$\rho(v) = \lambda v$, $\rho(v') = \lambda ' v'$ where $\lambda ' \neq \lambda\Rightarrow \langle v,v'\rangle = 0$

Question

Show that if $\rho$ is a normal operator and $\rho(v) = \lambda v$, $\rho(v') = \lambda ' v'$ where $\lambda ' \neq \lambda\Rightarrow \langle v,v'\rangle = 0$

My attempt:

First, $\langle \rho^*(v'),\rho^*(v')\rangle = \langle v',\rho(\rho^*(v'))\rangle = \langle v', \rho^*(\rho(v'))\rangle = \bar{\lambda'}\langle v', \rho^*(v')\rangle = \langle\bar{\lambda'} v', \rho^*(v')\rangle \Rightarrow \rho^*(v') = \bar{\lambda'} v'$

So, $\lambda\langle v, v'\rangle = \langle\lambda v, v'\rangle = \langle\rho(v), v'\rangle = \langle v, \rho^* (v')\rangle = \langle v, \lambda' v'\rangle = \lambda'\langle v, v'\rangle \Rightarrow (\lambda - \lambda')(\langle v, v'\rangle) = 0$ $\Rightarrow \langle v, v'\rangle = 0$

Answer 1

I think you have the right idea, but there is one missing piece: that $\rho^*(v')=\overline{\lambda'}v'$.

Step 1: one can show that, since $\rho$ is normal and $I$ is self-adjoint, the operator $T:=\rho-\lambda' I$ is normal (I do this below in the hidden box).

$$\begin{align}TT^*&=(\rho-\lambda'I)(\rho-\lambda'I)^*\\&=(\rho-\lambda'I)(\rho^*-\overline{\lambda'}I^*)\\&=\rho\rho^*-\lambda'\rho^*-\overline{\lambda'}\rho-\lambda'\overline{\lambda'}I\\&=\rho^*\rho-\overline{\lambda'}\rho-\lambda'\rho^*-\overline{\lambda'}\lambda'I\\&=(\rho^*-\overline{\lambda'}I^*)(\rho-\lambda'I)=T^*T.\end{align}$$

Step 2: since $T$ is normal, we have $\Vert Tx\Vert=\Vert T^*x\Vert $ for all $x$ (I show this below in the hidden box).

$$\Vert Tx\Vert^2=\langle Tx,Tx\rangle=\langle x,T^*Tx\rangle=\langle x,TT^*x\rangle=\langle T^*x,T^*x\rangle=\Vert T^*x\Vert^2.$$

Step 3: since $v'$ is a $\lambda'$ eigenvector of $\rho$, we have $\Vert Tv'\Vert=0$. At this point you should be able to conclude the desired result.

I will also note: in your proof you write $\langle v,\rho^*(v')\rangle=\langle v,\lambda'v'\rangle$, but you are forgetting the conjugate on $\lambda'$, then this goes away when using conjugate linearity in the second argument of the inner product.