- #1

balletomane

- 25

- 0

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.