A question about the characteristic equation

[Cygni]

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

 

[CFDFEAGURU]

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

 

[HallsofIvy]

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.

 

[Cygni]

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.

 

[aPhilosopher]

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.

 

[Cygni]

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?

 

[aPhilosopher]

Absolutely correct. Sorry if I missed something

 

[HallsofIvy]

It wasn't you. Cygni missed saying he was talking about vector spaces (or matrices) over the real numbers.

 

[Cygni]

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 :)