Let $\rho$ be a normal operator, show $\rho^*(v) = 0 \iff \rho(v) = 0$

Question

Let $\rho$ be a normal operator, show that $\rho^*(v) = 0 \iff \rho(v) = 0$

My attempt:

$\Rightarrow$

$\langle \rho(v),v\rangle = \langle v,\rho^*(v)\rangle = \langle v,0\rangle = 0 \Rightarrow \rho(v) = 0.$

$\Leftarrow$

$\langle \rho(v),v\rangle = \langle 0,v\rangle = \langle v,\rho^*(v)\rangle = 0 \Rightarrow \rho^*(v) = 0.$

Is it right or that also implies $v=0$?

Answer 1

$\langle \rho(v),v\rangle=0$ only means that $\rho (v)$ is perpendicular to $v $. Proceed as given below:

$$ \begin {align*} \langle \rho(v),\rho (v)\rangle &= \langle v,\rho^*\rho(v)\rangle\\ & = \langle v,\rho\rho^*(v)\rangle = 0. \end {align*}$$