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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.

flag	asked Jan 29 2012 at 16:29 Harish 11●2

Ira Gessel · Answer 1 · 2012-01-29 20:12:27Z UTC

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If setting $x=y$ makes the rank go down by $k$, then $(x-y)^k$ is a factor. Harald Helfgott and I used this idea in evaluating a determinant http://www.combinatorics.org/Volume_6/Abstracts/v6i1r16.html); actually the determinant was evaluated earlier by this method by Zavrotsky. The reference we gave for the fact relating the rank of the matrix and the multiplicity of $x-y$ as a factor is R. A. Frazer, W. J. Duncan, and A. R. Collar, Elementary Matrices and Some Applications to Dynamics and Differential Equations, Cambridge University Press, 1947, page 17.

edited Jan 29 2012 at 20:12

answered Jan 29 2012 at 20:05

Ira Gessel
2,361●5●15

	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. – Harish Jan 30 2012 at 3:06
	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 $x-y$ – Ira Gessel Jan 30 2012 at 3:25

Identifying factors of higher order in a determinant

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