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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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