TITLE: Existence of an $n\times n$ real matrix $A$ such that $A^2=-I$.
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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} $