A question about the characteristic equation

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Discussion Overview

The discussion revolves around the characteristic equation of a 2x2 matrix and the implications of obtaining complex roots. Participants explore the relationship between complex eigenvalues and the existence of eigenvectors, particularly in the context of real vector spaces.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that complex roots can arise from the characteristic equation, indicating the presence of complex eigenvalues.
  • One participant suggests that if the vector space is over the real numbers, complex eigenvalues imply that eigenvectors do not exist in that space.
  • Another participant counters that if the transformation operates on complex vectors, eigenvectors can exist corresponding to the complex eigenvalues.
  • There is a suggestion that complex roots in the characteristic equation indicate a lack of eigenvectors in Cartesian coordinates.
  • A later reply confirms that the initial misunderstanding was due to a lack of specification regarding the real number context.

Areas of Agreement / Disagreement

Participants express differing views on the implications of complex roots for eigenvectors, particularly regarding the context of real versus complex vector spaces. The discussion remains unresolved as multiple competing perspectives are presented.

Contextual Notes

There is ambiguity regarding the definitions of vector spaces and the conditions under which eigenvectors exist, particularly in relation to the underlying field of numbers.

Cygni
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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!

Cygni
 
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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.

Thanks
Matt
 
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.
 
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.
 
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.
 
aPhilosopher said:
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.


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?
 
Absolutely correct. Sorry if I missed something
 
It wasn't you. Cygni missed saying he was talking about vector spaces (or matrices) over the real numbers.
 
HallsofIvy said:
It wasn't you. Cygni missed saying he was talking about vector spaces (or matrices) over the real numbers.
Apologies for the vagueness in my question :)
 

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