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

[hoch449]

Homework Statement

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

Homework Equations

[A,B]= AB - BA= 0

Af=af

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

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..

 

[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?

 

[hoch449]

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

 

[Dick]

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

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