A determinant involving only cyclotomic factors - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T11:28:39Zhttp://mathoverflow.net/feeds/question/48729http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/48729/a-determinant-involving-only-cyclotomic-factorsA determinant involving only cyclotomic factorsRoland Bacher2010-12-09T08:40:36Z2010-12-09T08:52:32Z
<p>Let $\alpha:\mathbb Z\longrightarrow \mathbb Z$ be a quadratic polynomial taking only integral values on the integers and consider the sequence of square-matrices with coefficients
$x^{\alpha(i+j)}$ for $i,j\in {0,\dots,n}$. The determinant of such a matrix involves a (not nessecarily positive) power of $x$ and a polynomial which factors experimentally always into a product of cyclotomic polynomials. Is this always true or is
there a counterexample to this observation?</p>
http://mathoverflow.net/questions/48729/a-determinant-involving-only-cyclotomic-factors/48730#48730Answer by Gjergji Zaimi for A determinant involving only cyclotomic factorsGjergji Zaimi2010-12-09T08:52:32Z2010-12-09T08:52:32Z<p>You can always factor out $x^{\alpha(i)}$ from each row and $x^{\alpha(j)}$ from each column, then multiply by $x^{n\alpha(0)}$ and you reduce to $\det(x^{\beta ij})=\prod (x^{\beta i}-x^{\beta j})$ by Vandermonde, where $\beta$ is some integer.</p>