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January 1st, 2015, 02:45 AM  #1 
Newbie Joined: Jan 2015 From: UK Posts: 1 Thanks: 0  Eigenvalues of rotation matrix
how can it be proved that the eigenvalues of the 2x2 rotation matrix i.e. cos(x) sin(x) sin(x) cos(x) are cos(x)+isin(x) and cos(x)isin(x)? 
January 5th, 2015, 10:41 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2203 
You can use the definition of the eigenvalues to calculate them.

January 9th, 2015, 05:09 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Or use the fact that the eigenvalues of the matrix $\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ satisfy the "characteristic equation", $\displaystyle \left\begin{array}{cc}a\lambda & b \\ c & d \lambda\end{array}\right= (a \lambda)(d \lambda) bc= 0$. In this particular case, you have $\displaystyle \left\begin{array}{cc}cos(\theta) \lambda & sin(\theta) \\ sin(\theta) & cos(\theta) \end{array}\right= (cos(\theta) \lambda)^2+ sin^2(\theta)= 0$ $\displaystyle cos^2(\theta) 2\lambda cos(\theta)+ \lambda^2+ sin^2(\theta)= \lambda^2 2cos(\theta)\lambda+ 1= 0$. That is a quadratic equation. Use the quadratic formula or complete the square to solve it. Last edited by Country Boy; January 9th, 2015 at 05:11 AM. 

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