TITLE: Are there Complex eigenvalues in a given matrix?
QUESTION [6 upvotes]: If a matrix is Hermitian, then its eigenvalues are all real. But given any real matrix that is not Hermitian, how to determine whether there are complex eigenvalues or not?
Or the question can be re-formulated in this way: rather than Hermitian, are there any more general rules which can be used to determine whether the eigenvalues of a matrix are all real numbers?
EDIT:
I could think of another solution to this (it is only an indirect way compared with the Hermitian matrix):  

If a matrix is similar to a Hermitian matrix, then its eigenvalues are all real.

REPLY [2 votes]: Understanding eigenvalue problems for general matrices is just as difficult as solving general polynomial equations, as seen by the matrices of the form 
$$A_n=\begin{bmatrix}0 & 1  & 0 & \cdots & 0 \\ 0 & 0 & 1 &\cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & 1 \\-a_0 & -a_1 & -a_2 & \cdots & -a_{n-1}  \end{bmatrix}, $$
whose characteristic polynomial is $\lambda^n+a_{n-1} \lambda^{n-1}+\dots+a_1 \lambda+a_0$. Some information about the "realness" of the roots can be obtained via the discriminant.