Give an affine transformation T(x^→ ) = x0^→ + x^→A, where A = [-1/3 2/3 -2/3 2/3 2/3 1/3 2/3 -1/3 -2/3].Try to determine these x0^→ so that T is a rotation. In this case, determine the axis and the angle of the rotation, also the rotational plane.

Question

Give an affine transformation

[latex]T\left(\overrightarrow{x} ight)=\overrightarrow{x_{0}}+\overrightarrow{x}A,[/latex] where [latex]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} ight] .[/latex]

Try to determine these [latex]\overrightarrow{x_{0}}[/latex] so that T is a rotation. In this case, determine the axis and the angle of the rotation, also the rotational plane.

Accepted Answer

The three row (or column) vectors of A are of unit length and are perpendicular to each other. So [latex]A^{*}=A^{-1}[/latex] and A is orthogonal. Direct computation shows that

[latex]det A = \frac{1}{27}\left(4+4+4+8+8-1 ight)=\frac{1}{27}\cdot27=1.[/latex]

Therefore the associated T is a rotation of R³ if [latex]\overrightarrow{x_{0}}[/latex] is chosen from [latex]Im\left(I_{3}-A ight).[/latex]

The characteristic polynomial of A is

[latex]det \left(A-tI_{3} ight)=-\frac{1}{3}\left(t-1 ight)\left(3t^{2}+4t+3 ight)[/latex]

and thus A has a real eigenvalue 1 of multiplicity one. Solve

[latex]\overrightarrow{x} \left(A-I_{3} ight)=\left(x_{1},x_{2},x_{3} ight)\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} ight] =\overrightarrow{0}[/latex]

[latex]\Rightarrow 2x_{1}-x_{2}-x_{3}=0, 2x_{1}-x_{2}+5x_{3}=0[/latex]

[latex]\Rightarrow \overrightarrow{x}=x_{1}\left(1,2,0 ight) for x_{1} \in R.[/latex]

Hence, take an eigenvector [latex]\overrightarrow{v_{1}}=\frac{1}{\sqrt{5} }\left(1,2,0 ight).[/latex]

Let [latex]\overrightarrow{v}=\left(\alpha _{1},\alpha _{2},\alpha _{3} ight).[/latex] Then

[latex]\overrightarrow{v} \bot \overrightarrow{v_{1}}[/latex]

[latex]\Leftrightarrow \alpha _{1} +2\alpha _{2}=0.[/latex]

Therefore, we may choose [latex]\overrightarrow{v_{2}}=\frac{1}{\sqrt{5} }\left(-2,1,0 ight).[/latex] Again, solve [latex]\alpha _{1} +2\alpha _{2}= -2\alpha _{1} +\alpha _{2}=0,[/latex] we may choose [latex] \overrightarrow{v_{3}} =\left(0,0,1 ight)=\overrightarrow{e_{3}}.[/latex] Now, [latex]B = \left\{\overrightarrow{v_{1}},\overrightarrow{v_{2}},\overrightarrow{v_{3}} ight\}[/latex] forms an orthonormal basis for R³.
By computation,

[latex]\overrightarrow{v_{2}}A=\frac{1}{3\sqrt{5} }\left(4,-2,5 ight)=\frac{2}{3}\overrightarrow{v_{2}}+\frac{\sqrt{5}}{3 }\overrightarrow{v_{3}},[/latex]

[latex]\overrightarrow{v_{3}}A=\frac{1}{3}\left(2,-1,-2 ight)=-\frac{\sqrt{5}}{3 }\overrightarrow{v_{2}}+\frac{2}{3}\overrightarrow{v_{3}}[/latex]

[latex]\Rightarrow \left[A ight]_{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} ight] , 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} ight].[/latex]

To find [latex]Im \left(I_{3}-A ight), let \overrightarrow{y}=\overrightarrow{x}\left(I_{3}-A ight).[/latex]Then

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

[latex]\Leftrightarrow y_{1}=\frac{2}{3}\left(2x_{1}-x_{2}-x_{3} ight), y_{2}=\frac{1}{3}\left(-2x_{1}+x_{2}+x_{3} ight)[/latex] and

[latex]y_{3}=\frac{1}{3}\left(2x_{1}-x_{2}+5x_{3} ight)[/latex]

[latex]\Leftrightarrow y_{1}+2y_{2}=0.[/latex]

So the image subspace is [latex]x_{1}+2x_{2}=0[/latex] with a unit normal vector [latex]\overrightarrow{v_{1}}.[/latex]Note that, since [latex]A^{*}=A^{-1}, Im \left(A-I_{3} ight)^{\bot}= Ker \left(A^{*}-I_{3} ight)=Ker \left(A^{-1}-I_{3} ight)=Ker \left(A-I_{3} ight).[/latex] This is the theoretical reason why [latex]\overrightarrow{v_{1}}\bot Im \left(A-I_{3} ight)=Ker \left(A-I_{3} ight)^{\bot}= \ll \overrightarrow{v_{2}},\overrightarrow{v_{3}}\gg.[/latex]

Take 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 axis of rotation is [latex]\ll \overrightarrow{v_{1}}\gg,[/latex] the rotational plane is [latex]\ll \overrightarrow{v_{2}},\overrightarrow{v_{3}}\gg = Im \left(A-I_{3} ight)[/latex] and the angle of rotation is [latex] heta = an ^{-1} \frac{\sqrt{5}}{ 2}.[/latex] See Fig. 3.70.

We have two questions:

Q1 What is the image of the unit sphere [latex]x^{2}_{1}+x^{2}_{2}+x^{2}_{3}=1[/latex] under [latex]\overrightarrow{y}=\overrightarrow{x}A?[/latex]

Q2 What is the image of the cylinder [latex]x^{2}_{1}+x^{2}_{2}=1[/latex] under [latex]\overrightarrow{y}=\overrightarrow{x}A?[/latex]

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 [latex]\overrightarrow{e_{3}}A=\left( \frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} ight)[/latex] and the base circle lying on the plane [latex]2x_{1}-x_{2}-2x_{3}=0.[/latex] And computation shows that the image has the equation, a complicated one,

[latex]5x^{2}_{1}+8x^{2}_{2}+5x^{2}_{3}+4x_{1}x_{2}+8x_{1}x_{3}-4x_{2}x_{3}=9 \left(*_{} ight)[/latex]

in the eyes of [latex]N = \left\{\overrightarrow{0},\overrightarrow{e_{1}},\overrightarrow{e_{2}},\overrightarrow{e_{3}} ight\}.[/latex]

Hence, we raise a third question concerning Q2 and [latex]\left(*_{} ight).[/latex]

Q3 What is the equation of the image cylinder [latex]\left(*_{} ight)[/latex] in the orthonormal affine basis [latex]C = \left\{\overrightarrow{0},A_{1*},A_{2*},A_{3*} ight\}[/latex] where [latex]A_{i*}[/latex] denotes the [latex]i[/latex]th row vector of A for [latex]i = 1, 2, 3?[/latex]

In B, one needs to compute

[latex]\overrightarrow{x} \left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{matrix} ight]\overrightarrow{x^{*}} [/latex]

[latex]= \left(\overrightarrow{y}P^{-1} ight) \left(PAP^{-1} ight)P\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{matrix} ight]P^{-1}\left(PAP^{-1} ight)^{-1}\left(\overrightarrow{y}P^{-1} ight)^{*}=1,[/latex]

where [latex]\overrightarrow{y}=\overrightarrow{x}A and \overrightarrow{y}P^{-1}[/latex] is the coordinate vector of [latex]\overrightarrow{y}[/latex] in the basis B.

In C, one needs to compute

[latex]\overrightarrow{x} \left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{matrix} ight]\overrightarrow{x^{*}}=\left(\overrightarrow{y}A^{-1} ight)\left[\begin{matrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{matrix} ight]\left(\overrightarrow{y}A^{-1} ight)^{*}= \alpha^{2}_{1}+\alpha^{2}_{2}=1,[/latex]

where [latex]\overrightarrow{y}A^{-1}=\left(\alpha _{1} ,\alpha _{2} ,\alpha _{3} ight)[/latex] is the coordinate vector of [latex]\overrightarrow{y}=\overrightarrow{x} A[/latex] in the basis C. Hence, to view the image cylinder, under[latex]\overrightarrow{y}=\overrightarrow{x} A,[/latex] in C is the same as to view the original cylinder in N.

Question 3.34: Give an affine transformation T(x^→ ) = x0^→ + x^→A, where A...

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