} }\left(1,2,0\right).
Let \overrightarrow{v}=\left(\alpha _{1},\alpha _{2},\alpha _{3}\right). Then
\overrightarrow{v} \bot \overrightarrow{v_{1}}
\Leftrightarrow \alpha _{1} +2\alpha _{2}=0.
Therefore, we may choose \overrightarrow{v_{2}}=\frac{1}{\sqrt{5} }\left(-2,1,0\right). Again, solve \alpha _{1} +2\alpha _{2}= -2\alpha _{1} +\alpha _{2}=0, we may choose \overrightarrow{v_{3}} =\left(0,0,1\right)=\overrightarrow{e_{3}}. Now, B = \left\{\overrightarrow{v_{1}},\overrightarrow{v_{2}},\overrightarrow{v_{3}}\right\} forms an orthonormal basis for R³.
By computation,
\overrightarrow{v_{2}}A=\frac{1}{3\sqrt{5} }\left(4,-2,5\right)=\frac{2}{3}\overrightarrow{v_{2}}+\frac{\sqrt{5}}{3 }\overrightarrow{v_{3}},
\overrightarrow{v_{3}}A=\frac{1}{3}\left(2,-1,-2\right)=-\frac{\sqrt{5}}{3 }\overrightarrow{v_{2}}+\frac{2}{3}\overrightarrow{v_{3}}
\Rightarrow \left[A\right]_{B}=PAP^{-1}=\left[\begin{matrix} 1 & 0 & 0 \\ \\ 0 & \frac{2}{3} & \frac{\sqrt{5}}{3 } \\ \\ 0 & -\frac{\sqrt{5}}{3 } & \frac{2}{3} \end{matrix} \right] , where P = \left[\begin{matrix} \frac{1}{\sqrt{5} } & \frac{2}{\sqrt{5} } & 0 \\ \\ -\frac{2}{\sqrt{5} } & \frac{1}{\sqrt{5} } & 0 \\ \\ 0 & 0 & 1 \end{matrix} \right].
To find Im \left(I_{3}-A\right), let \overrightarrow{y}=\overrightarrow{x}\left(I_{3}-A\right).Then
\overrightarrow{y}=\overrightarrow{x}\left(I_{3}-A\right)
\Leftrightarrow y_{1}=\frac{2}{3}\left(2x_{1}-x_{2}-x_{3}\right), y_{2}=\frac{1}{3}\left(-2x_{1}+x_{2}+x_{3}\right) and
y_{3}=\frac{1}{3}\left(2x_{1}-x_{2}+5x_{3}\right)
\Leftrightarrow y_{1}+2y_{2}=0.
So the image subspace is x_{1}+2x_{2}=0 with a unit normal vector \overrightarrow{v_{1}}.Note that, since A^{*}=A^{-1}, Im \left(A-I_{3}\right)^{\bot}= Ker \left(A^{*}-I_{3}\right)=Ker \left(A^{-1}-I_{3}\right)=Ker \left(A-I_{3}\right). This is the theoretical reason why \overrightarrow{v_{1}}\bot Im \left(A-I_{3}\right)=Ker \left(A-I_{3}\right)^{\bot}= \ll \overrightarrow{v_{2}},\overrightarrow{v_{3}}\gg.
Take any point \overrightarrow{x_{0}} on Im \left(A-I_{3}\right), say \overrightarrow{x_{0}}=\overrightarrow{0} for simplicity. Then, the axis of rotation is \ll \overrightarrow{v_{1}}\gg, the rotational plane is \ll \overrightarrow{v_{2}},\overrightarrow{v_{3}}\gg = Im \left(A-I_{3}\right) and the angle of rotation is \theta = \tan ^{-1} \frac{\sqrt{5}}{ 2}. See Fig. 3.70.
We have two questions:
Q1 What is the image of the unit sphere x^{2}_{1}+x^{2}_{2}+x^{2}_{3}=1 under \overrightarrow{y}=\overrightarrow{x}A?
Q2 What is the image of the cylinder x^{2}_{1}+x^{2}_{2}=1 under \overrightarrow{y}=\overrightarrow{x}A?
For Q1 It is the unit sphere itself both by geometric intuition and by analytic proof.
For Q2 Geometric intuition tells as that it is still a cylinder with central axis along the vector \overrightarrow{e_{3}}A=\left( \frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right) and the base circle lying on the plane 2x_{1}-x_{2}-2x_{3}=0. And computation shows that the image has the equation, a complicated one,
5x^{2}_{1}+8x^{2}_{2}+5x^{2}_{3}+4x_{1}x_{2}+8x_{1}x_{3}-4x_{2}x_{3}=9 \left(*_{}\right)
in the eyes of N = \left\{\overrightarrow{0},\overrightarrow{e_{1}},\overrightarrow{e_{2}},\overrightarrow{e_{3}}\right\}.
Hence, we raise a third question concerning Q2 and \left(*_{}\right).
Q3 What is the equation of the image cylinder \left(*_{}\right) in the orthonormal affine basis C = \left\{\overrightarrow{0},A_{1*},A_{2*},A_{3*}\right\} where A_{i*} denotes the ith row vector of A for i = 1, 2, 3?
In B, one needs to compute
\overrightarrow{x} \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right]\overrightarrow{x^{*}}
= \left(\overrightarrow{y}P^{-1}\right) \left(PAP^{-1}\right)P\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right]P^{-1}\left(PAP^{-1}\right)^{-1}\left(\overrightarrow{y}P^{-1}\right)^{*}=1,
where \overrightarrow{y}=\overrightarrow{x}A and \overrightarrow{y}P^{-1} is the coordinate vector of \overrightarrow{y} in the basis B.
In C, one needs to compute
\overrightarrow{x} \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right]\overrightarrow{x^{*}}=\left(\overrightarrow{y}A^{-1}\right)\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right]\left(\overrightarrow{y}A^{-1}\right)^{*}= \alpha^{2}_{1}+\alpha^{2}_{2}=1,
where \overrightarrow{y}A^{-1}=\left(\alpha _{1} ,\alpha _{2} ,\alpha _{3}\right) is the coordinate vector of \overrightarrow{y}=\overrightarrow{x} A in the basis C. Hence, to view the image cylinder, under\overrightarrow{y}=\overrightarrow{x} A, in C is the same as to view the original cylinder in N.
