Unitary vector space/Isometry/Eigenvalues/Fact/Proof

Proof

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

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

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

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