Prove that $SL(n,\mathbb{Z})$ is generated by $(n^2-n)$ elements.

Question

Statement : Prove that $SL(n,\mathbb{Z})$ is generated by $(n^2-n)$ elements.

The determinant is a n linear function of the rows of the matrix. Given any matrix, if the determinant is nonzero, say $det(x_1, x_2, ...x_n) = \alpha$ then by simply scaling the last row of the matrix (i.e n elements) we can get , $det(x_1,x_2,...,\frac{x_n}{\alpha}) = 1$.

Am I correct? Please give some hint how to prove this?

Answer 1

As an example, $SL(3,\Bbb Z)$ is generated by the matrices $$\pmatrix{1&1&0\\0&1&0\\0&0&1},\quad \pmatrix{1&0&1\\0&1&0\\0&0&1},\quad \pmatrix{1&0&0\\1&1&0\\0&0&1},\quad \pmatrix{1&0&0\\0&1&1\\0&0&1},\quad \pmatrix{1&0&0\\0&1&0\\1&0&1},\quad \pmatrix{1&0&0\\0&1&0\\0&1&1}. $$

Can you prove this? And can you extend this to any $n$?

Answer 2

There is a nice and elementary article that $SL_n(\Bbb{Z})$ can even be generated by $2$ elements for all $n\ge 2$: A pair of generators for the unimodular group. For a discusssion on the minimal number of generators of $SL_n(\Bbb{Z})$ and $GL_n(\Bbb{Z})$ see here.