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The matrix is $ \begin{pmatrix}0&1\\-1&0\end{pmatrix}^n $ for $n=2 \implies \left(\begin{matrix}-1 & 0\\ 0 &-1\end{matrix}\right)$ for $n=3 \implies \begin{pmatrix}1&0\\0&1\end{pmatrix}$ for $n=4 \implies \begin{pmatrix}-1&0\\0&-1\end{pmatrix}$

so I assume that for every $n=2k $, where k is a natural number and bigger than $0$ the matrix will be $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ and for every $n=2k+1$ where k is a natural number and bigger than $0 $the matrix will be $\begin{pmatrix}1&0\\0&1\end{pmatrix}$

How can I prove it? probably with induction and how can I get easily the inverses of the matrices?

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    $\begingroup$ Could you write that first sentence out with more actual words, please? Your use of symbols is hard to understand -- it seems that you're definitely not using "$\Longrightarrow$" with its usual mathematical meaning, but it is difficult to figure out what you do mean by it. $\endgroup$
    – hmakholm left over Monica
    Sep 28, 2016 at 6:57
  • $\begingroup$ Try inducing over $k \in\mathbb{N}$, using $p(k): A^{4k}=\ldots \land A^{4k+1}=\ldots \land A^{4k+2}=\ldots \land A^{4k+3}=\ldots$. Also possible: prove only $q(k): A^{4k}=\ldots$ by induction on $k$. Once you know that $q$ is always true, deduce that $p$ is also always true. $\endgroup$
    – chi
    Sep 28, 2016 at 11:18

3 Answers 3

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You just have to calculate the first power:

$$A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}\;,\;\;A^2=\begin{pmatrix}\!-1&0\\0&-1\end{pmatrix}=-I$$

and that's all we need, since then

$$A^3=A\cdot A^2=-A\;,\;\;A^4=I\;,\;\;A^5=A;,\ldots\text{etc.}$$

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$${{\left( \begin{matrix} \cos \,\phi & -\sin \,\phi \\ \sin \,\phi & \cos \,\phi \\ \end{matrix} \right)}^{n}}=\left( \begin{matrix} \cos \,n\phi & -\sin \ n\phi \\ \sin \,n\phi & \cos \,n\phi \\ \end{matrix} \right)$$ We know $A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}$ then set $\phi=\frac{\pi}{2}$.

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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\pars{\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}}^{n}} = \ic^{n}\pars{\begin{array}{rr}0 & -\ic \\ \ic & 0\end{array}}^{n} = \color{#f00}{\left\{\begin{array}{rcl} \ds{\pars{-1}^{n/2}\pars{\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}}} & \mbox{if} & \ds{n}\ \mbox{is}\ even \\[2mm] \ds{\pars{-1}^{\pars{n - 1}/2}\pars{\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}}} & \mbox{if} & \ds{n}\ \mbox{is}\ odd \end{array}\right.} \end{align}

because $\ds{\sigma_{y} \equiv \pars{\begin{array}{rr}0 & -\ic \\ \ic & 0\end{array}}}$ satisfies $\ds{\sigma_{y}^{2} = \sigma_{y}}$.

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