From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: endomorphism
Date: 6 Jan 2000 23:23:42 GMT
Newsgroups: sci.math
Summary: [missing]

In article <852mou$78u$1@gannett.math.niu.edu>,
 rusin@vesuvius.math.niu.edu (Dave Rusin) writes:
> In article <02a4ec93.8f6f78b6@usw-ex0110-075.remarq.com>,
> ebru2w  <jean-claude.serlet@insee.fr> wrote:
> 
> >   U and V 2 endomorphism ( V,U:E--->E,E is one EV with
> >dimE=2) ,U different from 0,UoV-VoU=U.
> >  You have to show that U is not one bijection...
> 
> ...
> The problem here is that what is to be proved isn't really true!
> Over a field of characteristic  2, take  U = [[1,1],[0,1]]  and
> V = [[1,0],[1,0]]. Then  UV-VU = U ! 
> 
> So we will have to abandon purely geometric or axiomatic approaches
> and use something which uses algebra. Note that  trace(UV)=trace(VU)
> so trace(U) will have to be zero, and thus its characteristic roots
> are negatives of each other AND THUS distinct.... 


I think the best way to understand this is to rewrite the equation
as

  UV = U + VU = (I + V)U.

Thus there is a solution with invertible U precisely when there is
some V that is similar to I + V  (that is,  U V U^{-1} = I + V). 

This forces f(x) = f(x-1), where f is the characteristic polynomial
of V.  In characteristic p, there will be a p by p example with
minimal polynomial x^p - x.


William C. Waterhouse
Penn State