**Part 1.** Let $Av = \lambdav$ with $v \neq \mathbf{0}$. Then $\lambdav^\daggerv = v^\dagger(Av) = (A^\daggerv)^\daggerv = (Av)^\daggerv = \overline{\lambda}v^\daggerv$. Since $v^\daggerv = \sum |v_i|^2 > 0$, we can divide: $\lambda = \overline{\lambda}$, so $\lambda \in \mathbb{R}$. **Part 2.** Let $Av = \lambdav$ and $Aw = \muw$ with $\lambda \neq \mu$. Then $\lambdaw^\daggerv = w^\dagger(Av) = (Aw)^\daggerv = \muw^\daggerv$ (using $\overline{\mu} = \mu$ from Part 1). So $(\lambda - \mu)w^\daggerv = 0$, and since $\lambda \neq \mu$, we get $w^\daggerv = 0$.