ST and TS have the same eigenvals

[balletomane]

Hi.

I need to prove that for S,T linear operators on V. ST and TS have the same eigenvalues. I've gotten as far as (say g is the eigenvalue and u is a nonzero vector): STu=gu so TS(Tu)=g(Tu). So TS has eigenvalue g corresponding to eigenvector Tu. But I don't know how to guarantee that Tu is nonzero. (Or how to resolve the possibility that it is zero)

Thanks for any help.

 

[Office_Shredder]

If Tu is zero, what's STu?

 

[balletomane]

If Tu=0

STu=0=gu, but u is nonzero, so g=0?

So does zero as an eigenvalue become a special case? Then would I do the nonzero eigenvalue case separately?