Is there a $2 \times 2$ real matrix $A$ such that $A^2=-4I$? Does there exist a $2 \times 2$ matrix $A$ with real entries such that $A^2=-4I$ where $I$ is the identity matrix?
Some initial thoughts related to this question:


*

*The problem would be easy for complex matrices, we could simply take identity matrix multiplies by $2i$.

*There is another question on this site showing that this has no solution for $3\times3$ matrices, since the LHS has determinant $\det(A^2)=\det^2(A)$ which is a square of real number, but determinant of $-4I_3$ is negative. But the same argument does not work for $2\times2$ matrices, since $\det(-4I_2)=4$ is positive.


$A = \left [\begin{array}{ccc}
0 & 2  \\
-2 & 0  \\
\end{array} \right ]$.
