# binomial determinant

benoit abcloitre at wanadoo.fr
Mon Sep 15 19:19:01 CEST 2003

```For info. here an answer from Robert Israel  :

"Let M_n be the nxn matrix M_(i,j)=binomial(i*z_j,j) where
z_1,z_2,...,z_n are n indeterminates.  I noticed without proof that :
det(M_n)=prod(k=1,n,(z_k)^k).  Is there a simple way to prove
(disprove) this fact? "

Here of course we take binomial(i*z_j,j) = (i*z_j)(i*z_j - 1)...(i*z_j
-j+1)/j! so this makes sense when i*z_j is not a positive integer.
You can write binomial(x,j) = sum_{k=0}^j c_{k,j} x^k for constants
c_{k,j} with c_{j,j} = 1/j!.  Thus there is an upper triangular matrix
R with 1's on the diagonal such that MR has entries :

(MR)_{i,j} = (i*z_j)^j/j!

so that det(M) = det(MR) = det(A) prod_{j=1}^n (z_j)^j where A_{ij} =
i^j/j!.
When all z_j = 1 we have det(M) = det(A) with M_{ij} = binomial(i,j).
But this M is lower triangular with 1's on the diagonal, so det(A) = 1.

Robert Israel

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