binomial determinant

[Edwin Clark]

Benoit,

In looking at det([binomial(i*z_j,j)]) it seemed to me that it would help
to know det([binomial(a_i*z_j,j)] ) which appears to be related to the
vandermonde determinant. For example I get the following generalization
of your determinant using Maple:

det([binomial(a_i*z_j,j)]

= 1/(C(n))* prod(a_i,i=1..n)*prod(a_i-a_j, 1<=j<i <=n)*prod(z_j^j,j=1..n)

where C(n) = A000178(n) = the Vandermonde determinant of the numbers
1,2,..(n+1) (among other things).

I haven't proved it, but Maple verifies it symbolically up to n = 5 and
putting in specific numbers it checks out up to n = 8.

--Edwin

On Sat, 13 Sep 2003, benoit wrote:

> 
> Thanks for the confirmation. I wonder if the following Gessel and 
> Viennot paper on binomial determinants could help.
> 
> I.Gessel and X. Viennot, Binomial determinants, paths and hook length 
> formulae , Adv. in Math., 58 (1985), 300–321
> 
> I didn't read it but I know that's a nice interpretation for binomial 
> determinants.
> 
> Benoit.
> 
> >>
> >> let z be any function : N-->N, let M_n be the nxn matrix
> >> M_(i,j)=binomial(i*z(j) , j),  then I observed :
> >>
> >> det M_n = prod(k=1,n, z(k)^k)
> >>
> >> I'm looking for  a proof or anything related... Thanks.
> >>
> >> Benoit Cloitre
> >>
> >
> > Maple gives a proof for n from 1 to 5:  I replace your functional 
> > values
> > z(j) by inteterminates z_i (or z[i] in Maple notation). Also I convert 
> > the
> > binomial expressions to factorial expressions. Then it holds without
> > restriction on the variables:
> >
> >> with(LinearAlgebra):
> >> for n from 1 to 5 do
> >> M:=Matrix(n,n,(i,j)->convert(binomial(i*z[j],j),factorial)):
> >> simplify(Determinant(M));
> >> print(n,%);
> >> end do:
> >
> >                                1, z[1]
> >
> >
> >                                         2
> >                             2, z[1] z[2]
> >
> >
> >                                      2     3
> >                          3, z[1] z[2]  z[3]
> >
> >
> >                              4          2     3
> >                       4, z[4]  z[1] z[2]  z[3]
> >
> >
> >                           4     3     5     2
> >                    5, z[4]  z[3]  z[5]  z[2]  z[1]
> >
> >
> > With some effort there should be a proof of this. :-)
> >
> > --Edwin
> >
> >
> 

------------------------------------------------------------
    W. Edwin Clark, Math Dept, University of South Florida,
           http://www.math.usf.edu/~eclark/
------------------------------------------------------------