linear algebra - Field extension containing the eigenvectors of a Hermitian matrix

A special case is the following:

Pick:

An integer $n$ that is a square;

$$
H =F^{*} D F
$$

a matrix with $n$ lines and $n$ columns

where

$D$ is a diagonal square matrix with $n$ lines and with integer coefficients
$F$ is the Fourier matrix, with $n$ lines defined by

$$
F = (1/sqrt{n}) (s^{(i-1)(j-1)})
$$

where

$$
s = e^{-2 pi i/n}
$$

$*$ means conjugate-transpose.

You get

$H$ is hermitian with entries algebraic integers.

$H$ is also a circulant matrix.

Pick now:

$$
U =F^{*}
$$

so that

$$
Q(H) subseteq Q(U) = Q(s)
$$

while

$$
Q(U,D) = Q(F^{*},D) = Q(s).
$$

Observe that
$$
Q(s)
$$
is the classic extension of $Q$ containing the $n$-th roots of unity
so that it has degree

$$
varphi(n)
$$

over $Q$, where $varphi$ is the Euler totient's function.

Thus,

The extension $Q(U,D)$ over $Q(H)$ has degree $d$ bounded above by $varphi(n).$

Observe that this degree $d$ is substantially slower than $n !$ since

$$
d leq varphi(n) < n.
$$