Re: Non-Commutative calculations in Calc, Revisited

[Neon Absentius <absent@sdf.lonestar.org>]

On Thu, Oct 06, 2005 at 03:42:40PM +0200, David Kastrup wrote:
> Jay Belanger <belanger@truman.edu> writes:
> 
> > Neon Absentius <absent@sdf.lonestar.org> writes:
> >
> >> When in non-commutative (aka `matrix') mode calc calculates 
> >> (a b)^-1 to be a^-1 b^-1 instead of the correct b^-1 a^-1.
> 
> Uh, what's correct about b^-1 a^-1?
> 
> Set a=[1,1], b=[1;0], then (a b)^-1 = [1], and neither b^-1 nor a^-1
> exist.
> 
> This only works with a and b being square matrices.

Yes you are right.

However I am more interested in the case that there is a *globally*
defined associative multiplication and I am using the matrix mode as a
"poor man's no-commutative algebra mode".  In this context my remark 
makes sence.  I still believe that there is a need for such a mode,
perhaps a "sub-mode" of the matrix mode where one assumes that all
matrices are square.  I am not sure how hard this is to implement.
What do you think Jay?
 
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