Let $A(x)$ be a matrix with entries that are polynomials in say $\mathbb{Z}$. Suppose furthermore that $A$ is invertible. For any fixed $x$ we can find an eigenvalue of $A$ (in $\mathbb{C}$) and this eigenvalue varies smoothly with $x$.
Is there a rational function $R(x)$ that acts as an eigenvalue for $A(x)$ for any $x$? What about a Laurent series? I suspect the answer for the first question is false in general but the second question is true. I can't find a proof though.
With $A$ having entries in $\mathbb Z[x]$, we can find its eigenvalues in the usual way by solving the characteristic equation $\det(A- tI)=0$ for $t$, where we treat $x$ as an unknown (real number, let's say). In general then, this means solving a polynomial equation $p(t)=0$ where $p$ has coefficients in $\mathbb Z[x]$. In general you of course can't solve this for $t$ as a rational function of $x$, or even get any closed form expression.