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

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.

*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:

Is it true that $\mathrm{SL}(n,\mathbb{N})$ can be generated by elementary matrices, similarly as in the case $n=2$?

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:

- $\mathrm{Id}_n\in \mathrm{SL}^{\pm}(n,\mathbb{N})$
- 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 $\mathrm{SL}^{\pm}(n,\mathbb{N})$.
- If $A\in \mathrm{SL}^{\pm}(n,\mathbb{N})$ and we switch two columns of $A$, the result is still in $A$.

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

Notice that property $2$ is equivalent to

- 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) then $$A\cdot L_{i,j}(1)\in \mathrm{SL}^{\pm}(n,\mathbb{N}).$$

and property 3 is equivalent to

- 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

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 $T_{i,j}$.

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 $\mathrm{SL}(n,\mathbb{N})$.