Nth power of the following matrix 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?
 A: $${{\left( \begin{matrix}
   \cos \,\phi  & -\sin \,\phi   \\
   \sin \,\phi  & \cos \,\phi   \\
\end{matrix} \right)}^{n}}=\left( \begin{matrix}
   \cos \,n\phi  & -\sin \ n\phi   \\
   \sin \,n\phi  & \cos \,n\phi   \\
\end{matrix} \right)$$
We know $A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}$
then set $\phi=\frac{\pi}{2}$.
A: 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
$$A^3=A\cdot A^2=-A\;,\;\;A^4=I\;,\;\;A^5=A;,\ldots\text{etc.}$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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 $\ds{\sigma_{y}^{2} = \sigma_{y}}$.

