Find standard matrix of linear transformation for circle

Question

Circle $x^2 + y^2 = 1 \rightarrow x^2 - 4xy + 5y^2 = 1$

Given linear transformation above, I need to find the standard matrix of such linear transformation, but I do not know how to start. If there is no such transformation matrix, I need to explain why. Any advice will be appreciated.

Answer 1

1. Find the Homogeneous Coordinate Matrix for the curve

   $$ \begin{aligned} \mbox{[circle]} & = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} & \mbox{[curve]} & = \begin{vmatrix} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0 & 0 & 1 \end{vmatrix} \end{aligned} $$

   such that $\pmatrix{x & y & 1} \mbox{[curve]} \pmatrix{x \\ y \\ 1} =0$ produces $x^2 -4 x y + 5 y^2-1 = 0$.

2. Find the rotation that diagonalizes the matrix

   Consider the rotation matrix $\mathrm{R}(\theta) \equiv \begin{vmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{vmatrix} $

   Find $\theta$ such that the inverse rotation on the curve is a diagonal matrix.

   $$ \mathrm{R}(\theta) \begin{vmatrix} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0 & 0 & 1 \end{vmatrix} \mathrm{R}(-\theta) = \mbox{[diagonal]}$$

   _{Remember if a transformation is to be applied to the circle, its inverse needs to be applied to the curve.}

   I found a solution at $\theta = - \frac{\pi}{8}$, but there are other solutions also.

3. Find the scaling that makes the circle match the rotated curve

   Consider the scaling $\mathrm{S}(s_x,s_y) \equiv \begin{vmatrix} s_x & 0 & 0 \\ 0 & s_y & 0 \\ 0 & 0 & 1 \end{vmatrix} $

   Calculate the rotated matrix and match is with the inverse scaling $$\mathrm{M} = \mathrm{R}(-\frac{\pi}{8}) \begin{vmatrix} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0 & 0 & 1 \end{vmatrix} \mathrm{R}(\frac{\pi}{8}) = \begin{vmatrix} 3 - 2\sqrt{2} & 0 & 0 \\ 0 & 2 \sqrt{2}+3 & 0 \\ 0 & 0 & -1 \end{vmatrix} $$

   This should represent the result of the inverse scaling of the circle

   $$\mathrm{M} = \begin{vmatrix} \sqrt{2}+1 & 0 & 0 \\ 0 & \sqrt{2}-1 & 0 \\ 0 & 0 & 1 \end{vmatrix}^\top \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \begin{vmatrix} \sqrt{2}+1 & 0 & 0 \\ 0 & \sqrt{2}-1 & 0 \\ 0 & 0 & 1 \end{vmatrix} $$

   which means the scaling is $s_x = (\sqrt{2}+1)^{-1} =\sqrt{2}-1 $ and $s_y = (\sqrt{2}-1)^{-1} = \sqrt{2}+1$

4. Verify the result

   $$ \mbox{[curve]} = \mathrm{R}(-\theta) \mathrm{S}(s_x,s_y) \mbox{[circle]} \mathrm{S}(s_x,s_y) \mathrm{R}(\theta) $$

   $$\begin{vmatrix} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0 & 0 & 1 \end{vmatrix} = \mathrm{R}(\frac{\pi}{8}) \mathrm{S}(\sqrt{2}-1,\sqrt{2}+1) \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \mathrm{S}(\sqrt{2}-1,\sqrt{2}+1) \mathrm{R}(-\frac{\pi}{8}) \;\checkmark$$