Unitary vector space/Isometry/Eigenvalues/Fact/Proof

Proof

Let $\phi (v)=\lambda v$ with $v\ne 0$, that is, $v$ is an eigenvector for the eigenvalue $\lambda \in 𝕂$. Due to the isometry property, we have

$\Vert v\Vert =\Vert \phi (v)\Vert =\Vert \lambda v\Vert =|\lambda |\cdot \Vert v\Vert .$

Because of $\Vert v\Vert \ne 0$, this implies $|\lambda |=1$. In the real case, this means $\lambda =\pm 1$.

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