Matrix square roots of -I Since we can see matrices as generalizations of complex numbers, I asked myself if there is a way to classify those matrices which are the "Basis" for the complex part. That is, I would like to identify the set of $n\times n$ real valued matrices $M$ whose square $M^2$ is equal to $-I$, where $I$ is the $n\times n$-identity matrix. Is this set already classified?    
The matrix $J = \left( \begin{smallmatrix}0 & -1\\1 & 0 \end{smallmatrix} \right)$ satisfies $J^{2} = -I$ in the 2-dimensional case. 
EDIT:
Thanks to your comments and answers I have the following observation (if its wrong, please tell me): 
Suppose that $M$ satisfies that $M^t=-M$, then we can observe for two vectors $x,y\in\mathbb{R}^{2n}$ that

$y^tAx=(Ax)^{t}y=x^{t}A^ty=-x^tAy$

My interpretation is that if the angle between $x$ and $Ay$ is $\alpha$ then $\pi+\alpha$ is the angle between $Ax$ and $y$, since ${\displaystyle \cos \;x=-\cos(x+\pi )}$ (here we suppose that the length of the vectors are not relevant). If now $y=x$ and then we can see that $Ax$ is orthogonal to $x$. 
Is it true that $A^2$ will be a rotation of 180 degrees, right? Because of vector length we will have something like $A^2=-CI$, where $C$ is just a constant. Is that true?

If you want to make progress, you will need to read some books...
The matrices $M\in M_2(\mathbb{R})$ s.t. $M^2=-I_2$ are similar to your $J$ (there are an infinity of such matrices).
Yet, if you consider that $\mathbb{R}^2$ is euclidean (with an inner product), then it is natural to consider the solutions $M$ that are normal ($MM^T=M^TM$). Then you find only $2$ solutions: $\pm J=Rot(\pm \pi/2)$, that are skew-symmetric. If we consider the isomorphism $f:a+ib\in \mathbb{C}\rightarrow \begin{pmatrix}a&-b\\b&a\end{pmatrix}=aI+bJ\in M_2(\mathbb{R})$, then $\pm J=\pm f(i)$ (cf. the Joppy's post). Of course, if $M$ is skew-symmetric ($a=0$), then $M^2=-b^2I$; yet, it's false when $n>2$.
Now, in dimension $2n$, the matrices s.t. $M^2=-I_{2n}$ are similar to $diag(J_1,\cdots,J_n)$ where $J_k=J$.
In the euclidean case, where $M$ is normal, then the solutions are skew-symmetric and in the form $Pdiag(J_1,\cdots,J_n)P^T$ where $P$ is an orthogonal matrix;  that is to say $\mathbb{R}^{2n}$ is an orthogonal sum of $n$ planes $\Pi_k$ s.t., over each $\Pi_k$, the linear transformation is $Rot(\pm \pi/2)$.
