From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: eigenvalue-problem Date: 27 Nov 1995 21:39:29 GMT In article <199511192342.a21051@ac3.maus.de>, Claus Juergensen wrote: >Let G be a finite Group and F an algebraically closed field >with characteristic not dividing the order of G and be >X : G -> GL(n, F) an irreducible representation of G. >Be N(j,k) the n x n-matrix with 1 at position (j,k) and 0 at >all the other positions. >Be T the mapping > T : F^(n x n) -> F^(n x n) > N -> sum for g in G of gX N (gX)^-1. > >I'd like to know the eigenvalues of N(j,k)T. Each matrix NT is invariant under conjugation by an element of G. Thus NT is a matrix which commutes with all matrices gX and all matrices in their span. Since the representation is irreducible, this span is the set of _all_ matrices. The only matrices which commute with all others are the scalar matrices. Thus, NT is scalar for all N. The entry on the diagonal is of course (1/n)*trace(NT). On the other hand, the trace of NT is the sum of the traces of the gX N gX^(-1), each of which by necessity has the same trace as N. In particular, if N=N(j,j) then NT is order(G)/n times the identity matrix; otherwise, N(j,k) T is the zero matrix. dave (who never liked operators on the right)