Hi everybody, I've been doing linear algebra revision lately and one query I have has really been winding me up so I'd be really grateful for some help: If I have a linear operator that is not Hermitian or unitary, will its eigenvectors necessarily be orthogonal(in absence of degeneracy)? In either case why? Also, if the eigenvectors do have to be orthogon when there's no degeneracy would you be able to show how this is true?

 

Thanks a lot in advance:-)

Been a while since I have done any analysis! A theorem is in order before we continue.

 

Theorem

The spectrum of a bounded operator on a Hilbert space is non-empty.

 

As such, I think it makes sense to consider only bounded operators if we are going to lax the Hermitian property. Maybe, this is a bit technical, but we could consider operators that don't have any eigenvalues. (It is also known that the spectrum of a self-adjoint operator is also non-empty).

 

What I suggest you do is prove the below theorems first;

 

Theorem

If A is self-ajoint then the eigenvectors of A, belonging to distinct eigenvalues are orthogonal.

 

 

Theorem

If A is unitary then the eigenvectors of A, belonging to distinct eigenvalues are orthogonal.

 

Both are not hard to prove. If you can't do it I will post a proof later.

 

Now, can you modify the proofs to remove the self-adjoint or unitary conditions?

 

Yes, you can.

 

Theorem

If A is normal then the eigenvectors of A, belonging to distinct eigenvalues are orthogonal.

 

By normal I mean [math] A A^{\dag} = A^{\dag} A[/math]

 

You should be able to prove this also.

 

Can it be generalised any further? I am not sure. I am no expert in analysis.

Edited January 9, 200916 yr by ajb