User harish - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:39:12Z http://mathoverflow.net/feeds/user/20217 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86966/identifying-factors-of-higher-order-in-a-determinant Identifying factors of higher order in a determinant Harish 2012-01-29T16:29:10Z 2012-01-29T20:12:27Z <p>Consider a $n\times n$ matrix $A$ whose elements are some polynomials in the indeterminates $x_1, x_2,\ldots,x_m$. To calculate the determinant of such a matrix, one of the usual ways is to treat the determinant as a polynomial in $x_1,\ldots,x_m$ and identify its factors. The usual idea being if $x = y$ makes the determinant vanish then $x -y$ is one of the factors. What I however do not understand is how to identify its order, that is to identify the exact $k$ such that $(x - y)^k$ is the factor. </p> http://mathoverflow.net/questions/84427/regular-points-of-a-polynomial-system Regular Points of a polynomial system Harish 2011-12-28T02:50:33Z 2011-12-28T03:46:32Z <p>With $x\equiv (x_1, x_2, x_3, x_4),$ consider the polynomial system, $P(x_{1}, x_{2}, x_{3}, x_{4}) = 0$, given by $P_1(x) = x_1 * x_3,$<br> $P_2(x) = x_2 * x_4,$<br> $P_3(x) = x_1 + x_3 + \beta_1(x_1 * x4 + x2 * x3),$<br> $P_4(x) = x_2 + x_4 + \beta_2(x_1 * x4 + x2 * x3).$<br> I want to know if there is a simple way to show that the set of points where the Jacobian matrix is non-singular is an open dense set in $\mathbb{R} ^4$. My present ideas go something like this:<br> The determinant of the Jacobian matrix is clearly a polynomial, say $g:\mathbb{R}^4 \rightarrow \mathbb{R}.$ I try to show $g$ is NOT an identically zero polynomial and hence its set of zeros, or equivalently the set of critical points, has measure zero. Hence its complement, the set of regular points, is dense. Lastly, since $g$ is continuous, the set of regular points is open. </p> http://mathoverflow.net/questions/86966/identifying-factors-of-higher-order-in-a-determinant/86980#86980 Comment by Harish Harish 2012-01-30T03:06:42Z 2012-01-30T03:06:42Z I do not seem to have understood your idea of rank going down. For example, consider the matrix $\left( \begin{array}{ccc} (x + a_1)^2 &amp; (x + a_1)*(y + a_1) &amp; (y + a_1) \\ (x + a_2)^2 &amp; (x + a_2)*(y + a_2) &amp; (y + a_2) \\ (x + a_3)^2 &amp; (x + a_1)*(y + a_3) &amp; (y + a_3) \\ \end{array} \right)$. The determinant of this matrix is given by $(x - y)^2 * (a_1 - a_2) * (a_1 - a_3) * (a_2 - a_3).$ The rank initially is $3$. After substituting $x= y$, the rank becomes $2$. Am i missing something? The idea of derivatives given in the book that you suggested, however, is something I could work with.