From: Robin Chapman 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