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Could you help me please and give some tips on how should I start solving this problem. How can one prove, that this equation is right, when n from $\mathbb{Z}$ and $\alpha$ is from $\mathbb{R}$?

$\begin{pmatrix} \cos \alpha & -\sin \alpha\\ \sin \alpha & \cos \alpha \end{pmatrix}^{n} = \begin{pmatrix} \cos n\alpha & -\sin n\alpha\\ \sin n\alpha & \cos n\alpha \end{pmatrix}$

Should I use induction?

Thank you in advance.

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    $\begingroup$ Accepting answers is easy. Just click on the empty tick mark to the left of the most helpful answer $\endgroup$ – Dima Thefencingdude Lukhtai Nov 2 '11 at 2:04
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Yes, induction should work. ${}$

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  • $\begingroup$ Thank you for response. Can you give me please some more tips on how should I move on? I used to prove by induction when n is from natural numbers, so I knew, where the start is. I'm a little bit confused by integers. $\endgroup$ – Lissa Nov 2 '11 at 2:11
  • $\begingroup$ First prove it by induction for positive $n$. Then either prove it by induction for negative $n$, or use what you know about the form of the inverse of a 2 x 2 matrix to prove the negative result from the positive. And please learn how to accept answers, or else people will stop giving any. $\endgroup$ – Gerry Myerson Nov 2 '11 at 3:28
  • $\begingroup$ Thank you. I've done it per induction. Will try later to do it by "positive result --> negative result". $\endgroup$ – Lissa Nov 2 '11 at 16:30
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You can also see that your matrix is the rotation matrix of α radians. Hence, if you rotate n times α radians you would end up with effectively a rotation of nα radians. Little tricky but slightly shorter.

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  • $\begingroup$ Thank you response. I think, I'm not "ripe" enough at the moment to do it this way. But I'll come back to this method later =) $\endgroup$ – Lissa Nov 2 '11 at 16:32

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