Is it possible to find a $n\times n$ matrix $M$ such that $M^2 =- I_n$, where $-I_n$ is the identity matrix? Is it possible to find a $n\times n$ matrix $M$ such that $M^2 = -I_n$, where $I_n$ is the identity matrix?
For odd $n$, this cannot hold, since $(-1)^n = \det(-I_n) = \det (M^2) = \det(M)^2$ but what can one say about even $n$?
 A: Consider the $2 \times 2$ matrix
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 1$
we have
$J^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = -I_2, \tag 2$
as may be readily verified by an easy calculation.  We may then construct a $2n \times 2n$ matrix $M_0$ by placing $n$ copies of $J$ along the (block) diagonal:
$M_0 = \begin{bmatrix} J & 0 & \ldots & 0 \\ 0 & J & \ldots & 0 \\ \dots & \vdots & J & 0 \\ 0 & 0 & \ldots & J\end{bmatrix}; \tag 3$
it is equally easy to see that
$M_0^2 = -I_{2n}; \tag 4$
this establishes the existence of a desired $2n \times 2n$ matrix. An entire family of such matrices may be had by taking matrices $M$ similar to $M_0$:
$M = SM_0S^{-1}; \tag 5$
then
$M^2 = SM_0S^{-1} SM_0S^{-1} = SM_0 M_0S^{-1} = S(-I_{2n}) S^{-1} = -I_{2n}. \tag 6$
We see there are many matrices of the requisite form.
A: $\begin{pmatrix}
1 & 2\\
-1 & -1
\end{pmatrix}$
A: A classical example is
$$
M=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)
$$
This is a classical example as it allows to mimic (or represent) complex number computations using real arithmetic. 
$$
z=a+ib \leftrightarrow a\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right) + b\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)=\left(\begin{array}{cc}a & -b \\ b & a\end{array}\right)
$$
