TITLE: length of decompositions into elementary matrices
QUESTION [6 upvotes]: 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 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.

REPLY [4 votes]: 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.