Question 3.33: Let a0^→ = (1, 1, 0),a1^→ = (2, 0,−1) and a2^→ = (0,−1, 1). ...
Let \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). Try to construct a shearing with coefficient k ≠ 0 in the direction \overrightarrow{a_{1} }-\overrightarrow{a_{0} } with \overrightarrow{a_{0} } +\ll \overrightarrow{a_{2 }}-\overrightarrow{a_{0} }\gg ^{\bot } as the plane of invariant points. Note that \left(\overrightarrow{a_{1} }-\overrightarrow{a_{0} }\right) \bot \left(\overrightarrow{a_{2} }-\overrightarrow{a_{0} }\right) .
Since \overrightarrow{a_{1} }-\overrightarrow{a_{0} }=\left(1,-1,-1\right) has length \sqrt{3}, take the unit vector \overrightarrow{v_{1} }= \frac{1}{\sqrt{3}} \left(1,-1,-1\right). Since \overrightarrow{a_{2} }-\overrightarrow{a_{0} }=\left(-1,-2,1\right) happens to be perpendicular to \overrightarrow{a_{1} }-\overrightarrow{a_{0} }, i.e.
\left(\overrightarrow{a_{2} }-\overrightarrow{a_{0} }\right)\left(\overrightarrow{a_{1} }-\overrightarrow{a_{0} }\right)^{*}=\left(-1 -2 1\right) \left[\begin{matrix} 1 \\ -1 \\ -1 \end{matrix} \right] =-1+2-1=0,
so we can choose \overrightarrow{v_{2} } to be equal to \overrightarrow{a_{2} }-\overrightarrow{a_{0} } dividing by its length \sqrt{6}, i.e. \overrightarrow{v_{2} }=\frac{1}{\sqrt{6}} \left(-1,-2,1\right). Then, choose a vector \overrightarrow{v_{3} }=\left(\alpha _{1} ,\alpha _{2} ,\alpha _{3} \right) of unit length so that
\overrightarrow{v_{3} } \bot \overrightarrow{v_{1} } and \overrightarrow{v_{3} } \bot \overrightarrow{v_{2} }
\Rightarrow \alpha _{1} -\alpha _{2} -\alpha _{3} =0 and -\alpha _{1} -2\alpha _{2} +\alpha _{3}=0
\Rightarrow \alpha _{1}=\alpha _{3} and \alpha _{2}=0
\Rightarrow \overrightarrow{v_{3} }= \frac{1}{\sqrt{2}} \left(1,0,1\right).
In the orthonormal affine basis B = \left\{\overrightarrow{a_{0} },\overrightarrow{a_{0} }+\overrightarrow{v_{1} },\overrightarrow{a_{0} }+\overrightarrow{v_{2} },\overrightarrow{a_{0} }+\overrightarrow{v_{3} }\right\}, the required shearing T has the representation
\left[T\left(\overrightarrow{x} \right) \right]_{B}=\left[\overrightarrow{x} \right]_{B}\left[T\right]_{B}, where \left[T\right]_{B}= \left[\begin{matrix} 1 & 0 & 0 \\ k & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right], simply denoted as A.
While in N = \left\{\overrightarrow{0},\overrightarrow{e_{1} },\overrightarrow{e_{2} },\overrightarrow{e_{3} }\right\},
T\left(\overrightarrow{x} \right)=\overrightarrow{a_{0} }+\left(\overrightarrow{x}-\overrightarrow{a_{0} }\right)P^{-1}AP,
where
P = \left[\begin{matrix} \overrightarrow{v_{1} } \\ \overrightarrow{v_{2} } \\ \overrightarrow{v_{3} } \end{matrix} \right]= \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} \right] with P^{-1}=P^{*}.
By computation,
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} \right] , and
\overrightarrow{x_{0} }=\overrightarrow{a_{0} }-\overrightarrow{a_{0} }P^{-1}AP=\overrightarrow{a_{0} }P^{-1}\left(I_{3}-A\right)P=\frac{k}{\sqrt{2} }\left(1,-1,-1\right)
\Rightarrow T\left(\overrightarrow{x} \right)=\overrightarrow{x_{0} }+\overrightarrow{x} P^{-1}AP.
See Fig. 3.68.
For simplicity, take k = \sqrt{2} and consider the converse problem. Let
T\left(\overrightarrow{x} \right)=\overrightarrow{x_{0} }+\overrightarrow{x}B , where \overrightarrow{x_{0} }=\left(1,-1,-1\right) and
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} \right] .
We want to check if this T is a shearing. Follow the steps in (3.8.22).
1. B has characteristic polynomial
det \left(B-tI_{3}\right)=\frac{1}{27}\left(-27t^{3}+81t^{2}-81t+27\right)=-\left(t-1\right)^{3}.
So B has eigenvalue 1 of algebraic multiplicity 3. Furthermore, B-I_{3} \neq O but
\left(B-I_{3}\right)^{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} \right]^{2}=O_{3 \times 3}.
Hence B is not diagonalizable. Solve
\overrightarrow{x} \left(I_{3}-B\right)=\overrightarrow{x_{0} }=\left(1,-1,-1\right), where \overrightarrow{x}=\left(x_{1},x_{2},x_{3}\right)
\Rightarrow x_{1}+2x_{2}-x_{3}=3 has infinitely many solutions.
Pick any such a solution, say \overrightarrow{a_{0} } = (1, 1, 0). These information guarantee that T is a shearing.
2. Solve
\overrightarrow{x}\left(B-I_{3}\right)=\overrightarrow{0}
\Rightarrow x_{1}+2x_{2}-x_{3}=0.
Pick up two orthogonal eigenvectors of unit length, say \overrightarrow{v_{1} }=\frac{1}{\sqrt{3} } \left(1,-1,-1\right) and \overrightarrow{v_{3} }=\frac{1}{\sqrt{2} } \left(1,0,1\right).
3. To choose a unit vector \overrightarrow{v_{2} }= \left(\alpha _{1},\alpha _{2},\alpha _{3}\right) so that \overrightarrow{v_{2} }\bot \overrightarrow{v_{1} } and \overrightarrow{v_{2} }\bot \overrightarrow{v_{3} } hold, solve
\alpha _{1}-\alpha _{2}-\alpha _{3}=\alpha _{1}+\alpha _{3}=0 \Rightarrow \left(\alpha _{1},\alpha _{2},\alpha _{3}\right) = \alpha _{1} \left(1,2,-1\right).
Take \overrightarrow{v_{2} }=\frac{1}{\sqrt{6} } \left(1,2,-1\right). Notice that this \overrightarrow{v_{2} } is not an eigenvector associated to 1 since dim Ker\left(B-I_{3}\right)=2. By computation,
\overrightarrow{v_{2} }B=\frac{1}{\sqrt{6} } \left(1,2,-1\right)\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} \right] =\frac{1}{\sqrt{6} } \left(-1,4,1\right)
\Rightarrow \overrightarrow{v_{2} }B-\overrightarrow{v_{2} }=\frac{2}{\sqrt{6} } \left(-1,1,1\right)=-\sqrt{2}\overrightarrow{v_{1} }.
Therefore, −\sqrt{2} is the coefficient.
Hence, in C = \left\{\overrightarrow{a_{0} },\overrightarrow{a_{0} }+\overrightarrow{v_{1} },\overrightarrow{a_{0} }+\overrightarrow{v_{2} },\overrightarrow{a_{0} }+\overrightarrow{v_{3} }\right\} ,
\left[T\left(\overrightarrow{x} \right) \right]_{C}=\left[\overrightarrow{x} \right]_{C}\left[T\right]_{C}, where \left[T\right]_{C}= \left[\begin{matrix} 1 & 0 & 0 \\−\sqrt{2} & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right].
Notice that the \overrightarrow{v_{2} } in C is equal to – \overrightarrow{v_{2} } in B metioned above. This is the reason why we have −\sqrt{2} here instead of the original \sqrt{2} (see 2 in (3.8.20)).
Two questions are raised as follows.
Q1 What is the image plane of the plane x_{1}+x_{2}=0 under T? Where do they intersect?
Q2 Where is the image of the tetrahedron \Delta \overrightarrow{b_{0} }\overrightarrow{b_{1} }\overrightarrow{b_{2} }\overrightarrow{b_{3} }, where \overrightarrow{b_{0} }=\left(1,1,1\right),\overrightarrow{b_{1} }= \left(-1,1,1\right),\overrightarrow{b_{2} }=\left(1,-1,1\right) and \overrightarrow{b_{3} }=\left(1,1,-1\right)? How are their volumes related?
We need to use the inverse of T\left(\overrightarrow{x} \right) =\overrightarrow{x_{0} }+\overrightarrow{x}B, namely,
\overrightarrow{x}=\left(\overrightarrow{y} -\overrightarrow{x_{0} }\right)B^{-1}, where \overrightarrow{x_{0} }=\left(1,-1,-1\right) and \overrightarrow{y}=T\left(\overrightarrow{x} \right) and
B^{-1}=P^{-1}A^{-1}P= \frac{1}{3}\left[\begin{matrix} 4 & -1 & -1 \\ 2 & 1 & -2 \\ -1 & 1 & 4 \end{matrix} \right] .
For Q1
x_{1}+x_{2}= \overrightarrow{x} \left[\begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right] =0 if \overrightarrow{x}=\left(x_{1},x_{2},x_{3}\right)
\Rightarrow \left[\left(y_{1},y_{2},y_{3}\right)-\left(1,-1,-1\right)\right]\cdot \frac{1}{3}\left[\begin{matrix} 4 & -1 & -1 \\ 2 & 1 & -2 \\ -1 & 1 & 4 \end{matrix} \right]\left[\begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right] =0
\Rightarrow \left( replace y_{1},y_{2},y_{3} by x_{1},x_{2},x_{3}\right) x_{1}+x_{2}=0.
Hence, x_{1}+x_{2}=0 and its image plane are coincident. This is within our expectation because the normal vector (1, 1, 0) of x_{1}+x_{2}=0 is perpendicular to \overrightarrow{v_{1} }=\frac{1}{\sqrt{3} }\left(1,-1,-1\right), the direction of the shearing T (see 1 in (3.8.20)).
Any plane parallel to x_{1}+2x_{2}-x_{3}=3 has equation x_{1}+2x_{2}-x_{3}=c where c is a constant. Then
x_{1}+2x_{2}-x_{3}= \overrightarrow{x} \left[\begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] =c
\Rightarrow \left[\left(y_{1},y_{2},y_{3}\right)-\left(1,-1,-1\right)\right]\cdot \frac{1}{3}\left[\begin{matrix} 4 & -1 & -1 \\ 2 & 1 & -2 \\ -1 & 1 & 4 \end{matrix} \right]\left[\begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] =c
\Rightarrow y_{1}+2y_{2}-y_{3}=c,
which means that it is an invariant plane.
For Q2 By computation,
T\left(\overrightarrow{b_{0} }\right)=\left(1,-1,-1\right)+\left(1,1,1\right)B=\left(1,-1,-1\right)+\left(\frac{1}{3},\frac{5}{3},\frac{5}{3}\right)=\left(\frac{4}{3},\frac{2}{3},\frac{2}{3}\right),
T\left(\overrightarrow{b_{1} }\right)=\left(1,-1,-1\right)+\left(-1,1,1\right)B=\left(1,-1,-1\right)+\left(-1,1,1\right)=\left(0,0,0\right),
T\left(\overrightarrow{b_{2} }\right)=\left(1,-1,-1\right)+\left(1,-1,1\right)B
=\left(1,-1,-1\right)+\left(\frac{5}{3},-\frac{5}{3},\frac{1}{3}\right)=\left(\frac{8}{3},-\frac{8}{3},-\frac{2}{3}\right),
T\left(\overrightarrow{b_{3} }\right)=\left(1,-1,-1\right)+\left(1,1,-1\right)B
=\left(1,-1,-1\right)+\left(-\frac{1}{3},\frac{7}{3},\frac{1}{3}\right)=\left(\frac{2}{3},\frac{4}{3},-\frac{2}{3}\right).
These four image points form a tetrahedron \Delta T \left(\overrightarrow{b_{0} }\right) \ldots T \left(\overrightarrow{b_{3} }\right) . Thus
the signed volume of \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} \right|
= \frac{1}{6} \left|\begin{matrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{matrix} \right| =-\frac{4}{3},
the signed volume of \Delta T \left(\overrightarrow{b_{0} }\right) \ldots T \left(\overrightarrow{b_{3} }\right) = \frac{1}{6} \left|\begin{matrix}T \left(\overrightarrow{b_{1} }\right)-T \left(\overrightarrow{b_{0} }\right) \\ T \left(\overrightarrow{b_{2} }\right)-T \left(\overrightarrow{b_{0} }\right) \\ T \left(\overrightarrow{b_{3} }\right)-T \left(\overrightarrow{b_{0} }\right) \end{matrix} \right|
= \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} \right|
= \frac{1}{6}\cdot \left(-\frac{8}{27} \right)\cdot27=-\frac{4}{3}.
So both have the same volumes.
