Prove/disprove: $v\in V$ is an eigenvector of $T$ with an eigenvalue $\lambda\iff v$ is an eigenvector of $T^*$ with eigenvalue $\overline \lambda$.

Question

Prove/disprove: $v\in V$ is an eigenvector of $T$ with an eigenvalue $\lambda$ if and only if $v$ is an eigenvector of $T^*$ with eigenvalue $\overline \lambda$.

Hey all. I am looking to disprove this. We'll assume $Tv=\lambda v$. since $\langle Tv,v\rangle=\langle v,T^*v\rangle=\langle\lambda v,v\rangle=\lambda \|v\|^2 $, if $v$ is not necessarily an eigenvector then $T^*v=u$ where $u\notin \operatorname{Sp}\{v\}$ ($u\in V$), so we have $\lambda \|v\|^2 =\langle v,u\rangle$ and I see no reason this cannot happen. I don't really know where to look for an example to disprove this claim since I am not so good with complex numbers (unless of course this claim is true).

I would love to get your help, Thanks in advance :)

Answer 1

Let $V=l^2$ and $T : V \to V$ defined by $T(x_1,x_2,x_3,...)=(x_2,x_3,...)$.

Then $T^*$ is given by $T^*(x_1,x_2,x_3,...)=(0,x_1,x_2,...)$.

Let $v=(1,0,0,...)$, then $Tv=0$ , hence $v$ is an eigenvector of $T$ with eigenvalue $\lambda=0$.

But $T^*v=(0,1,0,0,...)$.