Logic behind finding a $ (2 \times 2) $-matrix $ A $ such that $ A^{2} = - \mathsf{I} $. I know the following matrix "$A$" results in the negative identity matrix when you take $A*A$ (same for $B*B$, where $B=-A$):
$$A=\begin{pmatrix}0 & -1\cr 1 & 0\end{pmatrix}$$
However, I am not certain how one would go about finding this matrix without guessing and checking. Is there some systematic way of doing so? I have tried by assuming the following:
$$\begin{pmatrix}a & b\cr c & d\end{pmatrix}^2 = \begin{pmatrix}-1 & 0\cr 0 & -1\end{pmatrix}$$
...and then get the following equations:
$$a*a+b*c=-1$$
$$a*b+b*d=0$$
$$a*c+c*d=0$$
$$b*c+d*d=-1$$
This only tells me that $a=-d$, and then as best I can tell leaves both $c$ and $d$ without a solution, so perhaps this isn't the best method, but its the only approach that's coming to mind.
Any suggestions?
 A: the negative identity matrix $B = \pmatrix{-1&0\\0&-1}$  represents the half rotation. so if you pick the matrix $A = \pmatrix{0&-1\\1&0}$ representing quarter rotation, then you should have $A^2 = B = -I.$
A: The most systematic way I know to attack this problem is to break the set of all possible such $A$, i.e., $2 \times 2$ matrices with 
$A^2 = -I \tag{1}$
into categories according to $\text{Tr}(A)$, the trace of $A$.  With
$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \tag{2}$
we have 
$A^2 = \begin{pmatrix} a^2 + bc & b(a + d) \\ c(a + d) & d^2 + bc \end{pmatrix}, \tag{2}$
as may be easily seen; note that
$\text{Tr}(A) = a + d. \tag{3}$
With
$A^2 = -I \tag{4}$
we have
$a^2 + bc = -1; \tag{5}$
$d^2 + bc = -1; \tag{6}$
$(a + d)b = b\text{Tr}(A) = 0; \tag{7}$
$(a + d)c = c \text{Tr}(A) = 0. \tag{8}$
If
$\text{Tr}(A) \ne 0, \tag{9}$
we see that
$b = c = 0, \tag{10}$
whence
$a^2 = d^2 = -1; \tag{11}$
thus
$a = \pm i \tag{12}$
and 
$d = \pm i; \tag{13}$
there are no real solutions; furthermore since $\text{Tr}(A) \ne 0$ we must in fact have $a = d$, thus in this case we may only have
$A = \pm i I. \tag{14}$
The case $\text{Tr}(A) = 0$ yields
$d = -a;  \tag{15}$
we can thus choose $a = \alpha \in \Bbb C$ freely and then $d = -\alpha$.  (15) implies
$a^2 = d^2 = \alpha^2, \tag{16}$
so that both (5) and (6) become
$\alpha^2 + bc = -1, \tag{17}$
$bc = -1 - \alpha^2; \tag{18}$
in the event $\alpha \ne \pm i$ we can choose
$0 \ne \beta \in \Bbb C \tag{19}$
then set
$b = \beta, \tag{20}$
and find
$c = -\dfrac{1 + \alpha^2}{\beta} \ne 0; \tag{21}$
thus
$A = \begin{pmatrix} \alpha & \beta \\ -\dfrac{1 + \alpha^2}{\beta} & -\alpha \end{pmatrix}, \;\; \alpha, \beta \in \Bbb C, \; \beta \ne 0. \tag{22}$
The case $\alpha = \pm i$ forces
$bc = 0; \tag{23}$
thus, at least one, and perhaps both, of $b, c = 0$.  Now $A$ takes either the form
$A = \begin{pmatrix} \pm i & \beta \\ 0 & \mp i \end{pmatrix}, \;\;  \beta \in \Bbb C, \tag{24}$
or the transposed form
$A = \begin{pmatrix} \pm i & 0 \\ \beta & \mp i \end{pmatrix}, \;\;  \beta \in \Bbb C. \tag{25}$
Of course, if we require real $A$, then (11)-(14) are denied to us; only cases with $\text{Tr}(A) = 0$ are permitted; we can however choose $\alpha \in \Bbb R$ freely as we can $0 \ne \beta \in \Bbb R$; $1 + \alpha^2 \ne 0$ in this case, but $A$ is still given by (22), though (24), (25) are ruled out. 
That's the most systematic analysis of this problem I know.  The observation that $\text{Tr}(A)$ provides a convenient classification of solutions is perhaps not a surprise in light of the importance of the trace, a similarity invariant, to a general understanding of matrices.  (Note that the equation $A^2 = -I$ is in fact preserved under similarity:  $(P^{-1}AP)^2 = A^2 = -I$.)  However, this line of analysis only works for matrices of size $2$.
A similar analysis of the case $A^2 = I$ may be found in my answer to this question; further commentary may there be found.
Finally, I for one would be very interested in seeing how these techniques carry over to matrices over other fields, such as the $GF(p^n)$ etc.; offhand, it seems as long as we are not in characteristic $2$, many of the arguments will fly.
A: If you assume that $A$ is invertible, then you can write:
$$
AA = -\mathbb{I}_2 \rightarrow A^{-1}AA = -A^{-1}\mathbb{I}_2 \rightarrow A = -A^{-1}\mathbb{I}_2
$$
But we know the inverse for a $2\times2$ matrix--and it's going to create a set of four linear equations for four variables:
$$
\begin{pmatrix}
a& b \\
c & d
\end{pmatrix} = \begin{pmatrix}
-d & b \\
c & -a
\end{pmatrix}
$$
This creates four equations:
\begin{align}
a = -d \rightarrow a + d = 0 \\
b = b \\
c = c \\
d = -a \rightarrow a + d = 0
\end{align}
This suggests that this is trues for all values of $a$ and $d$ such that $a + d = 0$ and any values of $b$ and $c$.  Let's check:
$$
\begin{pmatrix}
a & x \\
y & -a 
\end{pmatrix}
\times \begin{pmatrix}
a & x \\
y & -a 
\end{pmatrix}
=
\begin{pmatrix}
a^2 + xy & ax - ax \\
ay - ay & xy + a^2 
\end{pmatrix}
$$
Then it's a matter of solving the equation $a^2 + xy = -1$.  For instance, we could choose $a = 2$ and $xy = -5$:
$$
\begin{pmatrix}
2 & -5 \\
1 & -2
\end{pmatrix}
\text{ and its inverse }
\begin{pmatrix}
-2 & -5 \\
1 & 2
\end{pmatrix}
\text{ both work}
$$
...obviously there are many more you could choose.  A general way would be:
$$
\begin{pmatrix}
a & b \\
-\frac{a^2 + 1}{b} & -a
\end{pmatrix}
\text{ or }
\begin{pmatrix}
a & -\frac{a^2 + 1}{b} \\
b & -a
\end{pmatrix}
$$
It may be worth noting that if $b = 0$ then $a^2 + 1 = 0$ must also be true so that the other value, $\frac{a^2 + 1}{b}$, can take on any value.  In this case we get: $a^2 = -1 \rightarrow a = \pm i$ and gives the equation $a^2 + 2xy = -1 \rightarrow -1 + 2*x * 0 = -1 \rightarrow -1 = -1$ (which is true for any value of $x$--the non-zero off diagonal).  So this gives the other answer as well:
$$
\begin{pmatrix}
\pm i & 0 \\
x & \mp i
\end{pmatrix}
\text{ or }
\begin{pmatrix}
\pm i & x \\
0 & \mp i
\end{pmatrix}
$$
You can have arbitrarily complex components.  For instance choose $a = 5i$.  This gives: $a^2 + xy = -1 \rightarrow -25 + xy = -1 \rightarrow xy = 24$.  So we could choose 6 and 4, 3 and 8, 2 and 12...or any two real values (or any two complex values--although they would have to be conjugates).  Another complex example:
$$
\begin{pmatrix}
5i & 3 \\
8 & -5i
\end{pmatrix}
$$
or perhaps:
$$
\begin{pmatrix}
3i+2 & -4 \\
3i - 1 & -3i - 2
\end{pmatrix}
$$
A: One reason that it is hard to describe a systematic procedure that will let you find that solution to $A^2=-I$ is that there are so many other solutions to this equation. A solution cannot have real eigenvalues, so the image of the first standard basis vector $e_1$ (which give the firs column of your matrix) cannot be a multiple of $e_1$ itself, but otherwise it can be any vector$~v$ whatsoever. Then $[e_1,v]$ will be a basis of $\Bbb R^2$, and one can arrange that $A\cdot v=-e_1$ (since one can choose the images of a basis freely when defining a linear map), which will ensure that $A^2=-I$ (an easy check).
But the example you gave is so obtained for the simplest choice of$~v$, namely $v=e_2$.
