Hi, how would one approach proving this?
\(\displaystyle \mbox{Let }\, l_{\theta}\, \mbox{ be line in }\, \mathbb{R}^2\, \mbox{ going through }\, (0,\, 0)\, \mbox{ which makes}\)
\(\displaystyle \mbox{an angle }\, \theta\, \mbox{ above the positive }\, x\mbox{-axis. Consider the linear}\)
\(\displaystyle \mbox{transformation associated to the matrix:}\)
. . . . .\(\displaystyle \pi_{\theta}\, =\, \left(\, \begin{array}{cc}\cos^2(\theta)&\cos(\theta)\sin(\theta) \\ \cos(\theta)\sin(\theta)&\sin^2(\theta)\end{array}\,\right)\)
\(\displaystyle \mbox{For every point }\, v\, \in\, \mathbb{R}^2,\, \mbox{ show that }\, \pi_{\theta}(v)\, \mbox{ lies on the line }\, l_{\theta}.\)
My current understanding is that L theta is just the rotation transformation? Which is:
. . . . .\(\displaystyle R_{\alpha}\, =\, \left[\, \begin{array}{cc}\cos(\alpha)&-\sin(\alpha) \\ \sin(\alpha)&\cos(\alpha) \end{array}\, \right]\)
and then you apply a generic point (x,y) to both of these tranformations and prove they're equal? But I'm having trouble doing that.