User harish - MathOverflowmost recent 30 from http://mathoverflow.net2013-06-20T09:39:12Zhttp://mathoverflow.net/feeds/user/20217http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/86966/identifying-factors-of-higher-order-in-a-determinantIdentifying factors of higher order in a determinant Harish2012-01-29T16:29:10Z2012-01-29T20:12:27Z
<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-systemRegular Points of a polynomial systemHarish2011-12-28T02:50:33Z2011-12-28T03:46:32Z
<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#86980Comment by HarishHarish2012-01-30T03:06:42Z2012-01-30T03:06:42ZI 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 & (x + a_1)*(y + a_1) & (y + a_1) \\
(x + a_2)^2 & (x + a_2)*(y + a_2) & (y + a_2) \\
(x + a_3)^2 & (x + a_1)*(y + a_3) & (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.