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January 1st, 2015, 02:45 AM   #1
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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)?
arsenal15 is offline  
January 5th, 2015, 10:41 AM   #2
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You can use the definition of the eigenvalues to calculate them.
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January 9th, 2015, 05:09 AM   #3
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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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