Bill Allombert on Wed, 18 Dec 2002 11:15:57 +0100 |
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Re: don't understand inverseimage |
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...