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_
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