Question 2.25: Study the affine transformation T(x^→ ) = (−2, 2√3) + x^→A, ...

Although A is orthogonal, i.e. A^{*}=A^{-1} , but its determinant det A = −1. Therefore, the associated T is not a rotation.

The characteristic polynomial of A is

det\left(A-tI_{2} \right)=\left|\begin{matrix} \frac{1}{2}-t & \frac{\sqrt{3} }{2} \\ \frac{\sqrt{3} }{2} & -\frac{1}{2}-t \end{matrix} \right| =t^{2}-1,

so A has eigenvalues 1 and −1. According to 1 in (2.8.27), the associated T is a reflection.

Solve

\overrightarrow{x} \left(A-I_{2} \right)=\left(x_{1}   x_{2} \right)\begin{bmatrix} -\frac{1}{2} & \frac{\sqrt{3} }{2} \\ \\ \frac{\sqrt{3} }{2} & -\frac{3}{2} \end{bmatrix}= \left(0   0\right)

and get the associated eigenvectors \overrightarrow{v}=t\left(\sqrt{3} , 1\right), for t ∈ R and t ≠ 0. Pick up a unit vector   \overrightarrow{v_{1}}=\frac{1}{2} \left(\sqrt{3} , 1\right). On the other hand, solve

\overrightarrow{x} \left(A+I_{2} \right)=\left(x_{1}   x_{2} \right)\begin{bmatrix} \frac{3}{2} & \frac{\sqrt{3} }{2} \\ \\ \frac{\sqrt{3} }{2} & \frac{1}{2} \end{bmatrix}= \left(0   0\right)

and get the associated eigenvectors \overrightarrow{v}=t\left(-1,\sqrt{3}\right). Pick up a unit vector   \overrightarrow{v_{2}}=\frac{1}{2} \left(-1,\sqrt{3}\right). This \overrightarrow{v_{2}} is the direction of reflection. Note that \overrightarrow{v_{2}} is perpendicular to   \overrightarrow{v_{1}}, so that A is an orthogonal reflection with respect to the axis \ll \overrightarrow{v_{1}}\gg.

To see if the original affine transformation T\left(\overrightarrow{x} \right)=\overrightarrow{x_{0} }+\overrightarrow{x}A is an orthogonal reflection with respect to a certain axis, this is our problem. Suppose the axis is of the form   \overrightarrow{a_{0} }+\ll \overrightarrow{v_{1}}\gg, then

T\left(\overrightarrow{x} \right)=\overrightarrow{x_{0} }+\overrightarrow{x}A= \overrightarrow{x_{0} }+\overrightarrow{a_{0} }A+\left(\overrightarrow{x} -\overrightarrow{a_{0} }\right)A

\Rightarrow \overrightarrow{a_{0} }=\overrightarrow{x_{0} }+\overrightarrow{a_{0} }A

\Rightarrow \overrightarrow{x_{0} }=\overrightarrow{a_{0} }\left(I_{2}-A \right) which implies that \overrightarrow{x_{0} }\left(I_{2}+A \right)= \overrightarrow{0}, i.e. \overrightarrow{x_{0} }A=-\overrightarrow{x_{0} } or \overrightarrow{x_{0} } is an eigenvector of A corresponding to −1.     (2.8.36)

For the given \overrightarrow{x_{0} }=\left(-2,2\sqrt{3} \right), solve

  \overrightarrow{x} \left(I_{2}-A \right) =\overrightarrow{x}\left|\begin{matrix} \frac{1}{2} & -\frac{\sqrt{3} }{2} \\ -\frac{\sqrt{3} }{2} & \frac{3}{2} \end{matrix} \right|=\left(-2,2\sqrt{3} \right) ,  where \overrightarrow{x}= \left(x_{1}, x_{2}\right)

and the axis is  x_{1}-\sqrt{3} x_{2}=-4. Of course, any point on it, say (−4, 0), can be used as \overrightarrow{a_{0} } and the axis is the same as \overrightarrow{a_{0} }+\ll \overrightarrow{v_{1}}\gg. See Fig. 2.109.

We can reinterpret the transformation \overrightarrow{x}\rightarrow T\left(\overrightarrow{x} \right) as follows.

 \overrightarrow{x}=\left(x_{1},x_{2} \right)\overset{orthogonal  reflection}{\underset{with  axis  \ll \overrightarrow{e_{1}}\gg }{\longrightarrow}}  \overrightarrow{x}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}

=\left(x_{1},-x_{2} \right)\overset{rotation  with}{\underset{center  at   \overrightarrow{0} and  through  60^{\circ } }{\longrightarrow}}  \overrightarrow{x}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3} }{2} \\ -\frac{\sqrt{3} }{2} & \frac{1}{2} \end{bmatrix}

= \overrightarrow{x} \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3} }{2} \\ \frac{\sqrt{3} }{2} & -\frac{1}{2} \end{bmatrix}

= \overrightarrow{x} A \overset{translation}{\underset{along  \overrightarrow{x_{0}}= \left(-2,2\sqrt{3} \right) }{\longrightarrow}}\overrightarrow{x_{0}}+\overrightarrow{x}A=T\left( \overrightarrow{x}\right).

Fig. 2.110 illustrates all the steps involved.

What is the image of the unit circle x^{2}_{1}+x^{2}_{2}=1? As it might be expected beforehand, since A, as a mapping, does keep the circle invariant while the transformation \overrightarrow{x}\longrightarrow \overrightarrow{x_{0}}+\overrightarrow{x} preserves everything, the image is the circle \left| \overrightarrow{x}-\overrightarrow{x_{0}}\right|=1. Formally, we prove this as follows.

\lt \overrightarrow{x},\overrightarrow{x} \gt=\overrightarrow{x}\overrightarrow{x^{*} } =x^{2}_{1}+x^{2}_{2}=1

\Rightarrow \left(T\left(\overrightarrow{x}\right)- \overrightarrow{x_{0}}\right)A^{-1}\left[\left(T\left(\overrightarrow{x} \right)-\overrightarrow{x_{0} } \right)A^{-1} \right] ^{*}=1

\Rightarrow \left(T\left(\overrightarrow{x}\right)- \overrightarrow{x_{0}}\right)A^{-1}\left(A^{-1}\right)^{*}\left(T\left(\overrightarrow{x}\right) -\overrightarrow{x_{0}}\right)^{*}=1

\Rightarrow \left( Since A^{-1}\left(A^{-1}\right)^{*}=A^{*}A=A^{-1}A=I_{2}\right)\left(T\left(\overrightarrow{x}\right)-\overrightarrow{x_{0}}\right)\left(T\left(\overrightarrow{x}\right)- \overrightarrow{x_{0}}\right)^{*}=1

\Rightarrow \left|T\left(\overrightarrow{x}\right)-\overrightarrow{x_{0}}\right|=1  if \left|\overrightarrow{x}\right|=1.

In order to avoid getting confused, it should be mentioned and cautioned that the transformation T\left(\overrightarrow{x}\right)=\overrightarrow{x_{0}}+\overrightarrow{x}A in (2.8.26) is derived from T\left(\overrightarrow{x}\right)=\overrightarrow{a_{0}}+\left(\overrightarrow{x}-\overrightarrow{a_{0}}\right), where B=A^{-1}\left[T\right] _{N}A, so that

\overrightarrow{x_{0}}=\overrightarrow{a_{0}}-\overrightarrow{a_{0}}B=\overrightarrow{a_{0}}\left(I_{2}-B\right).

This implies explicitly why 3 in (2.8.27) and (2.8.29) should hold. But this is not the case in Example 2.25 where T\left(\overrightarrow{x}\right)=\overrightarrow{x_{0}}+\overrightarrow{x}A is given, independent of the form \overrightarrow{a_{0}}+\left(\overrightarrow{x}-\overrightarrow{a_{0}}\right)B. That is why we have to recapture \overrightarrow{a_{0}} from \overrightarrow{x_{0}} in (2.8.36).