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Let $\theta_0, \theta_1 \in [0, 2\pi)$ and $\theta_0 \ne \theta_1$. Consider the rotation matrices $$M_0 = \left[ \begin{matrix}\cos(\theta_0) & -\sin(\theta_0) \\ \sin(\theta_0) & \cos(\theta_0) \\ \end{matrix} \right],M_1 = \left[ \begin{matrix}\cos(\theta_1) & -\sin(\theta_1) \\ \sin(\theta_1) & \cos(\theta_1) \\ \end{matrix} \right] \in SO(2). $$ Prove that $M_0$ and $M_1$ are similar if and only if $\theta_0+\theta_1=2\pi$.

I think I've proved the $"<="$ direction. Using that $\sin(2\pi-\theta_0) = -\sin(\theta_0)$ and $\cos(2\pi-\theta_0) = \cos(\theta_0)$; and with $P:= \left[ \begin{matrix} 1&0\\0&-1 \end{matrix}\right]=P^{-1} $ I have found that $$P^{-1}M_0P = M_1,$$so that $M_0$ and $M_1$ are similar. However, for the $"=>"$ direction, I feel like I am missing something. What conclusions can we draw from the fact that $M_0$ and $M_1$ are similar?

A hint would be much appreciated.

Michael Hardy
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rehband
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2 Answers2

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Hint: if two matrices are similar, the have the same trace.

mookid
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Right ok, so then $cos(\theta_0) = cos(\theta_1)$ which is the case if and only if $\theta_1 = 2\pi - \theta_0 <=> \theta_1 + \theta_0 = 2\pi$, right?

If two matrices have the same trace, are they already similar? Is there a characterisation of matrix similarity for special orthogonal matrices?

rehband
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