Field extension containing the eigenvectors of a Hermitian matrix Let H be a (finite-dimensional) Hermitian matrix with algebraic numbers for its entries, all of which lie in some minimal field extension of the rational numbers; call this field ℚ(H) for short.  Let's assume that the eigenvalues of H are distinct, and let D be the diagonal matrix of eigenvalues of H in non-increasing order, say.  Since H is Hermitian with a non-degenerate spectrum, there is a unique unitary matrix U that diagonalizes H.  The normalized eigenvectors of H comprise the columns of U.  Finally, let ℚ(U,D) be the field extension of the rationals containing the matrix elements of U and D.
Is there any relationship between ℚ(H) and ℚ(U,D)?  In particular, I would like to know a bound on the degree of ℚ(U,D)/ℚ(H).  Also if possible, is there a way to take H (without diagonalizing it!) and calculate ℚ(U,D)? I'd also be happy with something that contains ℚ(U,D) and isn't that much bigger.  (I mean, its only bigger by a factor that is constant or depends on the dimension, not on the particular H.)
 A: Do you expect something smaller than $n!\cdot 2^{n-1}$, where $n$ is the size of our matrix $H$ ? We can get $n!\cdot 2^n$ the following way: In order to get the entries of $U$ and $D$ into our field, we extend our field step by step: First, we adjoin all the eigenvalues of the matrix (they are roots of a polynomial of degree $n$, namely the characteristic polynomial of $H$, so we need an extension of degree $n!$ in the worst case), then find the corresponding eigenvectors using Gauss' elimination algorithm (this requires no field extension), and then normalize these eigenvectors (requiring a field extension of degree $2$ for each eigenvector, as we have to divide by a square root). There is a subtle point here - we need to know that during each step of our stepwise field extension, the field that we get is symmetric with respect to the real axis (i. e., for every complex number $z$ lying in our field, its conjugate $z^{\ast}$ lies in the field as well). This is clear for our initial field $\mathbb{Q}\left(H\right)$ (in fact, the matrix $H$ is Hermitian, so for each entry it contains, it also contains its conjugate), and remains correct during all the extension steps that we perform (in fact, each of these steps consists of adjoining _all_roots of some given real polynomial (in particular, the characteristic polynomial of $H$ is real, and adjoining the square root of the length of a vector means adjoining all roots of a real quadratic), and it is clear that such a step won't destroy the symmetry with respect to the $x$-axis). Adding the observation that only the first $n-1$ eigenvectors need a quadratic extension of the base field to be normalized (because the $n$-th vector can then be taken as the (generalized to $n$ dimensions) cross product of the first $n-1$ ones), we can lower the $n!\cdot 2^n$ bound to $n!\cdot 2^{n-1}$. At least for $n=2$, this is also optimal. Any ideas on higher $n$?
A: 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
$$
d \leq \varphi(n) < n.
$$
