# Transvection matrices generate $\operatorname{SL}_n(\mathbb{R})$

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 $$\operatorname{SL}_{n+1}(\mathbb{R})$$

## 1 Answer

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 $A = (a_{ij}) \in {\rm SL}_n(K)$.

1. If $a_{12} = 0$ then, choose some $j$ with $a_{1j} \ne 0$ and add column $j$ to column $2$ to get $a_{12} \ne 0$.

2. Subtract $(a_{11}-1)/a_{12}$ times column $2$ from column $1$ to get $a_{11}=1$.

3. 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 $n-1$.