Spectrum of a generic integral matrix. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:03:38Z http://mathoverflow.net/feeds/question/21775 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21775/spectrum-of-a-generic-integral-matrix Spectrum of a generic integral matrix. Andrey Gogolev 2010-04-18T20:32:25Z 2010-04-19T22:53:56Z <p>My collaborators and I are studying certain rigidity properties of hyperbolic toral automorphisms.</p> <p>These are given by integral matrices A with determinant 1 and without eigenvalues on the unit circle.</p> <p>We obtain a result under two additional assumptions</p> <p>1) Characteristic polynomial of the matrix A is irreducible</p> <p>2) Every circle contains no more than two eigenvalues of A (i.e. no more than two eigenvalues have the same absolute values)</p> <p>We feel that the second assumption holds for a "generic" matrix. Is it true?</p> <p>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?</p> <p>Comments:</p> <ul> <li>Assumption 1) doesn't bother us as it is a necessary assumption.</li> <li>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.</li> </ul> http://mathoverflow.net/questions/21775/spectrum-of-a-generic-integral-matrix/21906#21906 Answer by David Speyer for Spectrum of a generic integral matrix. David Speyer 2010-04-19T22:53:56Z 2010-04-19T22:53:56Z <p>Yes, a generic integer matrix has no more than two eigenvalues of the same norm. More precisely, I will show that matrices with more than two eigenvalues of the same norm lie on a algebraic hypersurface in $\mathrm{Mat}_{n \times n}(\mathbb{R})$. Hence, the number of such matrices with integer entries of size $\leq N$ is $O(N^{n^2-1})$. </p> <p>Let $P$ be the vector space of monic, degree $n$ real polynomials. Since the map "characteristic polynomial", from $\mathrm{Mat}_{n \times n}(\mathbb{R})$ to $P$ is a surjective polynomial map, the preimage of any algebraic hypersurface is algebraic. Thus, it is enough to show that, in $P$, the polynomials with more than two roots of the same norm lie on a hypersurface. Here are two proofs, one conceptual and one constructive.</p> <p><strong>Conceptual:</strong> Map $\mathbb{R}^3 \times \mathbb{R}^{n-4} \to P$ by <code>$$\phi: (a,b,r) \times (c_1, c_2, \ldots, c_{n-4}) \mapsto (t^2 + at +r)(t^2 + bt +r) (t^{n-4} + c_1 t^{n-5} + \cdots + c_{n-4}).$$</code></p> <p>The polynomials of interest lie in the image of $\phi$. Since the domain of $\phi$ has dimension $n-1$, the Zariski closure of this image must have dimension $\leq n-1$, and thus must lie in a hyperplane.</p> <p><strong>Constructive:</strong> Let $r_1$, $r_2$, ..., $r_n$ be the roots of $f$. Let <code>$$F := \prod_{i,j,k,l \ \mbox{distinct}} (r_i r_j - r_k r_l).$$</code> Note that $F$ is zero for any polynomial in $\mathbb{R}[t]$ with three roots of the same norm. Since $F$ is symmetric, it can be written as a polynomial in the coefficients of $f$. This gives a nontrivial polynomial condition which is obeyed by those $f$ which have roots of the sort which interest you.</p>