Question 3.31: (a) (See Example 1(a).) Find the one-way stretch with scale ...
(a) (See Example 3.30(a).) Find the one-way stretch with scale factor k≠1 of \mathbb{R}^3 along the direction \overrightarrow{v_{3} }=\left(-1,1,-1\right) with respect to the plane \left(-1,1,1\right) + \ll \overrightarrow{v_{1} },\overrightarrow{v_{2} }\gg, where \overrightarrow{v_{1} }=\left(0,1,0\right) and \overrightarrow{v_{2} }=\left(0,-1,1\right).
(b) Show that
T\left(\overrightarrow{x }\right)=\overrightarrow{x_{0} }+\overrightarrow{x }A, where \overrightarrow{x_{0} }=\left(-2,4,2\right) and A = \left[\begin{matrix} -1 & 4 & 2 \\ -1 & 3 & 1 \\ -1 & 2 & 2 \end{matrix} \right]
represents a one-way stretch. Determine its direction and its plane of invariant points.
(a) In the affine basis \mathcal{B} = \left\{\left(-1,1,1\right),\left(-1,2,1\right),\left(-1,0,2\right),\left(-2,2,0\right) \right\} , the one-way stretch T is
\left[T\left(\overrightarrow{x}\right)\right]_{\mathcal{B} }= \left[\overrightarrow{x}\right]_{\mathcal{B} }\left[T\right]_{\mathcal{B} }, where \left[T\right]_{\mathcal{B} }=\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & k \end{matrix} \right] .
In the natural affine basis N,
T\left(\overrightarrow{x}\right)=\left(-1,1,1\right)+\left(\overrightarrow{x}-\left(-1,1,1\right)\right)P^{-1}\left[T\right]_{\mathcal{B} }P,
where
P=P^{B}_{\mathcal{N} }=\left[\begin{matrix}0 & 1 & 0 \\ 0 & -1 & 1 \\ -1 & 1 & -1 \end{matrix} \right] \Rightarrow P^{-1}=\left[\begin{matrix}0 & -1 & -1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{matrix} \right] .
Hence,
P^{-1}\left[T\right]_{\mathcal{B} }P=\left[\begin{matrix}0 & -1 & -1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{matrix} \right]\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & k \end{matrix} \right]\left[\begin{matrix}0 & 1 & 0 \\ 0 & -1 & 1 \\ -1 & 1 & -1 \end{matrix} \right]
=\left[\begin{matrix} k & 1-k & -1+k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right],
\left(-1,1,1\right)-\left(-1,1,1\right)P^{-1}\left[T\right]_{\mathcal{B} }P=\left(-1,1,1\right)-\left(-k,k,2-k\right)
=\left(-1+k,1-k,-1+k\right)
\Rightarrow T\left(\overrightarrow{x}\right)=\left(-1+k,1-k,-1+k\right)
+\overrightarrow{x} \left[\begin{matrix} k & 1-k & -1+k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] for \overrightarrow{x} \in \mathbb{R}^3, or
\left\{\begin{matrix} y_{1}=-1+k+kx_{1}, \\ y_{2}=1-k+\left(1-k\right)x_{1}+x_{2}, \\ y_{3}=-1+k+\left(-1+k\right)x_{1}+x_{3}, \end{matrix} \right.
where \overrightarrow{x} =\left(x_{1},x_{2},x_{3}\right) and \overrightarrow{y} =T\left(\overrightarrow{x}\right)=\left(y_{1},y_{2},y_{3}\right).
(b) The characteristic polynomial of A is det \left(A-tI_{3}\right)=-\left(t-1\right)^{2}\left(t-2\right).
So A has eigenvalues 1, 1 and 2. Since \left(A-I_{3}\right)\left(A-2I_{3}\right)=O_{3 \times 3} (see Example 3.11 of Sec. 3.7.6), A is diagonalizable and certainly, the corresponding T will represent a one-way stretch if \overrightarrow{x}\left(I_{3}-A\right)=\overrightarrow{x_{0}} has a solution. Solve
\overrightarrow{x}\left(I_{3}-A\right)=\overrightarrow{x_{0}}, i.e.
\left(x_{1},x_{2},x_{3}\right) \left[\begin{matrix} 2 & -4 & -2 \\ 1 & -2 & -1 \\ 1 & -2 & -1 \end{matrix} \right] =\left(-2,4,2\right)
\Rightarrow 2x_{1}+x_{2}+x_{3}=-2 does have a (in fact, infinite) solution.
Therefore, T is a one-way stretch with scale factor 2.
Note that \overrightarrow{x}\left(A-I_{3}\right)=\overrightarrow{0}\Leftrightarrow 2x_{1}+x_{2}+x_{3}=0. Take \overrightarrow{v_{1}}=\left(1,-2,0\right) and \overrightarrow{v_{2}}=\left(1,0,-2\right) as linearly independent eigenvectors corresponding to 1.
On the other hand, \overrightarrow{x}\left(A-2I_{3}\right)=\overrightarrow{0}\Leftrightarrow 3x_{1}+x_{2}+x_{3}=4x_{1}+x_{2}+2x_{3}=2x_{1}+x_{2}=0. Choose \overrightarrow{v_{3}}=\left(1,-2,-1\right) as an eigenvector corresponding to 2.
Since \overrightarrow{x_{0}}= -2\overrightarrow{v_{3}}, both \overrightarrow{x_{0}} and \overrightarrow{v_{3}} can be chosen as a direction of the one-way stretch T. To find an invariant point of T, let such a point be denoted as \alpha\overrightarrow{x_{0}}. By the very definition of a one-way stretch along \overrightarrow{x_{0}},
\alpha\overrightarrow{x_{0}}-T\left(\overrightarrow{0}\right)=\alpha\overrightarrow{x_{0}}-\overrightarrow{x_{0}}=k\left(\alpha\overrightarrow{x_{0}}-\overrightarrow{0}\right) with k = 2
\Rightarrow \left( since \overrightarrow{x_{0}} \neq \overrightarrow{0}\right)\alpha-1=k\alpha with k = 2
\Rightarrow \alpha=\frac{1}{1-k}=-1
\Rightarrow An invariant point \alpha\overrightarrow{x_{0}}=-\overrightarrow{x_{0}}=2\overrightarrow{v_{3}}
\Rightarrow The plane of invariant point is 2\overrightarrow{v_{3}}+\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}} \gg .
Notice that
\overrightarrow{x}=\left(x_{1},x_{2},x_{3}\right) \in 2\overrightarrow{v_{3}}+\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}} \gg
\Rightarrow x_{1}= 2+\alpha_{1}+\alpha_{2},x_{2}=-4-2\alpha_{1},x_{3}=-2-2\alpha_{2} for \alpha_{1},\alpha_{2} \in R
⇒ (by eliminating the parameters \alpha_{1} and \alpha_{2} ) 2x_{1}+x_{2}+x_{3}=-2
\Rightarrow \overrightarrow{x}\left(I_{3}-A\right)=\overrightarrow{x_{0}}
as we claimed before. In the affine basis C = \left\{2\overrightarrow{v_{3}},2\overrightarrow{v_{3}}+\overrightarrow{v_{1}},2\overrightarrow{v_{3}}+\overrightarrow{v_{2}},2\overrightarrow{v_{3}}+\overrightarrow{v_{3}}\right\},
\left[T\left(\overrightarrow{x}\right)\right]_{C}=\left[\overrightarrow{x}\right]_{C}PAP^{-1}=\left[\overrightarrow{x}\right]_{C}\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix} \right], where P = \left[\begin{matrix} 1 & -2 & 0 \\ 1 & 0 & -2 \\ 1 & -2 & -1 \end{matrix} \right].
See Fig. 3.62.
We propose the following questions:
Q1 Where is the image of a plane, parallel to 2x_{1}+x_{2}+x_{3} = −2, under T?
Q2 Where is the image of a plane or a line, nonparallel to 2x_{1}+x_{2}+x_{3} = −2, under T?Where do these two planes or lines intersect?
Q3 What is the image of the tetrahedron \Delta \overrightarrow{a_{0}} \overrightarrow{a_{1}} \overrightarrow{a_{2}} \overrightarrow{a_{3}} , where \overrightarrow{a_{0}}=\overrightarrow{0},\overrightarrow{a_{1}} =\left(1,0,-1\right),\overrightarrow{a_{2}}=2\overrightarrow{v_{3}}=\left(2,-4,-2\right) and \overrightarrow{a_{3}}=\left(-2,0,0\right), under T? What are the volumes of these two tetrahedra?
To answer these questions, we compute firstly that
A^{-1}=\frac{1}{2}\left[\begin{matrix} 4 & -4 & -2 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \end{matrix} \right].
This implies the inverse affine transformation of T is
\overrightarrow{x}=\left(T\left(\overrightarrow{x}\right)-\overrightarrow{x_{0}}\right)A^{-1}.
For Q1 A plane, parallel to 2x_{1}+x_{2}+x_{3}=-2, is of the form 2x_{1}+x_{2}+x_{3}=c which intercepts x_{1}-axis at the point \left(\frac{c}{2},0,0\right). Since 2x_{1}+x_{2}+x_{3}=-2 is the plane of invariant points of T, with scale factor 2, so the image plane (refer to Fig. 3.62) is
2x_{1}+x_{2}+x_{3}=-2+2 \left(c- \left(-2\right)\right)=2 \left(c+1\right),
where c is any constant.
We can prove this analytically as follows.
2x_{1}+x_{2}+x_{3}=\overrightarrow{x}\left[\begin{matrix} 2 \\ 1 \\ 1 \end{matrix} \right] =c
⇒ (let \overrightarrow{y}=T \left(\overrightarrow{x}\right)= \left(y_{1},y_{2},y_{3}\right) temporarily)
\left[\left(y_{1},y_{2},y_{3}\right)-\left(-2,4,2\right)\right]A^{-1}\left[\begin{matrix} 2 \\ 1 \\ 1 \end{matrix} \right] =c
⇒ ( since A^{-1}\overrightarrow{y^{*}_{0} }=\frac{1}{2}\overrightarrow{y^{*}_{0} } for \overrightarrow{y_{0}}=\left(2,1,1\right), see Note below )
\left[\left(y_{1},y_{2},y_{3}\right)-\left(-2,4,2\right)\right]A^{-1}\left[\begin{matrix} 2 \\ 1 \\ 1 \end{matrix} \right] =2c
\Rightarrow 2y_{1}+y_{2}+y_{3}-\left(-4+4+2\right)=2c
⇒ (change y_{1},y_{2},y_{3} back to x_{1},x_{2},x_{3} respectively)
2x_{1}+x_{2}+x_{3}=2\left(c+1\right)
as claimed above. Therefore, such a plane has its image a plane parallel to itself and, of course, to 2x_{1}+x_{2}+x_{3}=-2.
Note It is well-known that Ker \left(A-I_{3}\right)^{\bot} =Im \left(A^{*}-I_{3}\right) (see (3.7.32) and Ex. <A> 5 of Sec. 3.7.3). Since Ker \left(A-I_{3}\right)^{\bot} =\ll \left(2,1,1\right)\gg, so Im \left(A^{*}-I_{3}\right) =\ll \left(2,1,1\right)\gg holds. Since \left(A^{*}-I_{3}\right)\left(A^{*}-2I_{3}\right)=O_{3 \times 3}, so \overrightarrow{y_{0}} \left(A^{*}-2I_{3}\right)=\overrightarrow{0} where \overrightarrow{y_{0}}=\left(2,1,1\right). Therefore
\overrightarrow{y_{0}}A^{*}=2\overrightarrow{y_{0}}
\Rightarrow A \overrightarrow{y^{*}_{0} } =2\overrightarrow{y^{*}_{0} }
\Rightarrow A^{-1} \overrightarrow{y^{*}_{0} } =\frac{1}{2} \overrightarrow{y^{*}}
which can also be easily shown by direct computation.
For Q2 Take, for simplicity, the plane x_{1}-x_{2}-x_{3}=5. To find the image plane of this plane under T, observe that
x_{1}-x_{2}-x_{3}= \left(x_{1},x_{2},x_{3}\right) \left[\begin{matrix} 1 \\ -1 \\ -1 \end{matrix} \right] =5
⇒ (let \overrightarrow{y } = T\left( \overrightarrow{x } \right) temporarily) \left[\left(y_{1},y_{2},y_{3}\right)-\left(-2,4,2\right)\right] A^{-1}\left[\begin{matrix} 1 \\ -1 \\ -1 \end{matrix} \right] =5
\Rightarrow \left(y_{1}+2,y_{2}-4,y_{3}-2\right)\left[\begin{matrix} 5 \\ 1 \\ 1 \end{matrix} \right] =5\left(y_{1}+2\right)+y_{2}-4+y_{3}-2=5
⇒ (replace y_{1},y_{2},y_{3} by x_{1},x_{2},x ) 5x_{1}+x_{2}+x_{3} =1
which is the equation of the image plane.
The two planes x_{1}-x_{2}-x_{3}=5 and 5x_{1}+x_{2}+x_{3} =1 do intersect along the line x_{1}=1,x_{2}+x_{3}=-4 which obviously lies on the plane 2x_{1}+x_{2}+x_{3} =-2. Refer to Fig. 3.61.
For Q3 By computation,
T\left(\overrightarrow{a_{0}}\right)=T\left(\overrightarrow{0}\right) = \overrightarrow{x_{0}}=\left(-2,4,2\right),
T\left(\overrightarrow{a_{1}}\right)=\left(-2,4,2\right)+\left(1,0,-1\right)A=\left(-2,4,2\right)+\left(0,2,0\right)=\left(-2,6,2\right),
T\left(\overrightarrow{a_{2}}\right)=\left(-2,4,2\right)+\left(2,-4,-2\right)A=\left(-2,4,2\right)+\left(4,-8,-4\right)=\left(2,-4,-2\right),
T\left(\overrightarrow{a_{3}}\right)=\left(-2,4,2\right)+\left(-2,0,0\right)A=\left(-2,4,2\right)+\left(2,-8,-4\right)=\left(0,-4,-2\right).
These four points form a tetrahedron \Delta T\left(\overrightarrow{a_{0}}\right)T\left(\overrightarrow{a_{1}}\right)T\left(\overrightarrow{a_{2}}\right)T\left(\overrightarrow{a_{3}}\right) whose volume is equal to
det \left[\begin{matrix} T\left(\overrightarrow{a_{1}}\right)-T\left(\overrightarrow{a_{0}}\right) \\T\left(\overrightarrow{a_{2}}\right)-T\left(\overrightarrow{a_{0}}\right) \\ T\left(\overrightarrow{a_{3}}\right)-T\left(\overrightarrow{a_{0}}\right) \end{matrix} \right]= \left|\begin{matrix} 0 & 2 & 0 \\ 4 & -8 & -4 \\ 2 & -8 & -4 \end{matrix} \right| =16.
While \Delta \overrightarrow{a_{0}}\overrightarrow{a_{1}}\overrightarrow{a_{2}}\overrightarrow{a_{3}} has volume equal to
det \left[\begin{matrix} \overrightarrow{a_{1}}\\\overrightarrow{a_{2}} \\ \overrightarrow{a_{3}} \end{matrix} \right]=\left|\begin{matrix} 1 & 0 & -1 \\ 2 & -4 & -2 \\ -2 & 0 & 0 \end{matrix} \right| =8.
Therefore,
\frac{Volume of \Delta T\left(\overrightarrow{a_{0}}\right) \ldots T\left(\overrightarrow{a_{3}}\right) }{Volume of \Delta \overrightarrow{a_{0}} \ldots \overrightarrow{a_{3}}} =\frac{16}{8}=2=detA.
See Fig. 3.63.
