How many square roots of -1 (the four by four matrix) are there? How many square roots of the matrix:
$$A = \left[\begin{array}{c c c} -1 & 0 & 0 & 0\\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0  & 0 & 0 & -1 \end{array}\right]$$
where the square root $Q$ satisfies:
$$ QQ = A $$
exist?
Edit: I'm only considering the case of real numbers.
 A: Hint: if $Q$ is one square root, then $R Q R^{-1}$ is another, where $R$ is any invertible $4 \times 4$ matrix.  And (if $Q$ is not a multiple of the identity) there are lots of $R$ that don't commute with $Q$.
A: All eigenvalues of $Q$ must be either $i$ or $-i$. Furthermore, $Q$ must be diagonalizable (because the square of non-diagonalizable invertible matrices can be shown to be non-diagonalizable, but $A$ is diagonalizable). If we allow complex numbers, all matrices which are similar to a diagonal matrix $D$ with $i$ and $-i$ as diagonal elements (i.e. all matrices $RDR^{-1}$ with an invertible $R$) can be chosen as $Q$.
If we allow only real numbers in $Q$, then the eigenvalues of $Q$ must show up as conjugate pairs, i.e. $Q$ has the eigenvalue $i$ with multiplicity $2$ and the eigenvalue $-i$ with multiplicity $2$. Each real matrix, which is diagonalizable in $\mathbb{C}$ and has this configuration of eigenvalues, is similar to $$B = \left(\begin{array}{cccc} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array}\right)$$ which means that each real $Q$ can be written as $RBR^{-1}$ with an invertible matrix $R$.
