From: Robin Chapman <rchapman@mpce.mq.edu.au>
Subject: Re: Symmetric Toeplitz Determinants
Date: Sun, 11 Apr 1999 11:41:42 +1000
Newsgroups: sci.math.research
Keywords: Dodgson's condensation formula

Brian Stewart wrote:
>
> To define these let $t_0, t_1, \dots, t_n$ be a set of
> independent transcendents over some base field such as $\rational$.

The t_j could just as well be any elements of any commutative ring.

> Define
> $\mathbf{T}_{n+1}$ to be the $(n+1)\times (n+1)$ matrix whose
> $(i,j)$-th entry is $t_{|i-j|}$, and $T_{n+1}$ to be its
> determinant. For convenience let $T_0$ denote $1$.
> 
> The relationship conjectured by Vein and Dale is
>
> The determinant
> 
> |T_{n-1}       T_n|
> |T_n       T_{n-1}|
> 
> is minus the square of the determinant
> 
> t_1 t_0 t_1 t_2           t_{n-2}
> t_2 t_1 t_0 t_1           t_{n-1}
> ....
> .....
> t_n ....                       t_1

This follows *immediately* from Dodgson's condensation formula
applied to the largest determinant T_{n+1}.

Dodgson's rule states that given a square matrix A we have
|A||B| = |A_{00}||A_{11}| - |A_{01}||A_{10}|
where
A_{00} is obtained from A by deleting its first row and first column,
A_{01} is obtained from A by deleting its first row and last column,
A_{10} is obtained from A by deleting its last row and first column,
A_{11} is obtained from A by deleting its last row and last column,
B is obtained from A by deleting its first and last row and first and
last column.

For a recent proof see the article by Doron Zeilberger in the Electronic
Journal of Combinatorics:
http://www.combinatorics.org/Volume_4/Abstracts/v4i2r22ab.html .


I think Dodgson was quite a well-known Oxford mathematician :-)

-- 
Robin Chapman				+ "Going to the chemist in
Department of Mathematics, DICS		-  Australia can be more
Macquarie University			+  exciting than going to
NSW 2109, Australia			-  a nightclub in Wales."
rchapman@mpce.mq.edu.au			+   Howard Jacobson,
http://www.maths.ex.ac.uk/~rjc/rjc.html	-     In the Land of Oz