For $2\times 2$ matrices we have the following result.



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> Any matrix in $\mathrm{SL}(2,\mathbb{Z})$ with nonnegative entries can be obtained from $\mathrm{Id}\_2$ by repeatedly adding one column to another.
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*Proof:* It is enough to prove that if $$\begin{pmatrix} 
a & b \\
c & d 
\end{pmatrix}\in\mathrm{SL}(2,\mathbb{Z})\setminus \{\mathrm{Id}\_n\}$$ has nonnegative entries,
then either 
$$\begin{pmatrix} 
a-b & b \\
c-d & d 
\end{pmatrix} \text{ or 
}\begin{pmatrix} 
a & b-a \\
c & d-a 
\end{pmatrix}$$
has nonnegative entries as well. After this you can finish by induction. Now to prove that, suppose $a$ is the biggest entry of the matrix, if $a=1$ then we obtain the matrices 
$$\begin{pmatrix} 
1 & 0 \\
1 & 1 
\end{pmatrix}, \begin{pmatrix} 
1 & 1 \\
0 & 1 
\end{pmatrix} \tag{$\star$} $$
and we are done. Otherwise $a>1$, hence 
$$d-c\leq d-bc/a=(ad-bc)/a=1/a$$ from which $d-c\leq 0$ and we arrive in the first case. 
The cases in which the maximal entry is different from $a$ are done similarly $\blacksquare$


This can be restated as saying that the elementary matrices in ($\star$) generate the semigroup $\mathrm{SL}(2,\mathbb{N})$. Here $\mathbb{N}$ denotes the non-negative integers (including $0$).


My question is: 



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> Is it true that $\mathrm{SL}(n,\mathbb{N})$ can be generated by elementary matrices, similarly as in the case $n=2$?
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I would guess this has already been discussed in the literature so a good reference would be enough. 


**Edit:** Probably the question above is stated better in terms of $\mathrm{SL}^{\pm}(n,\mathbb{N})$, the set of square $n\times n$ matrices with nonnegative integral entries and determinant $1$ or $-1$.


This set has the following properties:


1. $\mathrm{Id}\_n\in \mathrm{SL}^{\pm}(n,\mathbb{N})$
2. If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and we change a column of $A$ by the addition of it with other column, the result is still on $\mathrm{SL}^{\pm}(n,\mathbb{N})$.
3. If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and we switch two columns of $A$, the result is still in $A$.



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> The problem is to show that the set $\mathrm{SL}^{\pm}(n,\mathbb{N})$ is the minimum set of matrices with this three properties.
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Notice that property $2$ is equivalent to 


2. If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and $L\_{i,j}(1)$ is the elementary matrix that acts by changing column $i$ by the addition of column $i$ and $j$ (see for example [Wikipedia](https://en.wikipedia.org/wiki/Elementary_matrix)) then $$A\cdot L\_{i,j}(1)\in \mathrm{SL}^{\pm}(n,\mathbb{N}).$$


and property 3 is equivalent to


3. If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and $T\_{i,j}$ is the elementary matrix that acts by switching column $i$ and $j$ then $$A\cdot T\_{i,j}\in \mathrm{SL}^{\pm}(n,\mathbb{N}).$$


As $L\_{i,j}(1), T\_{i,j}\in \mathrm{SL}^{\pm}(n,\mathbb{N})$, the problem above is equivalent to



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> Every matrix in $\mathrm{SL}^{\pm}(n,\mathbb{N})$ can be written as a multiplication of matrices of the form $L\_{i,j}(1)$ and $T\_{i,j}$.
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As operation 2 and 3 commute with each other. We can put all permutation matrices at the end, then by multiplying them we can use only one permutation matrix. If the final matrix has determinant 1 so will have this permutation matrix. In this way we see that this question is equivalent to the original one stated in terms of $\mathrm{SL}(n,\mathbb{N})$.



Generators for the semigroup $\mathrm{SL}(n,\mathbb{N})$
I believe that the answer is negative.


A positive answer would mean that for any $A\in\mathrm{SL}^\pm(n,\mathbb{N})$, $A\neq1$ there exists $i,j$ and some other element $B\in\mathrm{SL}^\pm(n,\mathbb{N})$ such that $A=L\_{ij}(1)\cdot B$ or $A=B\cdot L\_{ij}(1)$, possibly after some permutation of rows and/or columns of $A$.


If, however, one can find an $A$ such that $L\_{ij}(-1)\cdot A', A'\cdot L\_{ij}(-1) \notin \mathrm{SL}^\pm(n,\mathbb{N})$ for any $i,j$ and any permutation $A'$ of columns/rows of $A$, then $A$ can not be decomposed into the product of $L\_{ij}(1)$'s and $T\_{ij}$'s.


A quick computer search reveals that
$$ A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 3 & 1 \\ 2 & 0 & 1 \end{pmatrix} $$
is one such matrix.


