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For any real number $\Theta$, we say $$R(\Theta)=\begin{bmatrix} \cos\Theta &-\sin\Theta \\ \sin\Theta & \cos\Theta \end{bmatrix}$$

Show that $(R(\Theta))^n= R(n\Theta)$ for any $\Theta \in \mathbb{R}$ and for any $n\in \mathbb{N}$

I know it makes sense; I just don't know which linear property to use to prove it in a more generic way.

Hmm, I'm even more confused now. I tried $(R(\Theta))^3$, so I multiplied $$(R(\Theta))^3=\begin{bmatrix} \cos2\Theta &-\sin2\Theta \\ \sin2\Theta & \cos2\Theta \end{bmatrix}\begin{bmatrix} \cos\Theta &-\sin\Theta \\ \sin\Theta & \cos\Theta \end{bmatrix}$$

but I did not end up with the expected

$$\begin{bmatrix} \cos3\Theta &-\sin3\Theta \\ \sin3\Theta & \cos3\Theta \end{bmatrix}$$

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  • $\begingroup$ i don't think so $\endgroup$ – spexel Oct 15 '15 at 16:28
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You may Try by induction on $n$. Then use the formulas: $$ \cos(a+b)= \cos(a) \cos(b) - \sin(a)\sin(b)$$ and $$ \sin(a+b)= \cos(a) \sin(b) + \sin(a)\cos(b)$$ Noting that $ (n+1) \theta = n \theta + \theta $.

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  • $\begingroup$ I'm still very confused :S $\endgroup$ – spexel Oct 15 '15 at 18:31
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    $\begingroup$ for $n=1$ , the result clearly holds. Now suppose the result holds for $n$ ( i.e. $(R(\Theta))^n= R(n\Theta)$ ), and required to show it holds for $n+1$. we have $$R( \Theta)^{n+1}=R( \Theta)^{n}*R( \Theta)= R( n\Theta)*R( \Theta)= \begin{bmatrix} \cos n\Theta &-\sin n\Theta \\ \sin n\Theta & \cos n\Theta \end{bmatrix}\begin{bmatrix} \cos\Theta &-\sin\Theta \\ \sin\Theta & \cos\Theta \end{bmatrix}= \begin{bmatrix} \cos n\Theta \cos \Theta- \sin n\Theta \sin\Theta & ... \\ ... & ... \end{bmatrix}$$ = \begin{bmatrix} \cos (n\Theta + \Theta) &... \\ ... & ...\end{bmatrix} $\endgroup$ – Nizar Oct 15 '15 at 19:05
  • $\begingroup$ $$= \begin{bmatrix} \cos (n+1)\Theta &... \\ ... & ...\end{bmatrix} = (R((n+1)\Theta)) $$. I leave for you to fill in the place of the dots. Hope this made the idea clear. $\endgroup$ – Nizar Oct 15 '15 at 19:07
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    $\begingroup$ got it, thank you, still a long way to go before it becomes intuitive but at least i got through this problem with your help. $\endgroup$ – spexel Oct 15 '15 at 19:38
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Prove a 'harder' theorem first:

$$ R(\Theta) R(\Phi) = R(\Theta + \Phi) $$

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Without spoiling the answer, try to regard the matrix $R$ not as a function $\mathbb R^2\rightarrow\mathbb R^2$, but as a function $\mathbb C\rightarrow\mathbb C$.

There are plenty of ways to prove the general statement but I consider the complex variant a nice demonstration of the surprising efficiency of complex numbers.

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Your matrix $R(\theta)$ is a rotation matrix by $\theta$. $R(\theta)^n$ is the repeated application, $n$ times, of this rotation, meaning you are rotation by a total angle of $\theta+ \theta+...\theta$(n times), i.e., by doing $R(\theta)^n$ you are rotatiog by a total of $n\theta$, which is equal to $R(n\theta)$.

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