Re: Non-Commutative calculations in Calc, Revisited

[Neon Absentius]

On Thu, Oct 06, 2005 at 06:01:13PM +0000, Neon Absentius wrote:
> 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.
> 

OTH, now that I thought litle bit about it, there is a way in which my 
statement 
makes sence even for non-square matrices.  Here is the general setting:

Let (C,*) be a set with a partially defined associative multiplication
(called composition) such that each element has a left and a right
identity element (such a structure is usually known as a "category").
The set of all matrices satisfy these conditions.  Assume that for two
composable elements a,b the composition a*b has an (bi-sided) inverse
(ab)^-1.  Then a has a *right* inverse a^-1 and b has a *left* inverse
b^-1 and the identity

  (ab)^-1 = b^-1a^-1

holds.

Proof:  Since ab(ab)^-1 = (ab)^-1 ab = 1 we can take a^-1 = b(ab)^-1 
        and b^-1 = (ab)^-1 a.  Then 
        
           b^-1 a^-1 = [(ab)^-1 a] [b (ab)^-1]
                     = (ab)^-1,  by associativity.

        It is also straightforward to check that all elements that I composed 
        are indeed composable. QED

I leave it as an exersise to find a^-1 and b^-1 for your example :)).

I am not sure it is a good idea to implement the rule in general
though.  In any case I do think that is a good idea to have a command
that assumes that all variables are square matrices or, in the
abstract language, that any two elements are composable.  Of course
being square matrices, the variables will have more properties, eg
determinats etc. That might also be useful in some cases.

--
 There is no national science just as there is no national
multiplication table; what is national is no longer science.
         -- Anton Checov