Re: [Axiom-math] Writing packages

[Ralf Hemmecke]

On 05/13/2006 10:35 PM, Fabio S. wrote:
> > >  BTW, I also have another question on this topic: if I compute the 
> > >  characteristic polynomial of a matrix and then I evaluate the 
> > > polynomial  at the matrix, I would expect to have the zero matrix as 
> > > result. This does  not happens: try
> > >
> > >  m:=matrix([[1,2],[3,4]])
> > >  p:=characteristicPolynomial(m)
> > >  p(m)
> > >
> > >  Why?
> >
> > Because Axiom is not smart enough to figure out all the types for you.
> >
> > Does this make you happier?
> >
> > M := SquareMatrix(2,Integer)
> > m:M:=matrix([[1,2],[3,4]])
> > P:=SUP(M)
> > p: P := characteristicPolynomial(m)::P
> > x := first variables p
> > eval(p, x, m)

> I must say that this doesn't make me happier at all. I didn't expect 
> that evaluating a polynomial at a matrix was such a terrible task for 
> the type resolver in axiom: the result of characteristicPolynomial(m) is 
> a polynomial with integer coefficient in one variable. I can't see the 
> difficulty to convert it to UP SquareMatrix(2,Integer) (of course, just 
> from a mathematical point of view).

Well, as I said SUP(Integer) should have been enough. But now, what you 
actually want is NOT a polynomial. You want a *function*

charpol: SquareMatrix(2, Integer) -> SquareMatrix(2, Integer)

and that is something different. I am sure you know the difference.

The big trouble now is to convert the polynomial you get into a 
polynomial function with the above type. I guess the interpreter would 
have been able to find such a conversion if there were an implementation 
of an evaluation of a polynomial at a *square* matrix (arbitrary 
matrices would, of course, not work). However, my browsing through 
hyperdoc was unsuccessful. So I guess the non-existence of such an 
implementation makes the interpreter telling you about the problem.

> Moreover, given the Hamilton-Cayley, it doesn't seem to me a very 
> bizzarre idea to try to evaluate the char pol of a matrix at the matrix 
> itself...

No, of course, not. But as far as I can tell, that is not directly 
implemented.

> I would expect that this is the first example for every linear algebra 
> student, so I would expect that the implementation of the function 
> should take care of this possibility.

I somehow have the impression that is one of the reasons why Axiom 
failed commercially. Simple things must be expressed in a complicated 
manner. I agree that Axiom is not (yet) perfect. Just help and provide a 
implementation for such a function and make all the people who come 
after you happier.

Thank you

Ralf