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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~ $C\sim n^2$). 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.

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length of decompositions into elementary matrices

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 ($C~ n^2$). 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.