If $\lambda$ is a simple root of $A$, a characteristic vector $x$ associated with $\lambda$ can always be taken to be a vector whose components are polynomials in $\lambda$ and the elements of $A$. Original problem description.
The problem suggests using the fact that if $\lambda$ is a simple root of $A$ then $|A_k-\lambda I|$ is nonzero for some $k$ where $A_k$ is a $(N-1)\times(N-1)$ matrix obtained by removing the $k$th row and column.
But before I figure out why that is relevant. Why doesn't something like row reduction on $(A-\lambda I)x=0$ using the minimal polynomial of the particular $\lambda$ in question for simplification work to find a characteristic vector $x$ whose components are polynomials in $\lambda$ and the elements of $A$? An eigenvector with rational components should be obtained but the denominators can be removed by multiplying by a common denominator. What obvious error am I missing?