$\DeclareMathOperator\GL{GL}$Consider the unimodular group $\GL\_n(\mathbb{Z})$, consisting of integral matrices $A \in \mathbb{Z}^{n \times n}$ such that that $\det(A) =\pm 1$.


It is well known that any $A \in \GL\_n(\mathbb{Z})$ can be written as a product of signed permutation matrices and 'Gauss moves', the latter referring to matrices that have 1 on the diagonal and at most one non-zero entry away from the diagonal. This is proven, for example, in M. Newman's book 'Integral Matrices' (Theorem II.7).


My question is: is there an upper bound (depending on $n$, but independent of $A$) on the number of moves required? If not, can you provide a sequence of matrices that require an increasing number of factors in the above decomposition?


Thank you in advance for your help.



Diameter of the unimodular group with Gauss moves
They’re usually called “elementary matrices”, not Gauss moves. Your question is equivalent to asking whether the integer special linear group is boundedly generated by elementary matrices. The answer is no for $n=2$ (an easy exercise using the free product with amalgamation description of the group in that case), but amazingly it is “yes” for larger $n$. This is a theorem of Carter, Keller, and Paige. See [here](https://arxiv.org/abs/math/0503083).


