Calculating the matrix $M^{2006}$

Question

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?

Answer 1

You have $M^{2000}=(M^8)^{250}=I^{250}=I$, so you just need to find $M^6$

Answer 2

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

Answer 3

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 $({}^0_{1}\,{}^{-1}_{\;0})$.

Answer 4

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 $M^3$.

Answer 5

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\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}

Answer 6

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 $M^{2006}=M^{2006\bmod8}=M^6$.