How can we calculate the exponential form of a rotation matrix

Question

Considering the rotation matrix: $$ A(\theta) = \left( \begin{array}{cc} \cos\space\theta & -\sin \space\theta \\ \sin \space\theta & \cos\space\theta \\ \end{array} \right) $$

How can I calculate $(A(\theta))^n$ where n ≥ 1 ? I'm not sure what to do nor how to write it.

Answer 1

$A(\theta)$ is called the rotation matrix simply because it rotates a point in the plane by an angle $\theta$. What happens when I apply this matrix $n$ times? I simply rotate the point $n$ times. How much has it rotated in the end? By an angle $n\cdot\theta$. If you can get this sort of intuitive grasp, you can say the following:

Let $P_n$ be the property $$[A(\theta)]^n=A(n\theta),\ \ n\ge 1$$ Let's now prove by induction on $n$ that $P_n$ is true.

Base case

$P_1$ is trivial, so let's look at $P_2$: \begin{align}[A(\theta)]^2&=A(\theta)\cdot A(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\\ &=\begin{pmatrix}\cos^2\theta-\sin^2\theta&-2\sin\cos\theta\\2\sin\cos\theta&\cos^2\theta-\sin^2\theta\end{pmatrix}\\ &=\begin{pmatrix}\cos(2\theta)&-\sin(2\theta)\\\sin(2\theta)&\cos(2\theta)\end{pmatrix}\\ &=A(2\theta) \end{align} Therefore $P_1$ and $P_2$ are true.

Inductive Step

Assume $P_n$ is true, and consider the case of $P_{n+1}$: \begin{align} [A(\theta)]^{n+1}&=[A(\theta)]^n\cdot A(\theta)=A(n\theta)\cdot A(n\theta)\\ &=\begin{pmatrix}\cos(n\theta)&-\sin(n\theta)\\\sin(n\theta)&\cos(n\theta)\end{pmatrix}\cdot\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\\ &=\begin{pmatrix}\cos(n\theta)\cos\theta-\sin(n\theta)\sin(\theta)&-\cos(n\theta)\sin\theta-\sin(n\theta)\cos\theta\\\sin(n\theta)\cos\theta+\cos(n\theta)\sin\theta&-\sin(n\theta)\sin\theta+\cos(n\theta)\cos\theta\end{pmatrix} \end{align} Now we use the following four trigonometric identities: \begin{cases}\cos\theta\cos(n\theta)&=&\frac{1}{2}(\cos(\theta-n\theta)+\cos(\theta+n\theta))\\\sin\theta\sin(n\theta)&=&\frac{1}{2}(\cos(\theta-n\theta)-\cos(\theta+n\theta))\\ \cos\theta\sin(n\theta)&=&\frac{1}{2}(\sin(\theta+n\theta)-\sin(\theta-n\theta))\\ \sin\theta\cos(n\theta)&=&\frac{1}{2}(\sin(\theta+n\theta)+\sin(\theta-n\theta))\end{cases} Which allow us to write: $$[A(\theta)]^{n+1}=\begin{pmatrix}\cos(\theta+n\theta)&-\sin(\theta+n\theta)\\\sin(\theta+n\theta)&\cos(\theta+n\theta)\end{pmatrix}=A\left(\left(n+1\right)\theta\right)$$ Therefore $P_{n+1}$ is true.

Conclusion

We have shown that the property $P$ is true for $n=1$ (trivially; but also for $n=2$), and for $n+1$, if $P_n$ holds. Therefore, by the axiom of induction, $P_n$ is true for all $n\ge 1$. $$[A(\theta)]^n=A(n\theta),\ \ n\ge 1\quad\Box$$

Answer 2

Notice that $[A(\theta)]^n = A(n\theta)$, based on geometric interpretation of rotations. If you apply $A(\theta)$ twice, it is the same as rotating by $\theta$ twice, or $2\theta$. The same is true for an arbitrary number of rotations. That shows you have to write $A(\theta)^n$, namely as the same matrix with $\theta$ replaced by $n\theta$.

Answer 3

let us write $z = x+iy.$ then the differential equation can be written as $\frac{dz}{dt} = e^{i\theta}z, z(0) = z_0$ which has a solution $$z = e^{t\cos \theta}\left(\cos(t\sin\theta) + i\sin(t\sin\theta)\right)(x_0+iy_0)$$ splitting into real and imaginary we get $$x = e^{t\cos \theta}\left(x_0\cos(t\sin \theta) - y_0\sin (t\sin \theta)\right), \ y = e^{t\cos \theta}\left(x_0\sin(t\sin \theta) + y_0\cos (t\sin \theta)\right)$$ writing in matrix form $$e^{At} = e^{t\cos \theta} \pmatrix{ \cos(t\sin \theta) &- \sin (t\sin \theta)\\ \sin(t\sin \theta) & \cos (t\sin \theta) }$$