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 <Claus_Juergensen@ac3.maus.de> 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)