Timeline for Identifying factors of higher order in a determinant
Current License: CC BY-SA 3.0
4 events
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| Jan 30, 2012 at 3:25 | comment | added | Ira Gessel | It seems that this theorem does not always give the exact power of $x-y$ dividing the determinant: if setting $x=y$ makes the rank go down by $k$ then $(x-y)^k$ is a factor of the determinant, but the highest power of $x-y$ dividing the determinant could be greater than $k$. I don't know of a stronger result that gives the exact power of $x-y$ | |
| Jan 30, 2012 at 3:06 | comment | added | Harish | 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 & (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. | |
| Jan 29, 2012 at 20:12 | history | edited | Ira Gessel | CC BY-SA 3.0 |
added 2 characters in body; deleted 1 characters in body
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| Jan 29, 2012 at 20:05 | history | answered | Ira Gessel | CC BY-SA 3.0 |