# Problem of the Week #19 - August 6th, 2012

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#### Chris L T521

##### Well-known member
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Thanks to those who participated in last week's POTW!! Here's this week's problem (I'm going to give group theory another shot).

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Problem: (i) Prove, by induction on $k\geq 1$, that

$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}^k = \begin{bmatrix}\cos(k\theta) & -\sin(k\theta)\\ \sin(k\theta) & \cos(k\theta)\end{bmatrix}.$

(ii) Prove that the special orthogonal group $SO(2,\mathbb{R}) = \{A\in O(2,\mathbb{R}) : \det A=1\}$ is isomorphic to the circle group $S^1$.

Remark: For part (ii), recall that the orthogonal group is defined as $O(2,\mathbb{R}) = \{A\in GL(2,\mathbb{R}): A^TA=AA^T = I\}$. I'll also provide a hint for part (ii):

Consider the map $\varphi:\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}\mapsto (\cos\theta,\sin\theta)$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka. You can find his answer below.

i) For $$k=1$$ the statement is obviously true. Suppose that the statement is true for $$k=p\in\mathbb{Z}^{+}$$. That is,

$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}^p = \begin{bmatrix}\cos p\theta & -\sin p\theta\\ \sin p\theta& \cos p\theta\end{bmatrix}$

Then,

\begin{eqnarray}

\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}^{p+1}&=&\begin{bmatrix} \cos p\theta & -\sin p\theta\\ \sin p\theta & \cos p\theta\end{bmatrix}\begin{bmatrix}\cos \theta & -\sin\theta\\ \sin\theta & \cos \theta\end{bmatrix}\\

&=&\begin{bmatrix}\cos p\theta\cos\theta-\sin p\theta\sin\theta & -\cos p\theta\sin\theta-\sin p\theta\cos\theta\\ \sin p\theta\cos\theta+cos p\theta\sin\theta & -\sin p\theta\sin\theta+\cos p\theta\cos\theta\end{bmatrix}\\

&=&\begin{bmatrix}\cos (p+1)\theta & -\sin (p+1)\theta\\ \sin (p+1)\theta& \cos (p+1)\theta\end{bmatrix}

\end{eqnarray}

Therefore the statement is true for $$n=p+1$$. By mathematical induction,

$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}^k = \begin{bmatrix}\cos(k\theta) & -\sin(k\theta)\\ \sin(k\theta) & \cos(k\theta)\end{bmatrix}\mbox{ for }k\in\mathbb{Z}^{+}$

Q.E.D

ii) The circle group can be defined as $$S^{1}=\left\{(\cos\theta, \sin\theta)\,:\,\theta\in[0,2\pi)\right\}$$ where the binary operation is given by,

$(\cos\theta_1, \sin\theta_1)(\cos\theta_2, \sin\theta_2)=(\cos[\theta_1+\theta_2],\sin[\theta_1+\theta_2])$

Define, $$\varphi: SO(2,\mathbb{R}) \rightarrow S^{1}$$ by $$\varphi:\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}\mapsto (\cos\theta,\sin\theta)$$

Take any, $$\begin{bmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{bmatrix},\,\begin{bmatrix}\cos \theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{bmatrix}\in SO(2,\mathbb{R})$$ such that,

$\begin{bmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{bmatrix}=\begin{bmatrix} \cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{bmatrix}$

Then,

$\cos\theta_1=\cos\theta_2\mbox{ and }\sin\theta_1=\sin\theta_2$

$\Rightarrow (\cos\theta_1,\sin\theta_1)=(\cos\theta_2,\sin \theta_2)$

$\therefore \varphi\begin{bmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{bmatrix}=\varphi\begin{bmatrix} \cos \theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{bmatrix}$

Hence $$\varphi$$ is a well defined function.

Take any, $$(\cos\theta_1,\sin\theta_1), (\cos\theta_2,\sin \theta_2)\in S^{1}$$ such that,

$(\cos\theta_1,\sin\theta_1)=(\cos\theta_2,\sin \theta_2)$

$\Rightarrow \cos\theta_1=\cos\theta_2\mbox{ and }\sin\theta_1=\sin\theta_2$

$\Rightarrow\begin{bmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{bmatrix}=\begin{bmatrix} \cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{bmatrix}$

Therefore $$\varphi$$ is injective.

Take any, $$(\cos\theta,\sin\theta)\in S^{1}$$. Then there exist $$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}\in SO(2,\mathbb{R})$$ such that,

$\varphi\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}=(\cos\theta,\sin\theta)$

Thereofore $$\varphi$$ is surjective.

Consider $$\varphi\left[\begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{pmatrix}\begin{pmatrix} \cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{pmatrix}\right]$$

\begin{eqnarray}

\varphi\left[\begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{pmatrix}\begin{pmatrix} \cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{pmatrix}\right]&=&\varphi\begin{bmatrix}\cos(\theta_1+\theta_2) & -\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2) & \cos(\theta_1+\theta_2)\end{bmatrix}\\

&=&(\cos[\theta_1+\theta_2],\sin[\theta_1+\theta_2])\\

&=&(\cos\theta_1, \sin\theta_1)(\cos\theta_2, \sin\theta_2)\\

&=&\varphi\begin{bmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{bmatrix}\,\varphi\begin{bmatrix} \cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{bmatrix}

\end{eqnarray}

Therefore $$\varphi$$ is a homomorphism.

$\therefore SO(2,\mathbb{R})\cong S^{1}$

Q.E.D

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