Let a0^→ = (1, 1, 0),a1^→ = (2, 0,−1) and a2^→ = (0,−1, 1). Try to construct a shearing with coefficient k ≠ 0 in the direction a1^→ − a0^→ with a0^→ +⊥ as the plane of invariant points. Note that (a1^→ − a0^→ )⊥ (a2^→ − a0^→ ).

Question

Let [latex]\overrightarrow{a_{0} }=\left(1,1,0\right), \overrightarrow{a_{1} }=\left(2,0,-1\right) and \overrightarrow{a_{2} }=\left(0,-1,1\right).[/latex] Try to construct a shearing with coefficient k ≠ 0 in the direction [latex]\overrightarrow{a_{1} }-\overrightarrow{a_{0} }[/latex] with [latex]\overrightarrow{a_{0} } +\ll \overrightarrow{a_{2 }}-\overrightarrow{a_{0} }\gg ^{\bot }[/latex] as the plane of invariant points. Note that [latex]\left(\overrightarrow{a_{1} }-\overrightarrow{a_{0} }\right) \bot \left(\overrightarrow{a_{2} }-\overrightarrow{a_{0} }\right) .[/latex]

Accepted Answer

Since [latex]\overrightarrow{a_{1} }-\overrightarrow{a_{0} }=\left(1,-1,-1 ight) [/latex] has length [latex] \sqrt{3},[/latex] take the unit vector [latex]\overrightarrow{v_{1} }= \frac{1}{\sqrt{3}} \left(1,-1,-1 ight).[/latex] Since [latex]\overrightarrow{a_{2} }-\overrightarrow{a_{0} }=\left(-1,-2,1 ight) [/latex] happens to be perpendicular to [latex]\overrightarrow{a_{1} }-\overrightarrow{a_{0} },[/latex] i.e.

[latex]\left(\overrightarrow{a_{2} }-\overrightarrow{a_{0} } ight)\left(\overrightarrow{a_{1} }-\overrightarrow{a_{0} } ight)^{*}=\left(-1 -2 1 ight) \left[\begin{matrix} 1 \ -1 \ -1 \end{matrix} ight] =-1+2-1=0,[/latex]

so we can choose [latex]\overrightarrow{v_{2} }[/latex] to be equal to [latex]\overrightarrow{a_{2} }-\overrightarrow{a_{0} }[/latex] dividing by its length [latex]\sqrt{6},[/latex] i.e. [latex]\overrightarrow{v_{2} }=\frac{1}{\sqrt{6}} \left(-1,-2,1 ight).[/latex] Then, choose a vector [latex] \overrightarrow{v_{3} }=\left(\alpha _{1} ,\alpha _{2} ,\alpha _{3} ight)[/latex] of unit length so that

[latex]\overrightarrow{v_{3} } \bot \overrightarrow{v_{1} } and \overrightarrow{v_{3} } \bot \overrightarrow{v_{2} }[/latex]

[latex]\Rightarrow \alpha _{1} -\alpha _{2} -\alpha _{3} =0 and -\alpha _{1} -2\alpha _{2} +\alpha _{3}=0[/latex]

[latex]\Rightarrow \alpha _{1}=\alpha _{3} and \alpha _{2}=0 [/latex]

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

In the orthonormal affine basis [latex]B = \left\{\overrightarrow{a_{0} },\overrightarrow{a_{0} }+\overrightarrow{v_{1} },\overrightarrow{a_{0} }+\overrightarrow{v_{2} },\overrightarrow{a_{0} }+\overrightarrow{v_{3} } ight\}, [/latex] the required shearing T has the representation

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

While in [latex]N = \left\{\overrightarrow{0},\overrightarrow{e_{1} },\overrightarrow{e_{2} },\overrightarrow{e_{3} } ight\}, [/latex]

[latex]T\left(\overrightarrow{x} ight)=\overrightarrow{a_{0} }+\left(\overrightarrow{x}-\overrightarrow{a_{0} } ight)P^{-1}AP,[/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{3} } & -\frac{1}{\sqrt{3} } & -\frac{1}{\sqrt{3} } \ \ -\frac{1}{\sqrt{6} } & -\frac{2}{\sqrt{6} } & \frac{1}{\sqrt{6} } \ \ \frac{1}{\sqrt{2} } & 0 & \frac{1}{\sqrt{2} } \end{matrix} ight] with P^{-1}=P^{*}.[/latex]

By computation,

[latex]P^{-1}AP= \left[\begin{matrix} 1-\frac{k}{3\sqrt{2} } &\frac{k}{3\sqrt{2} } & \frac{k}{3\sqrt{2} } \ \ \frac{-2k}{3\sqrt{2} } & 1+\frac{2k}{3\sqrt{2} } &\frac{2k}{3\sqrt{2} } \ \ \frac{k}{3\sqrt{2} } & \frac{-k}{3\sqrt{2} } & 1-\frac{k}{3\sqrt{2} } \end{matrix} ight] ,[/latex] and

[latex]\overrightarrow{x_{0} }=\overrightarrow{a_{0} }-\overrightarrow{a_{0} }P^{-1}AP=\overrightarrow{a_{0} }P^{-1}\left(I_{3}-A ight)P=\frac{k}{\sqrt{2} }\left(1,-1,-1 ight)[/latex]

[latex]\Rightarrow T\left(\overrightarrow{x} ight)=\overrightarrow{x_{0} }+\overrightarrow{x} P^{-1}AP.[/latex]

See Fig. 3.68.

For simplicity, take [latex]k = \sqrt{2}[/latex] and consider the converse problem. Let

[latex]T\left(\overrightarrow{x} ight)=\overrightarrow{x_{0} }+\overrightarrow{x}B ,[/latex] where [latex]\overrightarrow{x_{0} }=\left(1,-1,-1 ight)[/latex] and

[latex]B=P^{-1}AP= \left[\begin{matrix} \frac{2}{3} & \frac{1}{3} & \frac{1}{3} \ \ -\frac{2}{3} & \frac{5}{3} & \frac{2}{3} \ \ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{matrix} ight] .[/latex]

We want to check if this T is a shearing. Follow the steps in (3.8.22).

1. B has characteristic polynomial

[latex]det \left(B-tI_{3} ight)=\frac{1}{27}\left(-27t^{3}+81t^{2}-81t+27 ight)=-\left(t-1 ight)^{3}.[/latex]

So B has eigenvalue 1 of algebraic multiplicity 3. Furthermore, [latex]B-I_{3} eq O[/latex] but

[latex]\left(B-I_{3} ight)^{2}= \left[\begin{matrix} -\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \ \ -\frac{2}{3} & \frac{2}{3} & \frac{2}{3} \ \ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \end{matrix} ight]^{2}=O_{3 imes 3}.[/latex]

Hence B is not diagonalizable. Solve

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

[latex]\Rightarrow x_{1}+2x_{2}-x_{3}=3[/latex] has infinitely many solutions.

Pick any such a solution, say [latex]\overrightarrow{a_{0} } [/latex] = (1, 1, 0). These information guarantee that T is a shearing.
2. Solve

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

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

Pick up two orthogonal eigenvectors of unit length, say [latex]\overrightarrow{v_{1} }=\frac{1}{\sqrt{3} } \left(1,-1,-1 ight) and \overrightarrow{v_{3} }=\frac{1}{\sqrt{2} } \left(1,0,1 ight).[/latex]

3. To choose a unit vector [latex]\overrightarrow{v_{2} }= \left(\alpha _{1},\alpha _{2},\alpha _{3} ight)[/latex] so that [latex]\overrightarrow{v_{2} }\bot \overrightarrow{v_{1} } and \overrightarrow{v_{2} }\bot \overrightarrow{v_{3} }[/latex] hold, solve

[latex]\alpha _{1}-\alpha _{2}-\alpha _{3}=\alpha _{1}+\alpha _{3}=0 \Rightarrow \left(\alpha _{1},\alpha _{2},\alpha _{3} ight) = \alpha _{1} \left(1,2,-1 ight).[/latex]

Take [latex]\overrightarrow{v_{2} }=\frac{1}{\sqrt{6} } \left(1,2,-1 ight).[/latex] Notice that this [latex]\overrightarrow{v_{2} }[/latex] is not an eigenvector associated to 1 since dim [latex]Ker\left(B-I_{3} ight)=2.[/latex] By computation,

[latex]\overrightarrow{v_{2} }B=\frac{1}{\sqrt{6} } \left(1,2,-1 ight)\left[\begin{matrix} \frac{2}{3} & \frac{1}{3} & \frac{1}{3} \ \ -\frac{2}{3} & \frac{5}{3} & \frac{2}{3} \ \ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{matrix} ight] =\frac{1}{\sqrt{6} } \left(-1,4,1 ight)[/latex]

[latex]\Rightarrow \overrightarrow{v_{2} }B-\overrightarrow{v_{2} }=\frac{2}{\sqrt{6} } \left(-1,1,1 ight)=-\sqrt{2}\overrightarrow{v_{1} }.[/latex]

Therefore, [latex]−\sqrt{2}[/latex] is the coefficient.

Hence, in [latex]C = \left\{\overrightarrow{a_{0} },\overrightarrow{a_{0} }+\overrightarrow{v_{1} },\overrightarrow{a_{0} }+\overrightarrow{v_{2} },\overrightarrow{a_{0} }+\overrightarrow{v_{3} } ight\} ,[/latex]

[latex]\left[T\left(\overrightarrow{x} ight) ight]_{C}=\left[\overrightarrow{x} ight]_{C}\left[T ight]_{C},[/latex] where [latex]\left[T ight]_{C}= \left[\begin{matrix} 1 & 0 & 0 \−\sqrt{2} & 1 & 0 \ 0 & 0 & 1 \end{matrix} ight].[/latex]

Notice that the [latex]\overrightarrow{v_{2} }[/latex] in C is equal to [latex]- \overrightarrow{v_{2} }[/latex] in B metioned above. This is the reason why we have [latex]−\sqrt{2}[/latex] here instead of the original [latex]\sqrt{2}[/latex] (see 2 in (3.8.20)).
Two questions are raised as follows.

Q1 What is the image plane of the plane [latex]x_{1}+x_{2}=0[/latex] under T? Where do they intersect?

Q2 Where is the image of the tetrahedron [latex]\Delta \overrightarrow{b_{0} }\overrightarrow{b_{1} }\overrightarrow{b_{2} }\overrightarrow{b_{3} },[/latex] where [latex]\overrightarrow{b_{0} }=\left(1,1,1 ight),\overrightarrow{b_{1} }= \left(-1,1,1 ight),\overrightarrow{b_{2} }=\left(1,-1,1 ight) and \overrightarrow{b_{3} }=\left(1,1,-1 ight)?[/latex] How are their volumes related?

We need to use the inverse of [latex]T\left(\overrightarrow{x} ight) =\overrightarrow{x_{0} }+\overrightarrow{x}B,[/latex] namely,

[latex]\overrightarrow{x}=\left(\overrightarrow{y} -\overrightarrow{x_{0} } ight)B^{-1}, [/latex] where [latex]\overrightarrow{x_{0} }=\left(1,-1,-1 ight) and \overrightarrow{y}=T\left(\overrightarrow{x} ight)[/latex] and

[latex]B^{-1}=P^{-1}A^{-1}P= \frac{1}{3}\left[\begin{matrix} 4 & -1 & -1 \ 2 & 1 & -2 \ -1 & 1 & 4 \end{matrix} ight] .[/latex]

For Q1

[latex]x_{1}+x_{2}= \overrightarrow{x} \left[\begin{matrix} 1 \ 1 \ 0 \end{matrix} ight] =0 if \overrightarrow{x}=\left(x_{1},x_{2},x_{3} ight)[/latex]

[latex]\Rightarrow \left[\left(y_{1},y_{2},y_{3} ight)-\left(1,-1,-1 ight) ight]\cdot \frac{1}{3}\left[\begin{matrix} 4 & -1 & -1 \ 2 & 1 & -2 \ -1 & 1 & 4 \end{matrix} ight]\left[\begin{matrix} 1 \ 1 \ 0 \end{matrix} ight] =0 [/latex]

[latex]\Rightarrow \left( replace y_{1},y_{2},y_{3} by x_{1},x_{2},x_{3} ight) x_{1}+x_{2}=0.[/latex]

Hence, [latex]x_{1}+x_{2}=0[/latex] and its image plane are coincident. This is within our expectation because the normal vector (1, 1, 0) of [latex] x_{1}+x_{2}=0[/latex] is perpendicular to [latex]\overrightarrow{v_{1} }=\frac{1}{\sqrt{3} }\left(1,-1,-1 ight),[/latex] the direction of the shearing T (see 1 in (3.8.20)).

Any plane parallel to [latex]x_{1}+2x_{2}-x_{3}=3[/latex] has equation [latex]x_{1}+2x_{2}-x_{3}=c[/latex] where c is a constant. Then

[latex]x_{1}+2x_{2}-x_{3}= \overrightarrow{x} \left[\begin{matrix} 1 \ 2 \ -1 \end{matrix} ight] =c[/latex]

[latex]\Rightarrow \left[\left(y_{1},y_{2},y_{3} ight)-\left(1,-1,-1 ight) ight]\cdot \frac{1}{3}\left[\begin{matrix} 4 & -1 & -1 \ 2 & 1 & -2 \ -1 & 1 & 4 \end{matrix} ight]\left[\begin{matrix} 1 \ 2 \ -1 \end{matrix} ight] =c [/latex]

[latex]\Rightarrow y_{1}+2y_{2}-y_{3}=c,[/latex]

which means that it is an invariant plane.

For Q2 By computation,

[latex]T\left(\overrightarrow{b_{0} } ight)=\left(1,-1,-1 ight)+\left(1,1,1 ight)B=\left(1,-1,-1 ight)+\left(\frac{1}{3},\frac{5}{3},\frac{5}{3} ight)=\left(\frac{4}{3},\frac{2}{3},\frac{2}{3} ight),[/latex]

[latex]T\left(\overrightarrow{b_{1} } ight)=\left(1,-1,-1 ight)+\left(-1,1,1 ight)B=\left(1,-1,-1 ight)+\left(-1,1,1 ight)=\left(0,0,0 ight),[/latex]

[latex]T\left(\overrightarrow{b_{2} } ight)=\left(1,-1,-1 ight)+\left(1,-1,1 ight)B[/latex]

[latex]=\left(1,-1,-1 ight)+\left(\frac{5}{3},-\frac{5}{3},\frac{1}{3} ight)=\left(\frac{8}{3},-\frac{8}{3},-\frac{2}{3} ight),[/latex]

[latex]T\left(\overrightarrow{b_{3} } ight)=\left(1,-1,-1 ight)+\left(1,1,-1 ight)B[/latex]

[latex]=\left(1,-1,-1 ight)+\left(-\frac{1}{3},\frac{7}{3},\frac{1}{3} ight)=\left(\frac{2}{3},\frac{4}{3},-\frac{2}{3} ight).[/latex]

These four image points form a tetrahedron [latex]\Delta T \left(\overrightarrow{b_{0} } ight) \ldots T \left(\overrightarrow{b_{3} } ight) .[/latex] Thus

the signed volume of [latex]\Delta \overrightarrow{b_{0} }\overrightarrow{b_{1} }\overrightarrow{b_{2} }\overrightarrow{b_{3} }= \frac{1}{6}\left|\begin{matrix} \overrightarrow{b_{1} }-\overrightarrow{b_{0} } \ \overrightarrow{b_{2} }-\overrightarrow{b_{0} } \ \overrightarrow{b_{3} }-\overrightarrow{b_{0} } \end{matrix} ight| [/latex]

[latex]= \frac{1}{6} \left|\begin{matrix} -2 & 0 & 0 \ 0 & -2 & 0 \ 0 & 0 & -2 \end{matrix} ight| =-\frac{4}{3},[/latex]

the signed volume of [latex]\Delta T \left(\overrightarrow{b_{0} } ight) \ldots T \left(\overrightarrow{b_{3} } ight) = \frac{1}{6} \left|\begin{matrix}T \left(\overrightarrow{b_{1} } ight)-T \left(\overrightarrow{b_{0} } ight) \ T \left(\overrightarrow{b_{2} } ight)-T \left(\overrightarrow{b_{0} } ight) \ T \left(\overrightarrow{b_{3} } ight)-T \left(\overrightarrow{b_{0} } ight) \end{matrix} ight| [/latex]

[latex]= \frac{1}{6} \left|\begin{matrix} -\frac{4}{3} & -\frac{2}{3} & -\frac{2}{3} \ \ \frac{4}{3} & -\frac{10}{3} & -\frac{4}{3} \ \ -\frac{2}{3} & \frac{2}{3} & -\frac{4}{3} \end{matrix} ight|[/latex]

[latex]= \frac{1}{6}\cdot \left(-\frac{8}{27} ight)\cdot27=-\frac{4}{3}.[/latex]

So both have the same volumes.

Question 3.33: Let a0^→ = (1, 1, 0),a1^→ = (2, 0,−1) and a2^→ = (0,−1, 1). ...

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