Hint for proving ${\rm SL}_n(\mathbb{R})$ is generated by matrices that are off-diagonal entries added to identity.

Question

Let $S = \{ I_n + a\cdot e_{i,j} \mid a\in\mathbb{R},\ i,j= 1,\ldots,n,\ i\neq j\}$, where $e_{i,j}$ is the matrix with $1$ at entry $(i,j)$ and zero elsewhere. I need a hint to help prove that $\langle S\rangle ={\rm SL}_n(\mathbb{R})$. That is, such matrices generate the multiplicative group of matrices with determinant one.

Answer 1

Hint: Consider the elementary row operations needed to reduce a matrix $M \in \mathrm{SL}_n(\mathbb R)$ to the identity. Multiplying by an element in $S$ corresponds to one of your row operations, so what the question is asking you to show is that in fact you don't need to use the other row operations when reducing $M$.

Answer 2

I've just now got the proof for $SL_2(\mathbb{R})$:

Let $M = \bigl(\begin{smallmatrix} a & b\\ c & d \end{smallmatrix}\bigr)$ and $\det(M) = 1$. Then $\det(M) = ad - bc = 1$ and either $c \neq 0 $ or $c = 0 \implies ad = 1$. Let's do the case that $c\neq 0$: rearranging the determinant formula we have that $b - ad/c = -1/c$. Multiply on the left of $M$ by $\bigl(\begin{smallmatrix} 1 & -a/c\\ 0 & 1 \end{smallmatrix}\bigr)$ to get $\bigl(\begin{smallmatrix} 0 & -1/c\\ c & d \end{smallmatrix}\bigr) = M'$. That was a row operation of the required type, but notice that we can also perform collumn operations (multiplying on the right of $M$) since the idea is to reduce $M$ to $I_2$ through $E_1\cdots E_k\cdot M \cdot D_1\cdots D_r = I_2$, then $M = E_r^{-1}\cdots E_1^{-1} D_r^{-1}\cdots D_1^{-1}$, where each inverted elementary matrix is also one of the required type. So let's continue knowing that column ops are allowed too. $M'\cdot \bigl(\begin{smallmatrix} 1 & 0\\ -c & 1 \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 1 & -1/c \\ c(1-d) & d \end{smallmatrix}\bigr) = M''$, and finally $\bigl(\begin{smallmatrix} 1 & 0 \\ -c(1-d) & 1 \end{smallmatrix}\bigr) \cdot M'' = \bigl(\begin{smallmatrix} 1 & -1/c \\ 0 & 1 \end{smallmatrix}\bigr)$ which can clearly be reduced further to $I_2$.

For the case $c = 0 \implies ad = 1$ we have $M = \bigl(\begin{smallmatrix} a & b \\ c & 1/a \end{smallmatrix}\bigr)$.

and a proof follows along the lines of this post.

Proof of the general case:

Let $A = (a_{ij})$ be an $n\times n$ matrix with determinant one. If there exists a nonzero entry $a$ in the first column (row) at row (column) $i$, other than $a_{11}$, then column(row)-reduce using $(I_n + (1-a_{11})/a \cdot e_{1i})$. There is now a one in position $(1,1)$ and the rest of the first column and the first row can be zeroed out with off-diagonal elementary matrices.

If the only nonzero entry in the first row or column is $a_{11}$, then multiply $A$ on the left by $(I_n + (1 - a_{11})/a_{11}\cdot e_{21})$. Then $a_{11}$ can be reduced to one and the rest of the first column and first row can be reduced to zero using only off-diagonal elementary matrices.

Since only matrices of the off-diagonal type were used the determinant of the resulting matrix equals the determinant of $A$ and since there is only a one at $(1,1)$ and the rest are zero, expansion of the determinant along the first row or first column is equal to the determinant of the matrix $A$ with the first row and first column deleted. By induction on $n$, the smaller $(n-1)\times(n-1)$ matrix is also reducible to to identity using only off-diagonal elementary matrices.