From: "Dr. Uwe Prells" Subject: Re: Matrix with zero trace Date: Fri, 05 Jan 2001 10:23:21 +0000 Newsgroups: sci.math Summary: Is every matrix of trace zero a comutator? (Yes) Hello Ben, The answer is yes. For even n=2k=dim(V) the keyword is symplectic algebra sometimes denoted by sp(2k,F) and its dimension is 2k^2+k. If n=k+1 (note that this does not exclude the first case) then we have sl(V) the set of trace zero matrices which has dimension k+(k+1)^2-k-1. For more information have a look for a book on Lie algebra (J.E.Humphreys: Introduction to Lie Algebra and Representation Theory, Springer, 3rd rev. ed. 1980). Regards Uwe ben_geffen@my-deja.com wrote: > If A is an n by n matrix over a field K and trace(A) = 0 then can we > always write A as an commutator A = BC - CB for some matrices B and C ? > > Thanks, > Ben > > Sent via Deja.com > http://www.deja.com/ ============================================================================== From: Pasha Zusmanovich Subject: Re: Matrix with zero trace Date: Wed, 10 Jan 2001 00:14:40 +0200 Newsgroups: sci.math Dr. Uwe Prells wrote: [see above --djr] so this remarkable fact may be formulated as follows: for the Lie algebra sl_n(K) each element can be represented as commutator of two elements. I think it is also proved to be true for other simple Lie algebras of classical type. It is interesting whether this is true in general, i.e. if for any finite-dimensional Lie algebra L over a field K (maybe here field K should be required to be infinite or perfect) each element from [L,L] can be represented as a commutator of two elements. -- /"\ Pasha Zusmanovich \ / ASCII RIBBON CAMPAIGN myself@pasha.cx X AGAINST HTML MAIL http://pasha.cx/ / \ ============================================================================== From: "moubinool.omarjee" Subject: Re: Matrix with zero trace Date: Fri, 5 Jan 2001 13:49:54 +0100 Newsgroups: sci.math a écrit dans le message : 93338s$a5f$1@nnrp1.deja.com... > If A is an n by n matrix over a field K and trace(A) = 0 then can we > always write A as an commutator A = BC - CB for some matrices B and C ? __________________________________________________________ this result is true here an elementary proof we need the following lemma: A a matrix with trA=0 then a is similar to a matrix W=(w_(i,j)) with w_(i,i) = 0 for 1<= i <=n Now we suppose A = (a_(i,j)) with a_(i,i) = 0 take C = diag(t_i) with t_i distincts and B = (b_(i,j)) then BC - CB = ( (t_j - t_i).b_(i,j) ) => b_(i,j) = a_(i,j)/(t_j - t_i) if A = BC-CB => PAP^(-1) = B'C' - C'B' where B' = PBP^(-1) , C'=PCP^(-1) so the propertie can extand to similar matrix , to matrix with 0 on the diagonal moubinool omarjee [deleia --djr