From: Robin Chapman Subject: Re: Inverse of a special matrix Date: Thu, 24 Jun 1999 19:59:02 GMT Newsgroups: sci.math To: franz.vrabec@david.co.at Keywords: Advanced determinant calculus In article , franz.vrabec@david.co.at (Franz Vrabec) wrote: > I am interested in the (m+1)*(m+1)- matrices Tm (m a natural number), > Tm = [t(j,k)], 0 <= j,k <= m, > > defined as t(j,k) = cos((j+m)*k*pi/(2*m)) > > In MATLAB you can easily generate this matrices by > T=cos([m:2*m]'*[0:m]*pi/(2*m)); > > This matrices are non-singular (I can prove this). What I can't prove, > but what is numerically verified until m=10 is the following > astonishing fact: > in the inverse of Tm, the element in the lower left corner seems to be > equal to 1/sqrt(m). > > Can anyone give me some hints how to prove this property ? In fact, I > would already be satisfied if somone can prove the weaker statement > that in the inverse of Tm the lower left element is positive ! It's convenient to start with some preliminary reductions. First of all if we reflect the matrix in a horizontal axis we get a new matrix S wih entries s(j,k) = cos(k(2m-j) pi/2m) = (-1)^k cos(jk pi/2m). The inverse of S is obtained from the inverse of T by reflection in a vertical axis. Thus the bottom left entry of T^{-1} is the bottom right entry of S^{-1}. Now define R by negating the even columns of S. Then R has entries r(j,k) = cos(jk pi/2m). Also R^{-1} is obtained from S^{-1} by negating even rows. Our task is thus to show the bottom right entry of R^{-1} is (-1)^{m+1} sqrt(m). By Cramer's rule this is the quotient of the determinant of the matrix R' by that of R where R' is obtained from R by deleting the last row and column. Define D_m(X) to be the determinant of the m by m matrix with entries X^{jk} + X^{-jk} for 0 <= j, k < n. Then det(R) = 2^{-m-1} D_{m+1}(eta) and det(R') = 2^{-m} D_m(eta) with eta = exp(pi i/2m). We thus need to show that D_{m+1}(eta)/D_m(eta) = 2(-1)^m sqrt(m). I claim that D_{m+1}(X)/D_m(X) = X^{-m^2}(X^m-1) product_{j=1}^{2m-1} (X^j-1). (*) Assume (*) for now and let a_m = D_{m+1}(eta)/D_m(eta). Then |a_m|^2 = 2 product_{j=1}^{2m-1}|eta^j - 1|^2 = 2 product_{j=1}^{2m-1}(1 - eta^j)(1 - eta^{4m-j}) = product_{j=1}^{4m-1}(1 - eta^j) since eta^{2m} = -1. It's well-known that for each positive integer n we have product_{j=1}^{m-1}(1 - exp(j 2pi/n)) = n since product_{j=1}^{m-1}(X - exp(j 2pi/n)) = (X^n - 1)/(X - 1). Thus |a_m|^2 = 2 sqrt(m). For non-zero complex numbers z define s(z) = z/|z|. Note that s(zw) = s(z)s(w). To complete the evaluation of a_m we need that s(a_m) = (-1)^{m+1}. For 0 < j < 2m we have eta^j - 1 = exp(j pi i/4m) 2i sin(j pi/4m) so that s(eta^j - 1) = i xi^j. Thus s(a_m) = eta^{-m^2} s(eta^m - 1) product_{j=1}^{2m-1}s(eta^j - 1) = i^{-m} i xi^m i^{2m-1} product_{j=1}^{2m-1} xi^j = i^m xi^m xi^{m(2m-1)} = i^m xi^{2m^2}. Now xi^{2m} = i so this equals i^{2m} = (-1)^m a required. To establish (*) we need a fairly well-known determinant expansion. Let M be the n by n matrix with entries X_j^k + X_j^{-k} for 0 <= j, k < n. Then det(M) = 2 (X_0 X_1 ... X_{n-1})^{1-n} . product_{0 <= j < k < n}(X_i - X_j)(1 - X_j X_k). This can be found in Christian Krattenthaler, `Advanced determinant calculus' http://cartan.u-strasbg.fr:80/~slc/wpapers/s42kratt.html To prove it one uses an agrument like the standard Vandermonde evaluation. One clears the denominators, shows that the determinant is divisible by the given factors, and that their product has the same "degree". Then one compares the "leading coefficients". Now one specializes to X_j = X^j and compares the n = m and n = m + 1 cases to get (*). -- Robin Chapman http://www.maths.ex.ac.uk/~rjc/rjc.html "They did not have proper palms at home in Exeter." Peter Carey, _Oscar and Lucinda_ Sent via Deja.com http://www.deja.com/ Share what you know. Learn what you don't.