Let $\mathbf{A}$ be two by two matrix [sqrt(3)/2, -1/2; 1/2, sqrt(3)/2]. Then what is $\mathbf{A}^{2018} \begin{pmatrix} 2 \\ 2 \end{pmatrix}?$
I am stuck and cannot think of a method of simplifying.
I know that A * \begin{pmatrix} 2 \\ 2 \end{pmatrix} is equal to (sqrt(3)-1;sqrt(3)+1). I tried to do A^2017(A*\begin{pmatrix} 2 \\ 2 \end{pmatrix}) but it doesn't lead me anywhere.
$$ A=\pmatrix{\sqrt3/2 & -1/2\\1/2 & \sqrt3/2\\}= \pmatrix{\cos\pi/6 & -\sin\pi/6\\\sin\pi/6 & \cos\pi/6 \\} $$ is a rotation matrix, corresponding to an angle of $30°$. Hence $A^6=-1$ and $A^{2018}=A^2$.