Let $A_{\theta}$ be rotation counterclockwise by $\theta$ as follows: $$A_\theta = \left[ \begin{matrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{matrix} \right]$$
We can check easy that $A^{-1}_{\theta}=A_{-\theta}.$
Indeed, using $\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin(\theta)$
$$A_{-\theta}=\left[ \begin{matrix} \cos(-\theta) & -\sin(-\theta)\\ \sin(-\theta) & \cos(-\theta)\end{matrix} \right]=\left[ \begin{matrix} \cos(\theta) & \sin(\theta)\\ -\sin(\theta) & \cos(\theta)\end{matrix} \right]=A^{-1}_{\theta}$$
- In general, could this hold for : $$ \forall n \in \mathbb{Z} ,\quad (A_{\theta})^{n}=A_{n\theta} (*) $$
i'll try by induction
Basis step; $n= 1 \implies A^{1}_{\theta}=A_{\theta}$
Induction step;Assume that, for n =k , (*) works,
show true for n=k+1
Indeed,
\begin{align*} (A_{\theta})^{n+1}&=(A_{\theta})^{n}(A_{\theta})^{1}\\ &=(A_{n\theta})(A_{\theta}) ( \text{ by assumption of induction })\\ \text{Since :} A_{\theta 1}A_{\theta 2}=A_{\theta1+\theta2} \\ &=(A_{(n+1)\theta}) \end{align*}
but this is just true for $n\in\mathbb{N}$ how can i prove it for $\mathbb{Z}$
Yes, it is true.
By defination of function $A_{n\theta}$ rotates $n\theta$ in counterclockwise direction.
And $A_\theta$ rotates with angle $\theta$ and $(A_\theta)^n$ is taking composion of $n$ times as composion of linear function is equivalent to multiplying the coressponding matrices.
Edit: Let $T$ be a linear transformation from $R^2$ to $R^2$. (which means that $T$ is such a function that $T(cu+v)=cT(u)+T(v)$ where $c\in R$ and $u,v\in R^2$).
And by linear algebra we know that linear transformation coresponds to matrix multiplication or vice versa i.e. $T(u)=A\vec{u}$.
From that point $$T\circ T \circ...T(u)=A^n \vec u$$ i.e composition of linear transformation corresponds to martix multiplication.
Now, Just notice that ratation with angle $n\theta$ is composion of $n$ times of rotation of $\theta$ and use the fact that rotation is a linear transformation.