Let $A, B$ be matrices with elements in $\mathbb{Z}_n$. If $A x = 0$ and $B x = 0$ have the same set of solutions, where the vectors also have elements in $\mathbb{Z}_n$, does this mean that there is an invertible matrix $M$ such that $A = MB$?
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$\begingroup$ Let $n=6$, $A= [2,3]^t$ and $B = [1]$. $\endgroup$– Joshua MundingerCommented Nov 9, 2023 at 17:52
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$\begingroup$ An example by @JoshuaMundinger hints that a real statement should look a bit differently: the rows of $A$ are linear combinations of those of $B$, and vice versa. This can be done easily after passing to the Smith normal form. $\endgroup$– Ilya BogdanovCommented Nov 10, 2023 at 8:36
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$\begingroup$ @IlyaBogdanov, it is not clear to me that every matrix over a ring with zero divisors has a Smith normal form, as the way they are usually constructed uses the property that every element has an inverse. So this seems trickier over a ring. It also seems it is not straightforward to prove that two matrices of same size with the same set of solutions would have the same Smith normal form. Maybe something similar has already been done? $\endgroup$– JoséCommented Nov 11, 2023 at 20:19
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$\begingroup$ The Smith normal form works, e.hg., over $\mathbb Z$, where not every element has an inverse. You may apply that and then reduce modulo $n$ (Uniquness is not needed!) And yes, I am almost sure this has already been done… $\endgroup$– Ilya BogdanovCommented Nov 12, 2023 at 6:55
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