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I would like to get an explicit expression for the determinant of a Jacobian $$J_{ij} = \delta_{ij}\left(\sigma^{2}+\sum_{k}w_{ik}y_{k}^{2}\right)-w_{ij}y_{i}y_{j},$$ where $i,j = 1,...,n$, $w_{ij}\ge 0$ and $\delta_{ij}$ is the Kronecker delta.

I was able to find explicit expression for special cases of $w_{ij}$ but not for the general case.

Any hint to a solution would be greatly appreciated.

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The eigenvalues can be found explicitly. Consider separately how $J_{ij}$ acts on the vector $y^{i}$ Euclidean dual to $y_{i}$ and how it acts on space of vectors annihilated by $y_{i}$. – Dan Fox Aug 29 '11 at 12:53
Dan Fox's technique works because the matrix is $\sigma^2 I + J_1$ where $I$ is the identity matrix and $J_1$ is of rank 1: in general if $M,M_1$ are square matrices of the same size and $M_1$ has rank 1 then $\det(M+xM_1)$ is a polynomial in $x$ of degree at most 1. – Noam D. Elkies Aug 29 '11 at 13:02
I am not sure whether I can follow your answers. If my matrix had the form $\sigma^2I+J_1$ with rank one $J_1$ I would use Sylvester's determinant theorem and be done. However, in my case both matrices can have full rank: the first one is a diagonal matrix and the second is the elementwise product of an outer product with the matrix $W$. – fabee Aug 29 '11 at 14:07
It's true that $J y = \sigma^2 y$, but I don't see anything useful about the action on vectors annihilated by $y_i$. – Robert Israel Aug 29 '11 at 18:08
Sorry, I (and apparently Dan) must have misread $w_{ij}$ as $w_i w_j$. As it stands, the matrix seems to be so general that there'll be nothing simpler than using general determinant formulas. – Noam D. Elkies Aug 29 '11 at 19:18

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