My collaborators and I are studying certain rigidity properties of hyperbolic toral automorphisms.

These are given by integral matrices A with determinant 1 and without eigenvalues on the unit circle.

We obtain a result under two additional assumptions

1) Characteristic polynomial of the matrix A is irreducible

2) Every circle contains no more than two eigenvalues of A (i.e. no more than two eigenvalues have the same absolute values)

We feel that the second assumption holds for a "generic" matrix. Is it true?

To be more precise, consider the set X of integral hyperbolic matrices which have determinant 1 and irreducible characteristic polynomial. What are the possible ways to speak of a generic matrix from X? Does assumption 2) hold for generic matrices?

Comments:

- Assumption 1) doesn't bother us as it is a necessary assumption.
- Probably it is easier to answer the question when X is the set off all integral matrices. In this case we need to know that hyperbolicity is generic, 2) is generic and how generic is irreducibility.