Find smallest $n \in \mathbb{N}$ s.t $A^n=I$ Let $A$ be $2 \times 2$ matrix: 
$$
\left(  
\begin{matrix} 
\sin\frac{\pi}{18} \\ \sin\frac{4\pi}{9} 
\end{matrix} 
\begin{matrix} 
-\sin\frac{4\pi}{9} \\ \sin\frac{\pi}{18} 
\end{matrix} \right) $$
Then find smallest  $n \in \mathbb{N}$ s.t $A^n=I$.
Looking at this question I seems I am missing some concept of of linear algebra without which it cannot be solved. How should one approach this problem?
 A: Hint : $\sin(\frac{\pi}{18})=\cos(\frac{4\pi}{9})$
Thus you can rewrite $A$ as
 $$A= \left(  \begin{matrix} \cos\frac{4\pi}{9} \\ \sin\frac{4\pi}{9} \end{matrix} \begin{matrix} -\sin\frac{4\pi}{9} \\ \cos\frac{4\pi}{9} \end{matrix} \right) $$
This type is a very typical rotation in $\mathbb R^2$. To illustrate this, you need just to suppose an arbitrary vector in $\mathbb R^2$, like 
$$\xi=\left( \begin{array} \\x \\y \end{array} \right)$$
And just think what would happen if you left multiply $\xi$ with $A$ ? Try to calculate it on your own, you'll definitely find that the result $A\xi$ , also a vector in $\mathbb R^2$, is exactly $\xi$ rotated anti-clockwise by the angle $\frac{4\pi}{9}$! (In fact, this naturally holds for any angle $\theta$ as you shall find in your calculation)
Thus, every time you right multiply an $A$ to a vector in $\mathbb R^2$, you rotate it anti-clockwise by $\frac{4\pi}{9}$. If you do it $n$ times, it will be $A^n$, which rotates the vector anti-clockwise by $\frac{4n\pi}{9}$. Now, think of the identical matrix $I$. Think about it-- What does "identical" imply? It must leave the vector unchanged after the operation. Then, what can do that? A $2k\pi$ anti-clockwise rotation of course! So what do we have to do at all? Just find a smallest $n$, such that
$$\frac{4n\pi}{9}=2k\pi$$
(need I add, $k\in \mathbb Z$ of course ). Therefore, the requested $n=9$ when $k=2$ (and $k=0$ has been automatically excluded.)
A: Computing directly gives that the eigenvalues of $A$ are $\lambda_{\pm} := \exp \left(\pm \frac{4 \pi i}{9}\right)$, so there is a some matrix $P$ such that
$$A := P^{-1} \begin{pmatrix} \lambda_+ & 0 \\ 0 & \lambda_-\end{pmatrix} P.$$
So, for any integer $n$,
$$A^n = (P^{-1} A P)^n = P^{-1} \begin{pmatrix} \lambda_+ & 0 \\ 0 & \lambda_-\end{pmatrix}^n P = P^{-1} \begin{pmatrix} \lambda_+^n & 0 \\ 0 & \lambda_-^n \end{pmatrix} P.$$ Now, $\lambda_{\pm}^n = \exp\left(\pm \frac{4 \pi i n}{9}\right)$, so the smallest $n$ for which $\lambda_{\pm}^n = 1$ is $n = 9$, and for this $n$, we thus have $A^n = P^{-1} I^n P = I$. The only matrix similar to $I$ is $I$ itself, so no smaller positive $n$ satisfies the equation.
