Thus line of reflection pass through the origin the transformation is linear and could be expressed by 2 x 2 matrix. [br]1. Reflect base vectors e1, e2.[br]2. Images e1' and e2' are columns of the matrix of linear transformation.[br]3. Check the representation by means of the picture "lion".
Solution: [br]M={{0.8, 0.6},{0.6,-0.8}}[br]ApplyMatrix(M, lev)
Write the determinant of reflection in line
Find the fixed points FP and fixed directions FD
FP: x - 3y = 0[br]FD: (1,-3) and (3,1)
[b]Equation for fixed points: X = MX,[/b] i.e.[br][i]0.8x + 0.6y = x ⇒ [color=#1e84cc]−0.2x + 0.6y = 0[/color][br]0.6x − 0.8y = y ⇒ [color=#1e84cc]0.6x − 1.8y = 0[/color][/i][br]gives one linear property [i]x = 3y [/i]for mirroring line OA. [br][br][b]Equation for fixed directions ([url=https://www.wolframalpha.com/input?i=%7B%7B0.8%2C+0.6%7D%2C%7B0.6%2C+-0.8%7D%7D]eigenvectors[/url]): Mv = λv[/b], i.e.[br][i]0.8x + 0.6y = λx ⇒ [color=#1e84cc](0.8 − λ)x + 0.6y = 0[/color][br]0.6x − 0.8y = λy ⇒ [color=#1e84cc]0.6x − (0.8 + λ)y = 0[/color][/i][br][br]Dependent equations yield nontrivial solution: [i][color=#1e84cc]det(M − λE)[/color][/i] [color=#6fa8dc]= 0[/color]. [br][i][color=#1e84cc]det(M−λE)[/color][/i] = [color=#3d85c6][i]−[/i] 0.8[sup]2[/sup] + [i]λ[/i][sup]2 [/sup][i]−[/i] 0.6[sup]2[/sup][/color][br]Zero determinant for [color=#3d85c6][i]λ[/i][sup]2 [/sup]= 1. [/color][br][color=#3d85c6][i]λ[/i][sup] [/sup]= 1[/color] yields direction [i]x = 3y [/i]for mirroring line OA and [color=#3d85c6][i]λ[/i][sup] [/sup]= -1[/color] yields perpendicular direction. [br]