Existence of an $n\times n$ real matrix $A$ such that $A^2=-I$. Let $A$ be a $n\times n$ real matrix $A$ such that $A^2=-I$. Such an $A$ cannot be,


*

*Orthogonal.

*Invertible.

*Skew-symmetric.

*Symmetric.

*Diagonalizable. 


I tried to figure out the answer by looking at the [possible] determinant of $A$, using the fact $\det(AB)=\det(A)\det(B)$. So $\det(A^2)=(\det(A))^2=(-1)^n\det(I)$. If $n$ is even, then $\det(A)=\pm1$ and if odd then $\det(A)=\pm i$. There I stuck, if $n$ is even, I can try some guessing, but when $n$ is odd, the determinant becomes complex and I have no logic to go forward. Is there any other approach that might be more correct? How can I do this? Help me out.
 A: We can use the following characterization of diagonalizable: 

A matrix or linear map is diagonalizable over the field F if and only
  if its minimal polynomial is a product of distinct linear factors over
  F.

$A^2+I=0$ implies that $A$ satisfies the polynomial $x^2+1$ which is the minimal polynomial since it is irreducibe over the reals. This minimal polynomial is not a product of distinct linear factors over $\mathbb{R}$, thus $A$ is not diagonalizable.
Invertible, skew-symmetric, orthogonal can be all ruled out with the example $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.
Symmetric is false, by stity's comment below, since symmetric matrices are diagonalizable.
Alternatively, you can observe that it is false for $n=2$ by the computation:
$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&c\\b&d\end{pmatrix}=\begin{pmatrix}a^2+b^2&\dots\\\dots&\dots\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ is not possible over the reals.
A: I experimented with rotation matrices and powers  and found the following solution for $\small n=4$ where $\small w = \sqrt{1/3}$ :
     0   w  -w  w
A=  -w   0   w  w
     w  -w   0  w
    -w  -w  -w  0

Then
      -3*w^2       .       .       .
           .  -3*w^2       .       .
  A^2=     .       .  -3*w^2       .  = - I
           .       .       .  -3*w^2

And also,    


*

*$\small A$ is invertible

*$\small A^{-1} = A^T$ (or $\small  A \cdot A^T=I $ ) which is a condition for
orthogonal/orthonormal matrices.  

*Also $\small A^T = - A$ which means, $\small A$ is skew-symmetric.        

*Using complex numbers, a diagonalization is possible and gives the diagonal $\small D = [1,1,-1,-1] \cdot \sqrt{-3} $

