I try to say it all in the title.

I'm wondering under what conditions a matrix will have complex eigenvectors and eigenvalues. That question, I think, reduces to whether the characteristic polynomial has complex roots.

So, how do I know when a very high order polynomial has complex roots?

(Perhaps it's obvious that I don't know much about analysis.)


Well, every polynomial of degree $n \geqslant 1$ has at least one complex zero by the fundamental theorem of algebra. When $n = 2$ you can look at the discriminant. For $n > 2$ it depends. If you can reduce it to $n = 2$ via substitution then you can again look at the discriminant. In general, if a polynomial has real coefficients, then either all roots are real or there is an even number of non-real complex roots because non-real complex roots come in conjugate pairs. That is all we can say. We also know that if a matrix is symmetric i.e. $A = {A^T}$ or Hermitian i.e. $A = {A^H}$ where $H$ denotes the conjugate transpose, then it must have real eigenvalues if it is neither then it depends on the matrix. So we can say that a matrix $A$ could have complex eigenvalues only when it is not symmetric and not Hermitian.

  • 1
    $\begingroup$ I think mac389 is asking about non-real roots (all real numbers being complex numbers too). $\endgroup$ – Chris Card Oct 24 '12 at 20:46
  • $\begingroup$ I think Sturm's theorem for finding real roots might be of use. Also, if the polynomial has all real coefficients, the complex roots come in complex conjugate pairs. Also$^2$, there is somebody's root-squaring method which might be relevant. $\endgroup$ – marty cohen Oct 24 '12 at 21:28
  • $\begingroup$ @ChrisCard you're right- I meant non-real roots. $\endgroup$ – mac389 Oct 25 '12 at 10:44
  • $\begingroup$ Are there any special cases where one can strengthen the statement to say that if a matrix is not symmetric, not Hermitian, and has property X then it must have complex eigenvalues? $\endgroup$ – mac389 Oct 25 '12 at 14:54
  • $\begingroup$ Well, it is not difficult to prove that the eigenvalues of an upper triangular (or lower triangular) and diagonal matrices are the diagonal entries. So if all diagonal entires are complex then so are the eigenvalues. $\endgroup$ – glebovg Oct 25 '12 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.