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$\textbf{Question}$: For $0<t<\pi$, the matrix

$$ \left( \begin{array}{cc} \cos t & -\sin t \\ \sin t & \cos t \\ \end{array} \right) $$ has distinct complex eigenvalues $\lambda_1$ and $\lambda_2$. For what value of $t$, where $0< t< \pi$, is $\lambda_1+\lambda_2=1$?

$\textbf{My work}$: I guessed and checked when $t=\pi/2$ (with the two eigenvalues being $\pm i$) and when $t=\pi/4$ (with the sum of the eigenvalues being $\sqrt{2}$).

How does one work through this problem? Is there any geometric insight in finding a unique $t$ which satisfies the above condition?

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  • $\begingroup$ Another way to think about this is that $\lambda_1 = e^{it}$ and $\lambda_2 = e^{-it}$ you want $e^{it}+e^{-it}=1$ divide by $2$ to see $\frac{1}{2}(e^{it}+e^{-it})=\cos(t) = 1/2$ which brings us back to the other solutions. Geometrically, the eigenvalues are on the unit-circle in the complex plane. The condition $0 < t < \pi$ forces $\lambda_1$ to be where $y>0$ and $\lambda_2$ to be where $y<0$. Adding the $t=\pi/3$ terms allows the real parts to sum to one and the imaginary parts cancel. $\endgroup$ – James S. Cook Jun 5 '13 at 21:51
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The sum of the eigenvalues is the trace. So we need $2\cos t=1\implies \cos t=\frac{1}{2}.$ $\frac{\pi}{3}$ is a solution.

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  • $\begingroup$ Thanks JLA! I think I will prove this fact rather than memorize it. $\endgroup$ – user81136 Jun 5 '13 at 21:19
  • $\begingroup$ Yes that's a good idea. $\endgroup$ – JLA Jun 5 '13 at 21:20
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The characteristic polynomial of the matrix is $x^2-2(\cos t )x +1$, so it's roots are:

$$\frac{2\cos t\pm\sqrt{4\cos^2t-4}}{2}=\cos t\pm\sqrt{\cos^2t-1}$$

Therefore, their sum is $2\cos t$. Solving the equation $2\cos t=1$, we see that $t=\frac{\pi}{3}$.

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  • $\begingroup$ Thanks Jared! I didn't think about using the characteristic polynomial. Of course! $\endgroup$ – user81136 Jun 5 '13 at 21:24

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