Re: Does matrix rank makes sense modulo composite $n$?

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

On Sat, Jul 18, 2026 at 09:13:40AM +0300, Georgi Guninski wrote:
> Both pari and sage fail to compute the rank of matrix modulo composite $n$:
> ? M=Mod(1,4)*[2,2;2,2]
> ? matrank(M)
>   ***   at top-level: matrank(M)
>   ***                 ^----------
>   *** matrank: impossible inverse in Fl_inv: Mod(2, 4).
> 
> Does matrix rank makes sense modulo composite $n$?

No it does not, but you can do
? matimagemod([2,2;2,2],4)
%1 = [2;2]

Over a field K the image is always isomorphic to K^r 
where r called the rank, 
but over a ring R, the image will not always be a free module.
Here is is isomorphic to 2Z/4Z and not to Z/4Z.

Cheers,
Bill.