Is there a formula in radicals for the eigenvalues, or at least the largest eigenvalue, of an $n\times n$ symmetric matrix, in terms of the entries? For what $n$ is there a formula? There obviously is if $n\leq 4$, and one suspects not if $n\geq 5$. However, the characteristic polynomial of a symmetric matrix is rather special (e.g. all roots are real) and could possibly be solvable by radicals for larger $n$, although I would guess not.

1$\begingroup$ Have you tried computing the Galois group of a random example? $\endgroup$ – Douglas Zare Apr 28 '17 at 3:20

5$\begingroup$ Edward Bender, Characteristic ppolynomials of symmetric matrices, proved that any totally real monic polynomial of odd degree with rational coefficients can occur as the characteristic polynomial of a symmetric rational matrix. Some discussion at math.stackexchange.com/questions/574443/… $\endgroup$ – Gerry Myerson Apr 28 '17 at 3:40

3$\begingroup$ Here us a formula literally "in radicals" for the largest eigenvalue, although probably not what you had in mind: $$ \lambda_1=\lim_{n\to\infty}\sqrt[n]{\A^n\}.$$ (Of course, if you want a purely algebraic expression, there is no way to distinguish the largest eigenvalue from the others.) $\endgroup$ – Victor Protsak Apr 28 '17 at 5:38

3$\begingroup$ About Gerry's comment. Given a totally real monic polynomial $P$ of degree $n$, a simple procedure constructs a "companion" in the form of a real symmetric tridiagonal matrix $T$. If $P$ has rational coefficients (or in a number field $k$), then $T$ has rational entries (resp. in $k$). See Exercise #92 in my list perso.enslyon.fr/serre/DPF/exobis.pdf . $\endgroup$ – Denis Serre Apr 28 '17 at 6:16

$\begingroup$ If one takes your question more broadly as a sign that you need to know more about maps (class of numbers to take the entries of the matrix from)$\rightarrow$(class of all numbers that arise as ev of symmetric matrices), then the following might help: for any finite symmetric matrix, if all entries come from a fixed realclosed field $K$, then then all eigenvalues of the matrix are still in $K$. $\endgroup$ – Peter Heinig Apr 28 '17 at 7:41
If I calculated correctly, $$\begin{pmatrix}2 & 1 & 0 & 0 & 0\newline 1 & 3 & 1 & 0 & 0\newline 0 & 1 & 1 & 1 & 0 \newline 0 & 0 & 1 & 1 & 1 \newline 0 & 0 & 0 & 1 & 1 \end{pmatrix}$$ has characteristic polynomial $x^5+8x^420x^3+15x^2+4x5$ which is irreducible mod $3$ and has $3$ linear factors and a quadratic factor mod $79$, so its Galois group is $S_5$, so its roots can't be expressed with just radicals.

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