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?

It helps to know that the set of $2\times 2$ real matrices of the form
$$ \begin{bmatrix}a & b \\ -b & a \end{bmatrix} $$
behave exactly like the complex numbers $a+bi$ under both addition and multiplication.
Your $M$ therefore corresponds to $\frac1{\sqrt2}+\frac1{\sqrt2}i$ which is $e^{\pi i/4}$.
The $2006$th power of this therefore corresponds to $e^{\frac{2006}{4}\pi i} = e^{\frac32\pi i} = -i $; in other words $({}^0_{1}\,{}^{-1}_{\;0})$.
