Calculating the matrix $M^{2006}$ Say you have the matrix $M$:
$$\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}$$
How do you find $M^{2006}$?  My thinking was that you can find that $M^8 = I$, so if $\frac{2006}{8} = 250\frac{3}{4}$, then $M^{2003} = I$, so if you multiply this by $M$, $3$ times, you would get $M^{2006}$.  Though, there seems to be something wrong with my arithmetic or else you cannot do this with matrix powers, as this is the incorrect answer.
The correct answer is:
$$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$
How do I get there?

$${{\left[ \begin{matrix}
   \cos \,\phi  & -\sin \,\phi   \\
   \sin \,\phi  & \cos \,\phi   \\
\end{matrix} \right]}^{n}}=\left[ \begin{matrix}
   \cos \,n\phi  & -\sin \ n\phi   \\
   \sin \,n\phi  & \cos \,n\phi   \\
\end{matrix} \right]$$
let $\phi=-\frac{\pi}{4}$ and $n=2006$
