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A Cauchy matrix is an $m$-by-$n$ matrix $A$ whose elements have the form $a_{i,j} = \frac{1}{x_i-y_j}$, with $x_i \neq y_j$ for all $(i, j)$, and the $x_i$'s and $y_i$'s belong to a field (http://en.wikipedia.org/wiki/Cauchy_matrix). Also it seems to be part of the definition that the $x_i$'s and $y_j$'s are all distinct (does anyone know why?). I am only interested in the case where the field is the real numbers, and all the $x_i$'s and $y_j$'s are positive integers. My question is, what is known about the eigenvalues of a square, real Cauchy matrix? There is a formula for its determinant, which gives you their product, and the trace of the matrix, which is their sum, is easy to find. I have Googled this extensively and found almost nothing.

I originally posted this on Math Stack Exchange but I got no answers so I removed the question and I am posting it here.

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If $x_i=x_j$ or $y_i=y_j$ for some $i\neq j$, then the Cauchy determinant would be zero, and the matrix singular. –  Dietrich Burde Aug 3 '13 at 18:36
    
@DietrichBurde : I realize that. Why would it be so terrible if the matrix were singular? –  Stefan Aug 3 '13 at 19:03
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It would not be terrible. You can define singular Cauchy matrices, if you want. For solving Cauchy systems of equations invertible matrices are preferred. –  Dietrich Burde Aug 3 '13 at 19:33

2 Answers 2

up vote 3 down vote accepted

Suppose $x_i > 0$ and $y_j -x_j$, then $c_{ij} = 1/(x_i+x_j)$. These matrices are infinitely divisible, i.e., $[c_{ij}^r]$ is also positive definite for all $r > 0$.

Spectral properties of Cauchy-like matrices and kernels are studied here. For additional information and an easier read, on the general case ($1/(x_i+y_j)$), you might enjoy looking at the recent book by Pinkus.

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Thanks. In your first sentence, do you mean $y_j - x_j > 0$ for all $j$? –  Stefan Aug 4 '13 at 17:10
    
in the case $x_i > 0$ and $c_{ij} = 1/(x_i+x_j)$, are any bounds known on the (positive) eigenvalues of the matrix? –  Stefan Aug 4 '13 at 17:20
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@Stefan: Have a look at my older answer here: mathoverflow.net/a/137093/8430 --- that bound (suitably modified to your case) will yield a lower bound on the eigenvalues; more refined analysis is also possible for upper bounds I think. –  Suvrit Aug 4 '13 at 17:32

In some cases the Cauchy matrix can be shown to be positive definite, see:

Horn, Roger A. The Hadamard product. Matrix theory and applications (Phoenix, AZ, 1989), 87–169, Proc. Sympos. Appl. Math., 40, Amer. Math. Soc., Providence, RI, 1990. 15-02 (15A42 15A45 15A69)

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