I've been asked to diagonalize a matrix A such that $A^n = MD^nM^{−1}$, where A is known and the eigenvalues of the matrix are: $e^{^+_- \pi/3}$ I already know A to be:
$$A=\begin{bmatrix} 0 & 1\\ -1 & 1\\\end{bmatrix}$$
and I know that the eigenvalues of the matrix are given by: $e^{\pm\pi/3}$, from this I've easily found that $$D = \begin{bmatrix} e^{- \pi/3} & 0 \\ 0 & e^{\pi/3}\\\end{bmatrix},$$ and therefore $$D^n= \begin{bmatrix} e^{- n\pi/3} & 0 \\ 0 & e^{+ n\pi/3}\\\end{bmatrix}.$$
But I am currently struggling to produce the associated eigenvectors as to make M and $M^{-1}$, would anyone be able to help me find the associated eigenvectors and as such also help me find M and $M^{-1}$?
In addition I'm then asked to show that $$A^n = \frac{1}{\sin(\pi/3)}\begin{bmatrix} -\sin(n-1)\pi/3 & \sin(n\pi/3) \\ -\sin(n\pi/3) & \sin(n+1)\pi/3\\\end{bmatrix}.$$ I'm sure once I know M and $M^{-1}$ this will be a simple case 0f working through the steps but if anyone can help me with this also it would be much appreciated as matrix manipulation is definitely one of my weakest topics.
Many thanks!!
 A: The characteristic equation of $A$ is found as
$$
\mbox{det}(A - \lambda I) = 0
$$
i.e.
$$
\lambda^2 - \lambda + 1 = 0 \tag{1}
$$
Solving (1), we get complex conjugate eigenvalues of $A$ :
$$
\lambda_1 = {1 \over 2} - {\sqrt{3} \over 2} j = e^{-{\pi  \over 3} j} , \ \ 
\lambda_2 = {1 \over 2} + {\sqrt{3} \over 2} j = e^{{\pi  \over 3} j}  \tag{2}
$$
A simple calculation yields the corresponding eigenvectors as
$$
\mathbf{v}_1 = \left[ \begin{array}{c}
     1 \\[2mm]
e^{-{\pi  \over 3} j}   \\[3mm]
\end{array} \right], \ \
\mathbf{v}_2 = \left[ \begin{array}{c}
     1 \\[2mm]
e^{{\pi  \over 3} j} \\[3mm]
\end{array} \right] \tag{3}
$$
The required modal matrix $P$ is
$$
P = \left[ \matrix{ \mathbf{v}_1, \mathbf{v}_2 \cr} \right]
$$
It is easy to verify that
$$
P^{-1} A P = D = \left[ \begin{array}{cc}
 \lambda_1 & 0 \\[2mm]
0 & \lambda_2 \\[2mm]
\end{array} \right] =  \left[ \begin{array}{cc}
e^{-{\pi  \over 3} j}  & 0 \\[2mm]
0 &   e^{{\pi  \over 3} j} \\[3mm]
\end{array} \right]  \tag{4}
$$
A: The eigenvalues of $A$ are not $e^{\pm\pi/3}$. The are the roots of the characteristic polynomial of $A$, which is $\lambda^2-\lambda+1$, which are $e^{\pm\pi i/3}$. If you solve the equation $A.v=e^{\pi i/3}v$, then a solution that you will get will be $\left(\frac{1}{2} \left(1-i\sqrt{3}\right),1\right)$, and if you solve the equation $A.v=e^{-\pi i/3}v$, then a solution that you will get will be $\left(\frac12\left(1+i   \sqrt{3}\right),1\right)$. So, take$$M=\begin{bmatrix}\frac{1}{2} \left(1-i\sqrt{3}\right)&\frac{1}{2} \left(1+i\sqrt{3}\right)\\1&1\end{bmatrix}.$$Then$$M^{-1}=\begin{bmatrix} \frac{i}{\sqrt{3}} & \frac{1}{6} \left(3-i\sqrt{3}\right) \\-\frac{i}{\sqrt{3}} & \frac{1}{6} \left(3+i\sqrt{3}\right)\end{bmatrix}$$and it turns out that$$M^{-1}.A.M=\begin{bmatrix}e^{\pi i/3}&0\\0&e^{-\pi i/3}\end{bmatrix},$$is one would have expected. And therefore\begin{align}(\forall n\in\Bbb N):A^n&=M.\begin{bmatrix}e^{\pi i/3}&0\\0&e^{-\pi i/3}\end{bmatrix}^n.M^{-1}\\&=M.\begin{bmatrix}e^{n\pi i/3}&0\\0&e^{-n\pi i/3}\end{bmatrix}.M^{-1}\\&=\begin{bmatrix}\cos \left(\frac{\pi n}{3}\right)-\frac{\sin \left(\frac{\pi n}{3}\right)}{\sqrt{3}} & \frac{2 \sin \left(\frac{\pi n}{3}\right)}{\sqrt{3}} \\-\frac{2 \sin\left(\frac{\pi  n}{3}\right)}{\sqrt{3}} & \cos\left(\frac{\pi  n}{3}\right)+\frac{\sin \left(\frac{\pi  n}{3}\right)}{\sqrt{3}}\end{bmatrix}.\end{align}
