Re: don't understand inverseimage

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

On Tue, Dec 17, 2002 at 11:10:42PM +0100, rcasero@tsc.uc3m.es wrote:
> Hi everybody. I'm new to pari, and I cannot understad the description of
> the [mat]inverseimage(x, y) function. I suppose it's quite clear for
> everybody else, but maybe it's that I'm also quite new to number theory.
> I've searched for more information with Google, but I couldn't find
> anything beside that.
> 
> I've been doing some tests with pg and octave, and it seems that if x =
> A, y = y, it does something similar to
> 
> z = A^(-1) * y
> 
> But then , why z `belongs to the inverse image of y' and not `z is the
> inverse image of y'?

Because A is not supposed invertible!

matinverseimage(A,Y) return a solution of AX=Y if it exists, and [;] else.

A^(-1)*Y will fail if A is not invertible.

If A is not invertible, the linear mapping f: X -> AX is generally not injective [1].
matinverseimage return an element in f^-1({Y}) which is called the inverse image of {Y}.

I hope I have made things clearer ?

Cheers,
Bill.

[1] if A is rectangular it may be injective...