Re: Echelon Form

[Bill Allombert <allomber@math.u-bordeaux.fr>]

On Thu, Jul 29, 2004 at 09:20:41PM -0500, Ariel Pacetti wrote:
> 
> Given a matrix (say with integer coefficients), is there a way to find its 
> reduced Echelon form? (I looked through the manual but couldn't find this 
> specific routine).
> To fixed conventions, the reduced Echelon form is the usual Gaussian 
> elimination with the condition that given a pivot position all entries 
> above it are zero. For example:
> 
> [1 1]
> [0 1]
> 
> is not Echelon form while
> 
> [1 0]
> [0 1]

What kind of basis change do you allow ?
It seems you can do what you want with either
matdiagonal(matsnf(x))
(for basis change over Z.)
or
r=matrank(x);matdiagonal(vector(length(x),i,i<=r))
(for basis change over C.)

(I am uncertain about the english terminology of linear algebra,sorry).

Cheers,
Bill