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: Every matrix satisfies its characteristic polynomial. So, any matrix with characteristic polynomial $\lambda^2+4$ would work. Taking the diagonal to be zero and the off diagonal entries to be $4$ and $-1$ would give you a solution. 
A: A 2-by-2 matrix is not too bad to work out the algebraic equation for its coefficients:
Let 
$
A=\begin{bmatrix} a & b \\c  & d\end{bmatrix} 
$. We end up with the matrix equation:
$$
A^2 = \begin{bmatrix} a^2+bc & b(a+d) \\c(a+d)  & d^2 + bc\end{bmatrix} 
= -4 \, I = \begin{bmatrix} -4 & 0 \\0  & -4\end{bmatrix}.
$$
So we have 4 equations
$$
\begin{equation}
a^2+bc = -4 \quad (1)\\
b(a+d) = 0 \quad (2) \\
c(a+d) = 0 \,\quad (3) \\
d^2+bc = -4 \quad (4).
\end{equation}
$$
It is easy to see that $a + d = 0$. Since if otherwise $a+d \not=0$, equations (2) and (3) it would imply that: $b=c=0$, then from equation (1): $a^2 = -4$ which can't happen.
Since $a+d=0$, we have $a^2 = d^2$. Thus equations (1) and (4) are identical. Therefore, the conclusion is that there are infinitely many solutions, as long as $a,b,c,d$ satisfy that
$$
\begin{cases}
a+d=0 \quad &(5)\\
a^2 = -(4+bc) \quad &(6).
\end{cases}
$$
Since there are 4 variables with 2 equation, one can choose 2 free variables and derive the others from the free ones. The simplest is from suggestion of user "zwim":
$$
A=\begin{bmatrix} a & -\frac{a^2+4}{c} \\c  & -a\end{bmatrix},
$$
where $a$ and $c$ are any real numbers.
Example: choose a = 1, c = -5 we have:
$
A=\begin{bmatrix} 1 & 1 \\-5  & -1\end{bmatrix}. 
$
A: Yes.  
$A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}; \tag 1$
There are others.  Note that if
$J =\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \tag 2$
then
$J^2 = -I, \tag 3$
so $J$ is a real $2 \times 2$ matrix "version" of $i \in \Bbb C$.  Here,
$A = 2J. \tag 4$
A: We have two matrix equations 
$A^2=-4I$  i.e.
$A^2+4I=0$   
and general equation from Cayley-Hamilton theorem  for $ 2 \times 2$ matrices
$A^2-\text{tr}(A)A+\det(A)I=0$.
Comparing both equations we obtain
$\text{tr(A)}=0$   , $ \det(A)=4$.
So if we denote $A=\begin{bmatrix} a & b \\c  & d\end{bmatrix} $  then $d=-a$ and consequently $-a^2-bc=4$.
These conditions are sufficient  to obtain an infinite number of solutions,   even with integer values.
Check for example
$A=\begin{bmatrix} 2 & -1 \\8  & -2\end{bmatrix} $. 
Factorizing $a^2+4$ (with assistance  for example Number Empire site   ) you can obtain even a less obvious integer solutions, for example $A=\begin{bmatrix} 23 & -13 \\41  & -23\end{bmatrix}$ where all absolute values of entries are prime numbers.
A: $A = \left [\begin{array}{ccc}
0 & 2  \\
-2 & 0  \\
\end{array} \right ]$.
