Re: divisors of zero

In article <cjnelson9-EB1684.11213702052006@xxxxxxxxxxxxxxxx>,
Clifford Nelson <cjnelson9@xxxxxxxxxxx> wrote:

Is my memory playing tricks on me or do I remember getting and verifying
answers in Mathematica of LinearSolve[M,v] for n by n matrices M and
length n vectors v even when M did not have an inverse and had a
determinant of zero?

It may not be playing tricks. Yes, you can sometimes solve equations
M x = v when M is not invertible. The general solution will be of the
form x_p + x_h where x_p is one particular solution and x_h is an
arbitrary solution of M x = 0. I don't know why you think this has
anything to do with the question at hand. Dave is just using the
fact that if M is an n x n matrix (over some field k), there are
nonzero solutions of M x = 0 (with x in k^n) if and only if
det(M) = 0. In the case at hand, with M corresponding to left
multiplication by (a,b,c,d) in your algebra, the existence of
such a solution x = (e,f,g,h) means that (if (a,b,c,d) is nonzero)
it is a zero-divisor.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.