Every polynomial is the characteristic polynomial of a matrix (see companion matrix). Therefore, the eigenvalues of a matrix over a field $\mathbb F$ must belong to $\mathbb F$ if, and only if, $\mathbb F$ is algebraically closed.
However, the eigenvalues of a real symmetric matrix are always real, despite $\mathbb R$ being not algebraically closed. What is up with that?
Typically, the eigenvalues of a symmetric matrix over a field $\mathbb F$ are not required to belong to $\mathbb F$. Indeed, the matrix $$X = \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix},$$ which can be taken over any field, have eigenvalues $0$ and $\pm\sqrt2$, so $\mathbb Q$ (as well as a bunch of finite fields) have not this property. Is there a nice way to characterize the fields with it? What is special about $\mathbb R$?