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I have a set of matrices defined over the ring of the integers, which items are using only coefficients -1, 0 and 1. For example:

$$ A = \left(\begin{matrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 0 \end{matrix}\right) \hspace{1em} \text{(over }\mathbb{Z}\text{)} $$

Now I'm wondering, if I transform my matrices into equivalent matrices in GF(3) (mapping 0 to 0, 1 to 1, and -1 to 2), is their rank preserved?

With my example, the transformed matrix is:

$$ A' = \left(\begin{matrix} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 1 & 0 \end{matrix}\right) \hspace{1em} \text{(over }GF(3)\text{)} $$

and $\operatorname{rank} A = \operatorname{rank} A' = 3$. Is this result true in general?

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No, the result is not true in general, because $\det(A)$ may be a nonzero multiple of $3$, which becomes $0$ over $\Bbb F_3$. For example, $$ A=\begin{pmatrix} 1 & -1 & 1\cr -1 & 0 & 1 \cr 1 & 1 & 0\end{pmatrix} $$ has determinant $\det(A)=-3$ over $\Bbb Z$, which is $0$ over $\Bbb F_3$.

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  • $\begingroup$ Is it possible to have $\det A = 3k$, if all the items of $A$ are in $\{-1,0,1\}$ though? $\endgroup$ – lennox Feb 21 at 17:25
  • $\begingroup$ Yes, it is, at least for $k=\pm 1$. $\endgroup$ – Dietrich Burde Feb 21 at 17:26
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Take the matrix $$ M =\begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 0 & -1 & 1 \end{pmatrix} $$ It is clearly of full rank over $\mathbb{Z}$ but the transformed is not. We also have that $\det(M) = 3$ hence it is not invertible $\pmod{3}$.

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