Perhaps you're looking for something like this:
The algebraic and geometric multiplicities of $\lambda$ are different iff
there is a vector $v$ such that $(A - \lambda I) v \ne 0 $ but $(A - \lambda I)^2 v = 0$.
You could write this condition (in the case $\lambda \ne 0$) as:
there is a vector $v$ such that $ A^2 v = 2 \lambda A v - \lambda^2 v \ne \lambda^2 v $.
EDIT:
A necessary condition is that the discriminant of the characteristic polynomial is $0$. If so, the $\lambda$'s to consider are the roots of the greatest common divisor (in the sense of polynomials) of the characteristic polynomial and its derivative. So you don't need all the eigenvalues, but to get a positive answer you do need to find some of them.