Which polynomials are characteristic polynomials of a symmetric matrix? Let $f(x)$ be a polynomial of degree $n$ with coefficients in $\mathbb{Q}$.  There are well-known ways to construct a $n \times n$ matrix $A$ with entries in $\mathbb{Q}$ whose characteristic polynomial is $f$.  My question is: when is it possible to choose $A$ symmetric?
An obvious necessary condition is that the roots of $f$ are all real, but it is not clear to me even in the case $n = 2$ that this is sufficient.  In degree $2$ this comes down to determining whether or not every pair $(p,q)$ which satisfies $p^2 > 4q$ (the condition that $x^2 + px + q$ has real roots) can be expressed in the form
$$p = -(a + c)$$
$$q = ac - b^2$$
where $a$, $b$, and $c$ are rational.  I have some partial results from just fumbling around and checking cases, but it seems clear that a more conceptual argument would be required to handle larger degrees.
 A: I found an answer to your question in the following paper:
Estes, Dennis R.(1-SCA); Guralnick, Robert M.(1-SCA)
Minimal polynomials of integral symmetric matrices.
Computational linear algebra in algebraic and related problems (Essen, 1992).
Linear Algebra Appl. 192 (1993), 83–99. 
In the introduction, the authors mention:
"Bender, in a series of papers, studied this problem for $\Bbb Q$, the
rationals. He proved that any totally real monic polynomial over $\Bbb Q$ of odd degree can occur as the characteristic polynomial of a symmetric rational matrix. 
This fails already for polynomials of degree 2. The problem of
determining precisely which polynomials can occur as characteristic polynomials of rational symmetric matrices is still not solved. The conjecture is that this is a local problem (i.e., if $f \in \Bbb Q[X]$ is monic of degree $n$, then $f$ is the characteristic polynomial of some element of $S_n(\Bbb Q)$ if and only if $f$ is the
characteristic polynomial of some element of $S_n(\Bbb R)$ and $S_n(\Bbb Q_p)$ for all primes $p$."
A: I guess that the following paper shows how, for any given polynomial, it can be determined (calculated) a certain symmetric matrix A for which said polynomial is the characteristic polynomial of A (see link below).
Given the demonstration presented in the paper, I would not say that given a certain polynomial, matrix A could "be chosen", since that would imply that the degrees of freedom allow for such choosing to be possible.
As the demonstration shows, for any given polynomial a certain matrix A can be calculated, which means that there always is a system of linear equations that can be solved (the determinant of the system of linear equations is != 0), so, it means that that the degrees of freedom is always zero, so, there is no way that matrix A could be "chosen", even though the solution of A could be guessed.
Kind regards, GEN
http://ac.els-cdn.com/0024379590903235/1-s2.0-0024379590903235-main.pdf?_tid=bd2c75aa-42cb-11e6-aaf5-00000aacb362&acdnat=1467735530_b38cd0adce0c4955dffb08b1a1bb8882
