I need to prove that the transvection matrices generate the special linear group $\operatorname{SL}_n \left(\mathbb{R}\right) $.
I want to proceed using induction on $n$.
I was able to prove the $2\times 2$ case, but I am having difficulty with the $n+1$ case.
I supposed that the elementary matrices of the first type generate $\operatorname{SL}_n(\mathbb{R})$. And I want to show that an elementary matrix of the first type of order $n+1$ can generate $\operatorname{SL}_{n+1}(\mathbb{R})$
Multiplying a matrix on the left or the right by a transvection matrix corresponds to adding a multiple of one row/column to another row/column. So we just need to prove that we can reduce any matrix in ${\rm SL}_n(K)$ to the identity matrix using row and column operations of this type. This works for any field $K$.
The proof is by induction on $n$ and there is nothing to prove for $n=1$, so assume that $n>1$. Let $A = (a_{ij}) \in {\rm SL}_n(K)$.
If $a_{12} = 0$ then, choose some $j$ with $a_{1j} \ne 0$ and add column $j$ to column $2$ to get $a_{12} \ne 0$.
Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.
For all $j>1$ subtract $a_{1j}$ times column $1$ from column $j$ to get $a_{1j}=0$. Similarly use row operations to get $a_{j1}=0$ for all $j>1$.
You have now reduced to the case $n-1$.
