Question 3.32: Give an affine transformation T(x ) = x0 + x A, where A =[-1...
Give an affine transformation
T\left(\overrightarrow{x}\right)=\overrightarrow{x_{0}}+\overrightarrow{x}A, where A =\left[\begin{matrix} -1 & 6 & -4 \\ 2 & 4 & -5 \\ 2 & 6 & -7 \end{matrix} \right].
Try to determine \overrightarrow{x_{0}} so that T is a two-way stretch. In this case, determine the line of invariant points and the invariant plane.
A has characteristic polynomial det \left(A-tI_{3}\right)=-\left(t-1\right)\left(t+2\right)\times\left(t+3\right), so A has eigenvalues −2,−3 and 1 and thus A is diagonalizable.
Since
I_{3}-A=\left[\begin{matrix} 2 & -6 & 4 \\ -2 & -3 & 5 \\ -2 & -6 & 8 \end{matrix} \right]
has rank equal to 2, the range of I_{3}-A is of dimension two. The range of I_{3}-A is
\overrightarrow{x}\left(I_{3}-A\right)=\overrightarrow{y}, where \overrightarrow{x}=\left(x_{1},x_{2},x_{3}\right) and \overrightarrow{y}=\left(y_{1},y_{2},y_{3}\right)
\Leftrightarrow y_{1}=2x_{1}-2x_{2}-2x_{3} , y_{2}=-6x_{1}-3x_{2}-6x_{3}, y_{3}=4x_{1}+5x_{2}+8x_{3}
\Leftrightarrow y_{1}+y_{2}+y_{3}=0
\Leftrightarrow \left(replace y_{1},y_{2},y_{3} by x_{1},x_{2},x_{3} respectively\right)
Im \left(I_{3}-A\right)=\left\{\overrightarrow{x}=\left(\overrightarrow{x_{1}},\overrightarrow{x_{2}},\overrightarrow{x_{3}}\right)\mid x_{1}+x_{2}+x_{3}=0 \right\}.
So, any point \overrightarrow{x_{0}} so that \overrightarrow{x_{0}} \in Im \left(I_{3}-A\right) will work.
Solving \overrightarrow{x}\left(A-I_{3}\right)=\overrightarrow{0}, get the corresponding eigenvector \overrightarrow{v_{3}} = \left(1,4,-3\right). Solving \overrightarrow{x}\left(A+2I_{3}\right)=\overrightarrow{0}, get \overrightarrow{v_{1}} = \left(0,1,-1\right) and solving \overrightarrow{x}\left(A+3I_{3}\right)=\overrightarrow{0}, get \overrightarrow{v_{2}} = \left(1,0,-1\right).
For any \overrightarrow{x_{1}} such that \overrightarrow{x_{1}} \left(I_{3}-A\right)=\overrightarrow{x_{0}} holds, T is a two-way stretch with
the line of invariant points: \overrightarrow{x_{1}}+\ll \overrightarrow{v_{3}}\gg, and
the invariant plane: \overrightarrow{x_{1}}+\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg.
For example, take \overrightarrow{x_{0}} = \left(2,-3,1\right) and solve \overrightarrow{x}\left(I_{3}-A\right)=\left(2,-3,1\right) so that
\overrightarrow{x}=\left(x_{1},-3+4x_{1},2-3x_{1}\right)=\left(0,-3,2\right)+x_{1}\left(1,4,-3\right), x_{1} \in R
\Rightarrow \left(0,-3,2\right)+\ll \left(1,4,-3\right)\gg, where \overrightarrow{x_{1}}=\left(0,-3,2\right)
is the line L of invariant points. On the other hand, the plane
\left(0,-3,2\right)+\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg=\left\{\left(\alpha _{2},-3+\alpha _{1},2-\alpha _{1}-\alpha _{2}\right)\mid \alpha _{1},\alpha _{2} \in R\right\}
= \left\{\overrightarrow{x}=\left(x_{1},x_{2},x_{3}\right)\mid x_{1}+x_{2}+x_{3}=-1 \right\}
is an invariant plane.
Note Careful readers might have observed in Example 3.31 that, since \left(A-2I_{3}\right)\left(A-I_{3}\right)=O, Im \left(A-2I_{3}\right), Ker \left(A-I_{3}\right) and \ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg are the same plane 2x_{1}+x_{2}+x_{3}=0 and hence, the plane of invariant points is just a translation of it, namely 2x_{1}+x_{2}+x_{3}=-2. Refer to Fig. 3.62.
This is not accidental. This does happen in Example 3.32 too. Since
\left(A-I_{3}\right)\left(A+2I_{3}\right)\left(A+3I_{3}\right)=O_{3 \times 3}
\Rightarrow \left(since r\left(A-I_{3}\right)=2\right)Im\left(A-I_{3}\right)=Ker\left(A+2I_{3}\right)\left(A+3I_{3}\right)
\Rightarrow \left(since Ker \left(A+2I_{3}\right) \oplus Ker \left(A+3I_{3}\right) \subseteq Ker \left(A+2I_{3}\right) \left(A+3I_{3}\right)\right)
Im \left(A-I_{3}\right) = Ker \left(A+2I_{3}\right) \oplus \left(A+3I_{3}\right)=\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg.
This justifies that both Im \left(A-I_{3}\right) and \ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg are the same plane x_{1}+x_{2}+x_{3}=0 as indicated above. As a consequence, all the planes parallel to it are invariant planes. See Fig. 3.65.
In the affine basis C = \left\{\overrightarrow{x_{1}},\overrightarrow{x_{1}}+\overrightarrow{v_{1}},\overrightarrow{x_{1}}+\overrightarrow{v_{2}},\overrightarrow{x_{1}}+\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} -2 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 1 \end{matrix} \right] .
Try to answer the following questions:
Q1 What is the image of a plane parallel to the line of invariant points?
Q2 Is each plane containing the line of invariant points an invariant plane?
Q3 Where is the intersecting line of a plane, nonparallel to \overrightarrow{x_{1}}+ \ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg, with its image?
Q4 Let \overrightarrow{a_{0}} =\overrightarrow{x_{1}}=\left(0,-3,2\right), \overrightarrow{a_{1}} =\left(-1,-1,-1\right),\overrightarrow{a_{2}} =\left(1,-2,1\right) and \overrightarrow{a_{3}}=\overrightarrow{x_{1}}+\overrightarrow{v_{3}} =\left(1,1,-1\right). Find the image of the tetrahedron \Delta \overrightarrow{a_{0}}\overrightarrow{a_{1}}\overrightarrow{a_{2}}\overrightarrow{a_{3}} and compute its volume.
For these, note that the inverse transformation of T is
\overrightarrow{x}=\left(\overrightarrow{y}-\left(2,-3,1\right)\right)A^{-1}, where \overrightarrow{y}=T\left(\overrightarrow{x}\right) and A^{-1} =\frac{1}{6}\left[\begin{matrix} 2 & 18 & -14 \\ 4 & 15 & -13 \\ 4 & 18 & -16 \end{matrix} \right].
For Q1 Geometric intuition (see Fig. 3.65) suggests that the image of any such plane would be parallel to the plane. This can be proved analytically as follows.
A plane Σ parallel to the line L of invariant points has equation like \left(-4a_{2}+3a_{3}\right)x_{1}+a_{2}x_{2}+a_{3}x_{3}=b, where b \neq -3a_{2}+2a_{3},a_{2},a_{3} \in R
The condition b \neq -3a_{2}+2a_{3} means that L is not coincident on Σ. Then
∑ \parallel L
\Leftrightarrow \overrightarrow{x}\left[\begin{matrix} -4a_{2}+3a_{3} \\ a_{2} \\ a_{3} \end{matrix} \right]=b
\Leftrightarrow \left[\overrightarrow{y}-\left(2,-3,1\right)\right]A^{-1}\left[\begin{matrix} -4a_{2}+3a_{3} \\ a_{2} \\ a_{3} \end{matrix} \right] = \frac{1}{6}\left[\overrightarrow{y}-\left(2,-3,1\right)\right]\left[\begin{matrix} 10a_{2}-8a_{3} \\-a_{2}-a_{3}\\ 2a_{2}-4a_{3} \end{matrix} \right]=b
\Leftrightarrow \left(10a_{2}-8a_{3}\right)y_{1}-\left(a_{2}+a_{3}\right)y_{2}+\left(2a_{2}-4a_{3}\right)y_{3}=6b+25a_{2}-17a_{3}
\Rightarrow \left(let \alpha _{2}=-a_{2}-a_{3}, \alpha _{3}=2a_{2}-4a_{3}\right)
\left(-4 \alpha _{2}+3 \alpha _{3}\right) x_{1}+\alpha _{2} x_{2}+\alpha _{3} x_{3}=6b+25a_{2}-17a_{3},
where 6b+25a_{2}-17a_{3} \neq -18 a_{2}+12a_{3}+25a_{2}-17a_{3}=-3\alpha _{2}+2\alpha _{3}.Hence, the image plane T\left(∑\right)\parallel ∑\parallel L. See Note below.
Note (refer to Note in Example 3.31) It is easy to see that
∑ \parallel \ll \overrightarrow{v_{3}},\overrightarrow{v_{1}}\gg \Leftrightarrow a_{2}=a_{3}.
Hence, Σ has equation -ax_{1}+ax_{2}+ax_{3}=b, where a≠ 0, or equivalently, x_{1}-x_{2}-x_{3}=b with -\frac{b}{a} replacing by b ≠ 1. Let \overrightarrow{n} = \left(1,-1,-1\right) be the normal to these parallel planes. Then
\overrightarrow{v_{3}} \in Ker \left(A-I_{3}\right)=Im \left(A^{*}-I_{3}\right) ^{\bot} and
\overrightarrow{v_{1}} \in Ker \left(A+2I_{3}\right)=Im \left(A^{*}+2I_{3}\right) ^{\bot}
\Rightarrow \overrightarrow{n} \in Im \left(A^{*}-I_{3}\right) \cap Im \left(A^{*}+2I_{3}\right) = Ker \left(A^{*}+3I_{3}\right)
\Rightarrow \overrightarrow{n}A^{*}=-3\overrightarrow{n}
\Rightarrow A\overrightarrow{n^{*} } =-3\overrightarrow{n^{*} }
\Rightarrow A^{-1}\overrightarrow{n^{*} } =-\frac{1}{3}\overrightarrow{n^{*} }.
Similarly,
∑ \parallel \ll \overrightarrow{v_{3}},\overrightarrow{v_{2}}\gg \Leftrightarrow 2a_{2}-a_{3}=0.
\Rightarrow ∑ has equation 2x_{1}+x_{2}+2x_{3}=b, b ≠ 1.
\Rightarrow A^{-1}\overrightarrow{u^{*} }=-\frac{1}{2}\overrightarrow{u^{*} }, where \overrightarrow{u }= \left(2,1,2\right), the normal vector.
For Q2 The answer is negative in general except the planes \overrightarrow{x_{1}}+\ll \overrightarrow{v_{3}},\overrightarrow{v_{1}}\gg and \overrightarrow{x_{1}}+\ll \overrightarrow{v_{3}},\overrightarrow{v_{2}}\gg which are invariant planes. But each such plane and its image plane intersect along the line L of invariant points. Q_{1}answers all these claims.
For Q3 For simplicity, take x_{3}=0 as a sample plane. The plane x_{3}=0 intercepts the line L at the point \left(\frac{2}{3},-\frac{1}{3},0\right) and intersects the plane \overrightarrow{x_{1}}+\ll \overrightarrow{v_{1}},\overrightarrow{v_{2}}\gg along the line x_{3}=0, x_{1}+x_{2}+x_{3}=-1. To find its image under T,
x_{3}= \overrightarrow{x}\left[\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]=0
\Leftrightarrow \left[\overrightarrow{y}-\left(2,-3,1\right)\right]A^{-1}\left[\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right]=\frac{1}{6} \left[\overrightarrow{y}-\left(2,-3,1\right)\right]\left[\begin{matrix} -14 \\ -13 \\ -16 \end{matrix} \right]=0
\Rightarrow \left(replace \overrightarrow{y} by \overrightarrow{x}\right) 14x_{1}+13x_{2}+16x_{3}=5.
This image plane does intercept the line L at the invariant point \left(\frac{2}{3},-\frac{1}{3},0\right) and intersects the original plane x_{3}=0 along the line x_{3}= 0.14x_{1}+13x_{2}+16x_{3}=5.
For Q4 By computation,
T\left(\overrightarrow{a_{0}}\right)=\overrightarrow{x_{1}}=\left(0,-3,2\right),
T\left(\overrightarrow{a_{1}}\right)=\left(0,-3,2\right)+\left(-1,-1,-1\right)A=\left(-3,-19,18\right),
T\left(\overrightarrow{a_{2}}\right)=\left(0,-3,2\right)+ \left(1,-2,1\right)A=\left(-3,1,1\right),
T\left(\overrightarrow{a_{3}}\right)=\overrightarrow{x_{1}}+ \overrightarrow{v_{3}}=\left(1,1,-1\right).
Then,
The signed volume of \Delta \overrightarrow{a_{0}}\overrightarrow{a_{1}}\overrightarrow{a_{2}}\overrightarrow{a_{3}}
= \left|\begin{matrix} \overrightarrow{a_{1}}-\overrightarrow{a_{0}} \\ \overrightarrow{a_{2}}-\overrightarrow{a_{0}}\\ \overrightarrow{a_{3}}-\overrightarrow{a_{0}} \end{matrix} \right|=\left|\begin{matrix} -1 & 2 & -3 \\ 1 & 1 & -1 \\ 1 & 4 & -3 \end{matrix} \right| =-6, and
the signed volume of \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)
=\left|\begin{matrix} -3 & -16 & 16 \\ -3 & 4 & -1 \\ 1 & 4 & -3 \end{matrix} \right| =-36.
\Rightarrow \frac{ The signed volume of \Delta T\left(\overrightarrow{a_{0}}\right) \ldots T\left(\overrightarrow{a_{3}}\right)}{ The signed volume of \Delta\overrightarrow{a_{0}}\ldots \overrightarrow{a_{3}} } = \frac{-36}{-6}=6=detA.
