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.

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
