# Eigenvalues of Hermitian opertors

1. Jun 29, 2011

### migwing007

I'm looking for a proof of the fact that orthogonal eigenfunctions of a Hermitian operator have distinct eigenvalues. I know the proof the converse: that eigenfunctions belonging to distinct eigenvalues are orthogonal.
thanks alot!

2. Jun 30, 2011

### Matterwave

I don't believe that it is true that orthogonal eigenfunctions have distinct eigenvalues. Certainly in terms of degeneracy, one can employ the Gram-Schmidt orthogonalization procedure to get orthogonal eigenfunctions which have the same eigenvalue...

I'm pretty bad at math though, so I wouldn't take my word on it lol.

3. Jun 30, 2011

### George Jones

Staff Emeritus
As Matterwave has said, this it not true. Consider the matrix
$$\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$$

$\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ and $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ are orthogonal eigenvectors that both are associated with the eigenvalue 1 (as is any linear combination of these vectors).

4. Jul 8, 2011

### migwing007

This is really bothering me

I'm remembering from a while ago having to prove the converse of a theorem from Griffiths, but I can't remember which one of the following it is:

Prove converse of:

1.) The eigenfunctions of a Hermitian operator are complete,

2.) Eigenfunctions of Hermitian operators belonging to distinct eigenvalues are orthogonal, or

3.) Eigenvalues of Hermitian operators are real

thanks alot!

Last edited: Jul 8, 2011
5. Jul 9, 2011

### strangerep

Re: This is really bothering me

migwing007,

It's not clear (to me anyway) what you're asking. Maybe try stating the
theorem or its converse in exactly the wording that you want to prove?

6. Jul 9, 2011

### migwing007

Re: This is really bothering me

Well, the converse of the first one would be: if the eigenfunctions of an operator are complete then the operator is hermitian.

the second would be: if the eigenfunctions of an operator belonging to distinct eigenvalues are orthogonal then the operator is hermitian.

and the third is: If the eigenvalues of an operator are real then the operator is hermitian

Actually I don't think 2 and 3 are correct so that leaves #1.

I've looked online but can't seem to find it

Hope that clarifies it

Last edited: Jul 9, 2011
7. Jul 9, 2011

### samalkhaiat

Re: This is really bothering me

8. Jul 10, 2011

### A. Neumaier

Re: This is really bothering me

9. Jul 11, 2011

### bcrowell

Staff Emeritus
Migwing007 started four separate threads on the same topic. I've deleted two and merged two.