Unitary vector space/Isometry/Eigenvalues/Fact/Proof

Let ${\displaystyle {}\varphi (v)=\lambda v}$ with ${\displaystyle {}v\neq 0}$, that is, ${\displaystyle {}v}$ is an eigenvector for the eigenvalue ${\displaystyle {}\lambda \in {\mathbb {K} }}$. Due to the isometry property, we have

${\displaystyle {}\Vert {v}\Vert =\Vert {\varphi (v)}\Vert =\Vert {\lambda v}\Vert =\vert {\lambda }\vert \cdot \Vert {v}\Vert \,.}$

Because of ${\displaystyle {}\Vert {v}\Vert \neq 0}$, this implies ${\displaystyle {}\vert {\lambda }\vert =1}$. In the real case, this means ${\displaystyle {}\lambda =\pm 1}$.