In Artin's Algebra 2011 we have Lemma 6.3.5:

"An isometry $f$ that has the form $m=t_a\rho_\theta$, with $\theta\neq 0$, is a rotation through the angle $\theta$ about a point in the plane."

Earlier we have the definitions of $t_a$ and $\rho_\theta$:

translation $t_a$ by a vector $a$: $t_a(x)=x+a=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}+\begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$

rotation $\rho_{\theta}$ by an angle $\theta$ about the origin: $\rho_{\theta}(x)=\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$

So now I am trying to figure out how to write a rotation about a given point, and by a given angle, as a product $t_a\rho_\theta$.

If I have a rotation of $\frac{\pi}{2}$ (counterclockwise) about the point $(1,1)^{t}$, applied to some point $x=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$, then

$t_a = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}+\begin{bmatrix} 1 \\ 1 \end{bmatrix}=\begin{bmatrix} x_1+1 \\ x_2+1 \end{bmatrix}$ correct?


$\rho_{\theta}=\begin{bmatrix} \cos(\frac{\pi}{2}) & -\sin(\frac{\pi}{2}) \\ \sin(\frac{\pi}{2}) & \cos(\frac{\pi}{2}) \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} -x_2 \\ x_1 \end{bmatrix}$ correct?

So $t_a\rho_\theta=\begin{bmatrix} x_1+1 \\ x_2+1 \end{bmatrix}\begin{bmatrix} -x_2 \\ x_1 \end{bmatrix}$

But I have definitely understood something incorrectly since I can't multiply these together. Any hints?

  • $\begingroup$ Translation is not a linear operator, so you can't write $t_a$ as multiplication by a matrix. Thus, the composition $t_a\circ\rho_\theta$ cannot be written as matrix multiplication. Try the problem using composition notation (i.e., $t_a\circ\rho_\theta(x_1,x_2)^t=t_a(\rho_\theta(x_1,x_2)^t)=\ldots$). $\endgroup$ – D Wiggles Nov 8 '13 at 19:38
  • $\begingroup$ The problem statement says "write the formula for the rotation as a product $t_a\rho_\theta$", is the statement incorrect then? So it's actually composition, and not a product? $\endgroup$ – user101600 Nov 8 '13 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.