D**u
发帖数: 204 | 1 Suppose x_1,...,x_n are n positive numbers, prove that the n*n matrix
(1/(x_i+x_j)) (1<=i,j<=n) is positive definite.
Background: this is called Cauchy matrix, and the determinant can be
directed
computed (ref: http://en.wikipedia.org/wiki/Cauchy_determinant ).
What I am asking is: can you still prove the problem by not using
any polynomial-factoring method. Probabilistic/statistical method are
especially welcome. | x******i
发帖数: 3022 | 2 define
f_j(t) := exp(-x_j*t)
define inner product
(a,b) = \int_0^{\infty} a(t)*b(t)*dt
then
the matrix is
C_ij = (f_i,f_j)
which is obviously positive definite, because
sum a_i*a_j*C_ij = (sum a_i*f_i, sum a_j*f_j) >=0
【在 D**u 的大作中提到】
: Suppose x_1,...,x_n are n positive numbers, prove that the n*n matrix
: (1/(x_i+x_j)) (1<=i,j<=n) is positive definite.
: Background: this is called Cauchy matrix, and the determinant can be
: directed
: computed (ref: http://en.wikipedia.org/wiki/Cauchy_determinant ).
: What I am asking is: can you still prove the problem by not using
: any polynomial-factoring method. Probabilistic/statistical method are
: especially welcome.
| D**u
发帖数: 204 | 3 Great. Here is a "probabilistic" solution assuming that we do not notice to
use f_j(t) := exp(-x_j*t) right away.
Assume Y_i is a Poisson arrival process with arrival rate x_i, then we know
that
E(Y_i) = 1/x_i, and E(min(Y_i,Y_j)) = 1/(x_i+x_j). Suppose the CDF of Y_i
and Y_j are P_i(t) and P_j(t), then easy to see the CDF of min(Y_i,Y_j) is
1-(1-P_i(t))*(1-P_j(t)). Then with a little computation, we see that
E(min(Y_i,Y_j)) = \int_0^{\infty} (1-P_i(t))*(1-P_j(t)) dt
= \int_0^{\infty} exp(-x_
【在 x******i 的大作中提到】
: define
: f_j(t) := exp(-x_j*t)
: define inner product
: (a,b) = \int_0^{\infty} a(t)*b(t)*dt
: then
: the matrix is
: C_ij = (f_i,f_j)
: which is obviously positive definite, because
: sum a_i*a_j*C_ij = (sum a_i*f_i, sum a_j*f_j) >=0
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