Question

1. Answer the following questions

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a) Consider the linear transformation T(x) = proj_u(x), where u =  0

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Find the matrix for T.

b) Find the matrix for the linear transformation which reflects every vector in R² across the x-axis and then rotates every vector through an angle of 𝝅/6. (Show all working)

Answer 1

$a:\\T\left( x \right) =proj_u\left( x \right) =\frac{\left( u,x \right)}{\left\| u \right\| ^2}u=\frac{1x_1+0x_2+3x_3}{1^2+0^2+3^2}\left[ \begin{array}{c} 1\\ 0\\ 3\\\end{array} \right] =\\=\left[ \begin{array}{c} 0.1x_1+0.3x_3\\ 0\\ 0.3x_1+0.9x_3\\\end{array} \right] \\Matrix: A=\left[ \begin{matrix} 0.1& 0& 0.3\\ 0& 0& 0\\ 0.3& 0& 0.9\\\end{matrix} \right] \\b:\\The\,\,reflection: R\left( x,y \right) =\left( x,-y \right) \\The\,\,rotation: T\left( x,y \right) =\left( x\cos \frac{\pi}{6}-y\sin \frac{\pi}{6},x\sin \frac{\pi}{6}+y\cos \frac{\pi}{6} \right) =\\=\left( \frac{\sqrt{3}}{2}x-\frac{1}{2}y,\frac{1}{2}x+\frac{\sqrt{3}}{2}y \right) \\The\,\,composition:\\T\left( R\left( x,y \right) \right) =\left( \frac{\sqrt{3}}{2}x-\frac{1}{2}\left( -y \right) ,\frac{1}{2}x+\frac{\sqrt{3}}{2}\left( -y \right) \right) =\\=\left( \frac{\sqrt{3}}{2}x+\frac{1}{2}y,\frac{1}{2}x-\frac{\sqrt{3}}{2}y \right) \\Matrix: A=\left[ \begin{matrix} \frac{\sqrt{3}}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{\sqrt{3}}{2}\\\end{matrix} \right]$