I have to prove that if $\lambda$ is an eigenvalue of T then $\bar\lambda$ is eigenvalue of $T^*$ (adjoint)
I know that $<Tv,u> = <\lambda v,u> = \bar\lambda<v,u>=<v,\bar\lambda u>$
and that $<Tv,u>=<v,T^*u>$
but does this imply that there is a $u$ such that $T^*u=\bar\lambda u ?$
I believe something is wrong here. Any help?