$3 \times 3$ matrices with imaginary eigenvalues I know that for a $2 \times 2$ matrix $A$, having $\operatorname{trace}(A)=0$ and  $\det(A)>0 $ will guarantee pure imaginary eigenvalues.    
My question is, is there anything similar for a $3 \times 3$ matrix $B$?
 A: Up to a sign, the determinant of any $n\times n$ matrix is the free coefficient of the characteristic polynomial, and the trace is the coefficient of $x^{n-1}$ (the leading coefficient is always $1$). This is why for $2\times 2$ matrices the determinant and trace determine the characteristic polynomial completely, giving rise to the result you mention by elementary considerations of quadratic polynomials. 
But for polynomials of degree $3$ and above, knowledge of just three coefficients can't be sufficient to determine so much about the roots of the polynomial. Seeking an example, consider the polynomial $p(x)=x^3+ax+b$. By the above, if $p(x)$ is the characteristic polynomial of a matrix, then that matrix has $0$ trace and determinant equal to $\pm b$ (by the way, can you figure out if it's $b$ or $-b$?). So, we just want $p(x)$ not to have purely imaginary roots. Simple trial and error, playing with the values of $a$ and $b$, will quickly lead to an example. Finally, the companion matrix construction guarantees that there is indeed a matrix having $p(x)$ as characteristic polynomials. 
