How Can Matrix Powers and Group Isomorphisms Illuminate Group Theory?

[Chris L T521]

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):

Spoiler

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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[Chris L T521]

This week's question was correctly answered by Sudharaka. You can find his answer below.

Spoiler

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.Dii) 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