Find all $2\times2$ matrices that have pure imaginary eigenvalues. 
Find all $2 \times 2$ matrices that have pure imaginary eigenvalues. That is, determine conditions on the entries of a matrix that guarantee the matrix has pure imaginary eigenvalues.
  \begin{bmatrix}a & b\\
c & d \\
\end{bmatrix} 

$$(a - \lambda)(d-\lambda)-bc = 0  $$
$$\lambda = \frac{a+d \pm\sqrt{(a+d)^2-4(ad-bc)}}{2}$$
Focusing on the things inside the square root:
$$(a-d)^2 \lt - 4bc $$
One case is when $a=d$, $b\gt0$ and $c\lt0$ or $b\lt0$ and $c\gt0$. I am wondering what other cases lie in there?
 A: I'm going to address this for real matrices; our OP afsdf dfsaf's use of "$>$" and "$<$" in his question suggests, or even tacitly implies, that this is the case of primary interest.  
For any $2 \times 2$ real matrix 
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \tag{1}$
it is well-known, and easy to see, that the characteristic pilynomial $p_A(x)$ is
$p_A(x) = x^2 - \text{Tr}(A) x + \det(A), \tag{2}$
where $\text{Tr}(A)$ and $ \det(A)$ are the trace and determinant of $A$, respectively.  Applying the quadratic formula to
$p_A(x) = x^2 - \text{Tr}(A) x + \det(A) = 0, \tag{3}$
we obtain
$\lambda_\pm = -\dfrac{1}{2}(-\text{Tr}(A) \pm \sqrt{(\text{Tr}(A))^2 - 4\det(A)}(, \tag{4}$
so in order that the $\lambda_\pm$ be purely imaginary we must have $\text{Tr}(A) = 0$ and from this $\det(A) > 0$.  $\text{Tr}(A) = 0$ implies $d = -a$, whence
$A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}, \tag{5}$
with 
$\det(A) = -a^2 -bc > 0, \tag{6}$
or 
$a^2 + bc < 0. \tag{7}$
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: If you are looking for real $2\times 2$ matrices, (with eigenvalues $\lambda_1,\lambda_2$) then the characteristic polynomial should be of the form $p(\lambda)=\lambda^2+s$, $s\ge 0$. Hence, if
$$
A=\left(\begin{matrix} a&b\\c&d\end{matrix}\right),
$$
then $\lambda_1+\lambda_2=a+d=0$, (i.e. $d=-a$) and $\lambda_1\lambda_2=ad-bc\ge 0$ or
$-bc\ge a^2$.
A: A sufficient and concise condition is:
$$A^{*}=-A$$
That is: for all $a,b \in \mathbb{R}, z \in \mathbb{C}$:
$$A=\begin{bmatrix}ia & z \\ -\overline{z} &ib \end{bmatrix}$$
