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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.

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  • $\begingroup$ Every square matrix has complex eigen values by the Fundamental theorem of algebra (eigen values are roots of the characteristic polynomial). $\endgroup$ – Prahlad Vaidyanathan Jun 20 '17 at 2:04
  • $\begingroup$ @Batman You are wrong. Normal matrices can have complex eigenvalues. For example, $A=\begin{bmatrix}3&1\\-1&3\end{bmatrix}$ is a normal matrix but has complex conjugate eigenvalues. $\endgroup$ – winston Jun 20 '17 at 2:45
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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.

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  • $\begingroup$ Thank you. But for polynomials of higher degrees ($\ge 4$), the information given by the discriminant is still limited. Let $D$ denotes the discriminant. if $D=0$, then there are no complex eigenvalues, while there are at least two complex eigenvalues if $D<0$. However, we cannot determine in the case where $D>0$. $\endgroup$ – winston Jun 20 '17 at 4:07

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