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?
 A: $${{\left[ \begin{matrix}
   \cos \,\phi  & -\sin \,\phi   \\
   \sin \,\phi  & \cos \,\phi   \\
\end{matrix} \right]}^{n}}=\left[ \begin{matrix}
   \cos \,n\phi  & -\sin \ n\phi   \\
   \sin \,n\phi  & \cos \,n\phi   \\
\end{matrix} \right]$$
let $\phi=-\frac{\pi}{4}$ and $n=2006$
A: 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 $({}^0_{1}\,{}^{-1}_{\;0})$.
A: You have $M^{2000}=(M^8)^{250}=I^{250}=I$, so you just need to find $M^6$
A: The overall approach is right, and quite a nice approach. But your issue is that $M^{2003} \neq I$ (I cannot tell why the division made you think that; hopefully you can find your misunderstanding there). In fact, $$M^{2003} = M^{2000 + 3} = M^{2000}M^3 =  (M^8)^{250}M^3 = I^{250} M^3 = M^3.$$ 
This means you can modify your approach: rather than starting at $I \neq M^{2003}$ and multiplying by $M^3$, you can start at $M^3 = M^{2003}$ and multiply by $M^3$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
M & = \pars{\begin{array}{cc}
\ds{1 \over \root{2}} & \ds{1 \over \root{2}}
\\
\ds{-\,{1 \over \root{2}}} & \ds{1 \over \root{2}}
\end{array}} & =
{1 \over \root{2}}\braces{\overbrace{\pars{\begin{array}{cc}
\ds{1} & \ds{0}
\\
\ds{0} & \ds{1}
\end{array}}}^{\ds{\sigma_{0}}} + \ic\ \overbrace{%
\pars{\begin{array}{cc}
\ds{0} & \ds{-\ic}
\\
\ds{\ic} & \ds{0}
\end{array}}}^{\ds{\sigma_{y}}}} =
2^{-1/2}\pars{\sigma_{0} + \ic\sigma_{y}}
\end{align}
Note that
$$
\sigma_{0}\sigma_{y} = \sigma_{y}\sigma_{0}\quad\mbox{and}\quad
\sigma_{0}^{2} = \sigma_{y}^{2} = \sigma_{0}
$$
such that
\begin{align}
\exp\pars{2^{-1/2}\pars{\sigma_{0} + \ic\sigma_{y}}\lambda} & =
\exp\pars{2^{-1/2}\lambda}\exp\pars{2^{-1/2}\ic\sigma_{y}\lambda}
\\[3mm] & =
\exp\pars{\lambda \over \root{2}}\bracks{\cos\pars{\lambda \over \root{2}} +
\sin\pars{\lambda \over \root{2}}\ic\sigma_{y}}
\\[3mm] & =
\half\pars{1 + \sigma_{y}}\exp\pars{{1 + \ic \over \root{2}}\,\lambda} +
\half\pars{1 - \sigma_{y}}\exp\pars{{1 - \ic \over \root{2}}\,\lambda}
\\[3mm] & =
\half\pars{1 + \sigma_{y}}\exp\pars{\expo{\pi\ic/4}\lambda} +
\half\pars{1 - \sigma_{y}}\exp\pars{\expo{-\pi\ic/4}\lambda}
\end{align}

\begin{align}\color{#f00}{M^{2006}} & =
\bracks{2^{-1/2}\pars{\sigma_{0} + \ic\sigma_{y}}}^{2006} =
2006!\bracks{\lambda^{2006}}
\exp\pars{\vphantom{\LARGE A}2^{-1/2}\bracks{\sigma_{0} + \ic\sigma_{y}}\lambda}
\\[3mm] & =
\half\,\pars{1 + \sigma_{y}}\bracks{\exp\pars{{\pi \over 4}\,\ic}}^{2006} +
\half\,\pars{1 - \sigma_{y}}\bracks{\exp\pars{-\,{\pi \over 4}\,\ic}}^{2006}
\\[3mm] & =
\half\pars{\begin{array}{cc}\ds{1} & \ds{-\ic}\\ \ds{\ic} & \ds{1}\end{array}}
\pars{-\ic} +
\half\pars{\begin{array}{cc}\ds{1} & \ds{\ic}\\ \ds{-\ic} & \ds{1}\end{array}}
\ic =
\color{#f00}{\pars{\begin{array}{cc}\ds{0} & \ds{-1}\\ \ds{1} & \ds{0}\end{array}}}
\end{align}
A: Square the matrix once to get
$$M^2=\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$$
Then,
$$M^4=\begin{bmatrix}0&1\\-1&0\end{bmatrix}^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}.$$
$$M^6=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}=-M^2.$$
$$M^8=-M^4.$$
$$M^{10}=M^2.$$
and so on with period $8$.
Then $M^{2006}=M^{2006\bmod8}=M^6$.
