Problems

Solution:

1. Suppose \begin{align*} && \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \\ \Rightarrow && \binom{x_1}{y_1} &= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \binom{x_2}{y_2} \end{align*} Therefore if \(\frac{x_1^2}9+\frac{y_1^2}{4} = 1\) we must have \begin{align*} \frac{(\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2 }{9} + \frac{(-\frac{x_2}{\sqrt{2}}+\frac{y_2}{\sqrt{2}})^2}{4} = 1 \end{align*} but this is precisely the statement that \((x_1, y_1)\) is on \(G_1\) is equivalent to \((x_2,y_2)\) being on the \(G_2\). Since the point \((x_2,y_2)\) is a \(45^{\circ}\) rotation of \((x_1,y_1)\) anticlockwise about the origin, this means \(G_2\) is a \(45^{\circ}\) anticlockwise rotation of \(G_1\).
2. 1. \begin{align*} && \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -0.6 x + 0.8y \\ 0.8x + 0.6y \end{pmatrix} &= \begin{pmatrix} x \\ y \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} -1.6 x + 0.8y \\ 0.8x -0.4y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \Rightarrow && y &=2 x \end{align*}
   2. 

      + - ↺
3. Consider the transformation \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) which is a shear, leaving the \(x\)-axis invariant. Then we must have:
   

   + - ↺
   Since the shear leaves lines of the form \(y = k\) invariant, the points where \(\frac{\d y}{\d x} = 0\) must also map to points where this is true, ie \((\tfrac{\pi}{2}, 1), (\tfrac{3\pi}{2}, -1)\) map to points \((\tfrac{\pi}{2}+2,1), (\tfrac{3\pi}{2} -2,-1)\) where the tangent is horizontal. The line \(x = c\) map back to lines \(\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix} \begin{pmatrix} c \\ t\end{pmatrix} = \begin{pmatrix}c - 2t \\ t \end{pmatrix}\), ie \(y = -\frac12 x- \frac{c}{2}\). Therefore we are interested in points on the original curve where the gradient is \(-\frac12\), ie \((\frac{2\pi}{3}, \frac{\sqrt{3}}{2}), (\frac{4\pi}{3}, -\frac{\sqrt{3}}{2})\), these map to \((\frac{2\pi}{3}+\sqrt{3},\frac{\sqrt{3}}{2}), (\frac{4\pi}{3}-\sqrt{3}, -\frac{\sqrt{3}}{2})\)

Solution:

1. Suppose \((x,y)\) is on the line of invariant points, then \begin{align*} &&\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &&&= \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} \\ \Rightarrow && \begin{cases} (a-1)x + by = 0 \\ (cx + (d-1)y = 0 \end{cases} \tag{*} \end{align*} Therefore either \(x = 0, y = 0\) or \((a-1)(d-1)-bc = 0\) \(\Rightarrow ((a-1)(d-1)-bc)xy = 0\). We also know this is true for all values \(x,y\) on the line of invariant points. If there is one where both \(x \neq 0, y \neq 0\) we are done, otherwise the line of invariant points must be one of the axes. ie but then one of \(\begin{pmatrix} a \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) or \(\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) is true and we'd also be done. If the line doesn't go through the origin then there are points on every line, not equal to the origin which are fixed. But then every point on those lines is fixed (since \(\mathbf{A}\) is a linear operator) and so every point is fixed. ie \(\mathbf{A} = \mathbf{I}\).
2. Suppose \((a-1)(d-1) -bc = 0\) and \(b \neq 0\) then I claim that \(y = \frac{1-a}{b}x\) is a line of invariant points. It's clear that the first equation will be satisfied in \((*)\) so it suffices to check the second, but the first condition is equivalent to the equations being linearly dependent, ie both equations are satisfied. If \(b = 0\) then \((a-1)(d-1) = 0\), so our matrix must look like \(\begin{pmatrix} 1 & 0 \\ c & d\end{pmatrix}\) (if \(d \neq 1\))or \(\begin{pmatrix} * & 0 \\ * & 1\end{pmatrix}\). In the first case, the line \(y = \frac{c}{1-d}x\) and in the second \(x = 0\) is an invariant line.
3. Suppose the invariant line is \(y = mx+k\) then we must have that \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} &= \begin{pmatrix} (a + mb)x + bk \\ (c+dm)x + dk \end{pmatrix} \end{align*} and \((c+dm)x + dk = m((a + mb)x + bk) +k \Rightarrow k(d-mb-1) = x(-c+(a-d)m+m^2b)\) Since this equation must be true for all values of \(x\), and \(k \neq 0\) we can say that \(mb = d-1\) and \(-c+(a-d)m+m^2b = 0\), ie \(-c + (a-d)m + m(d-1) = 0 \Rightarrow (a-1)m-c = 0\) if \(m \neq 0\) then \((a-1)\frac{(d-1)}{b} - c = 0\) ie our desired relation is true. If \(m = 0\) then we must have that \(y = k\) is an invariant line, ie \(d-1=0\) and \(c=0\) which also satisfies our relation.