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

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

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