# Eigenvalue proof. (2nd opinion if my proof is right please)

1. Jun 19, 2009

### hoch449

1. The problem statement, all variables and given/known data

Prove that if two linear operators A and B commute and have non-degenerate eigenvalues then the two operators have common eigenfunctions.

2. Relevant equations

$$[A,B]= AB - BA= 0$$

$$Af=af$$

$$Bg=cg,\ let\ g=(f+1) --> B(f+1)=c(f+1)\ where\ a\neq c$$

3. The attempt at a solution

$$Af[B(f+1)]-B(f+1)[Af]=0$$
$$Af(Bf+B)-(Bf+B)Af=0$$

I have stopped here because I feel that I am on the wrong track. I have not used the fact that the eigenvalues are non-degenerate in this proof. Although continuation of what I am doing should show that the two operators commute..

2. Jun 19, 2009

### Dick

You don't want to show they commute. You already know that. You want to show they have common eigenfunctions. Look at [A,B]f where Af=af. That's ABf-BAf=A(Bf)-a(Bf)=0. That tells you Bf is also an eigenfunction of A corresponding to the eigenvalue a. Just like f. Now what does 'nondegenerate' tell you?

3. Jun 19, 2009

### hoch449

Ah that totally makes sense! Non-degenerate eigenvalues means different ie. (not the same) eigenvalues does it not?

4. Jun 19, 2009

### Dick

Sort of. It means there is only one eigenvector (up to a constant multiple) corresponding to every eigenvalue.