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A question about the characteristic equation

  1. Sep 19, 2009 #1
    Hi PF readers,

    When trying to establish [tex]\lambda[/tex] values by solving a characteristic equation (for simplicity of 2x2 matrix) can one produce solution that contains complex roots? If yes, what does that show about the eigenvectors?

    Thanks in advance!

  2. jcsd
  3. Sep 19, 2009 #2
    Yes, you can have complex roots. I leave the second part of your question to someone else, as I am not sure how to answer it.

  4. Sep 19, 2009 #3


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    Well, it shows that the eigenvalues are complex numbers! Which means that, since the eigenvalues of a linear transformation from vector space V to itself must belong in the underlying field, if your vector space is over the real numbers that particular linear transformation does not actually have eigenvalues.
  5. Sep 20, 2009 #4
    Thanks guys, that helped.

    I pretty much had an idea that if u get complex solutions in characteristic equation the eigen vectors would not exist as they'd require argand diagram in order to be displayed.
  6. Sep 20, 2009 #5
    That's not quite right. If you let it operate on vectors composed of complex numbers, it would have eigenvectors and the eigenvalues of those vectors would be the roots you have found.
  7. Sep 21, 2009 #6

    Yes, but if we are dealing with cartesian coordinates, the complex roots obtained from characteristic equation would indictae that there are no possible eigenvectors in that plane. Isn't that correct?
  8. Sep 21, 2009 #7
    Absolutely correct. Sorry if I missed something
  9. Sep 21, 2009 #8


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    It wasn't you. Cygni missed saying he was talking about vector spaces (or matrices) over the real numbers.
  10. Sep 23, 2009 #9
    Apologies for the vagueness in my question :)
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