The Gaussian algorithm tells us, that for any field $k$ a $n\times n$-matrix over $k$ can written as a product of at most $C$ elementary matrices ($C\sim n^2$).
I am wondering, whether such a constants also exists for other rings - like $\mathbb{Z}$.
Given a matrix $A\in SL\_2(\mathbb{Z})$, one can basically use the Euclidean algorithm to find such a decomposition. However if we take a the following [matrix](http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form) involving the Fibonacci numbers, the algorithm takes about $n$-steps and hence we get a decomposition in $\sim n$ factors. But there might still be a better decomposition.


So is there for every $n$ a matrix $A \in SL\_2(\mathbb{Z})$, that cannot be written as a product of elementary matrices ?


I guess the construction with the Fibonacci numbers might be a candidate, I don't know how to prove, that it is impossible to decompose it in a better way.



length of decompositions into elementary matrices
Yes. The group $SL\_2(\mathbb Z)$ is virtually free and is not boundedly generated by any finite generating set because of that. You can look at the (very nice) slides of Dave Witte-Morris' talks on bounded generation [here.](http://people.uleth.ca/~dave.morris/talks.shtml) 


