Unitary vector space/Isometry/Eigenvalues/Fact

Let ${}V$ be a finite-dimensional ${}{\mathbb {K} }$-vector space, and let

$\varphi \colon V\longrightarrow V$

denote a linear isometry.

Then the

modulus of every eigenvalue

of ${}\varphi$ is ${}1$.

In case ${}{\mathbb {K} }=\mathbb {R}$, only the eigenvalues ${}1$ and ${}-1$ are possible.

Proof, Write another proof