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$H$ is the set of the matrices $A$ of the form $$A= \begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}$$ where $0\leq \theta < 2\pi$ is a group with respect to matrix multiplication.

Question part (1) Prove by induction that for any $\theta$ and any positive integer $n$,

$$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{n} = \begin{pmatrix} \cos(n\theta) & \sin (n\theta) \\ -\sin (n\theta) & \cos (n\theta)\end{pmatrix}$$

When $n =1$ we have $$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}$$

and clearly this holds for the base case.

Now I'm going to assume true when $n=k$ and try and prove for $k+1$. So we have

$$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{k}\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(k\theta + \theta) & \sin (k\theta + \theta) \\ -\sin (k\theta + \theta) & \cos (k\theta + \theta)\end{pmatrix}$$

Now I'm sure I can cheese this, but is there some clever way to pull this apart? I want to just be able to say something like the determinant of the left side is $1^{k+1}$ and its the same on the right ...

Question part (2) Find an element of $H$ of order 5.

Here I'm totally lost, I seem to think that every element is of order 1 cause the determinant is 1.

Question part (3) Find an element of $H$ of infinite order.

Again I'm lost; I want to somehow make the determinant 0 for this but it can never be 0 for any $\theta$.

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In your last equation write the first matrix in terms of $k\theta$ and just multiply the two matrices and apply trig identities for sum an difference. – Maesumi Mar 3 '13 at 5:36
These are rotation matrices, order 5 should make you go around the circle once. Infinite order would be angle that no matter how many times you rotate by it you never come back where you started. Can you guess which angles are like that? – Maesumi Mar 3 '13 at 5:38
up vote 3 down vote accepted


For part 1, you didn't actually use the inductive hypothesis yet. Remember that you are assuming that the claim is true when $n=k$, i.e. that $$\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{k} = \begin{pmatrix} \cos(k\theta) & \sin (k\theta) \\ -\sin (k\theta) & \cos (k\theta)\end{pmatrix}.$$ Thus, what you want to prove can be rewritten as $$\begin{pmatrix} \cos(k\theta) & \sin (k\theta) \\ -\sin (k\theta) & \cos (k\theta)\end{pmatrix}\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(k\theta + \theta) & \sin (k\theta + \theta) \\ -\sin (k\theta + \theta) & \cos (k\theta + \theta)\end{pmatrix}.$$ You'll want to just multiply this out and use some trig identities.

For parts 2 and 3, you should note that multiplying a vector $\begin{pmatrix} a \\ b\end{pmatrix}$ in the plane $\mathbb{R}^2$ by the matrix $\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}$ produces the vector that is $\begin{pmatrix} a \\ b\end{pmatrix}$, except now rotated counterclockwise by an angle of $\theta$.

Can you think of an angle $\theta$ such that rotating the plane around the origin by $\theta$ 5 times in a row is equivalent to doing nothing?

Can you think of an angle $\theta$ such that no matter how many times you rotate the plane by it, it never comes back to its original orientation?

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For part 2 i guess i can think of an angle that does that $2\pi/5$ and $\theta$ =0 for the second part cause it doesn't do anything no matter how many times u multiply by it. Interesting it reminds me of the idea of index's in a closed orbit in a dynamical system. just that this idea is alot more complicated haha – Faust7 Mar 3 '13 at 6:10
You're correct for part 2, but not quite for part 3; since $\theta=0$ corresponds to doing nothing, already applying it once produces the plane in its original orientation. In other words, the matrix $$\begin{pmatrix} \cos(0) & \sin(0)\\ -\sin(0) & \cos(0)\end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$$ is the identity element of the group, and therefore has order 1, not order $\infty$. For part 3, try instead taking an angle $\theta$ that is not equal to $\frac{a}{b}\cdot \pi$ for any rational number $\frac{a}{b}$. – Zev Chonoles Mar 3 '13 at 6:19
lol wut? you want me to use an irrational, i think thats because it will make it impossible to for nl$\pi$ n is an rational number an l is irrational so nl/2 cannont be an element of the integers. but that leads me to an intresting point of confusion 2$\pi$/5 is not an integer is there something of order 5 that is? – Faust7 Mar 3 '13 at 6:36
Sorry for the confusion, my suggestion was intended for part 3, not for part 2; I've edited my comment above now. However, there is a very big difference between $\theta$'s which are rational multiples of $\pi$ (i.e. numbers of the form $\frac{a}{b}\cdot \pi$ where $\frac{a}{b}$ is a rational number) and numbers which are not. Note that if I rotate by $\theta$ $n$ times in a row, that corresponds to rotating once by $n\theta$. Then note that a rotation by an angle $\alpha$ only returns you to the starting orientation if $\alpha$ is a multiple of $2\pi$. – Zev Chonoles Mar 3 '13 at 6:41
Thus, if rotating by $\theta$ $n$ times returns you to the starting orientation, then... – Zev Chonoles Mar 3 '13 at 6:41

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