Unitary vector space/Isometry/Eigenvalues/Fact

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

${\displaystyle \varphi \colon V\longrightarrow V}$

denote a linear isometry.

Then the

modulus of every eigenvalue

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

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