Dimostrazione.
Sia $\mathbf{v} \neq 0$ tale che $f(\mathbf{v}) = \lambda \mathbf{v}$. Avremo
$ \ll f(\mathbf{v}),\mathbf{v} \gg \, = \, \ll \lambda \mathbf{v}, \mathbf{v} \gg \, = \, \lambda \ll \mathbf{v},\mathbf{v} \gg$, e
$ \ll f(\mathbf{v}),\mathbf{v} \gg \, = \, \ll \mathbf{v}, f(\mathbf{v}) \gg \, = \, \overline{\lambda} \ll \mathbf{v},\mathbf{v} \gg$;
allora, per $\mathbf{v} \neq 0; \, \ll \mathbf{v},\mathbf{v} \gg \neq 0 \, $ implica che $ \, \lambda = \overline{\lambda} \, \Rightarrow \, \lambda \in \mathbf{R}$.
c.v.d.