Every elementary matrix of the first kind is a commutator of invertible matrices

Statement
Suppose $$k$$ is a field and $$n$$ is a natural number. For $$\lambda \in k$$ and $$i \ne j$$ elements of $$\{ 1,2,3, \dots, n \}$$, denote by $$E_{ij}(\lambda)$$ the matrix with $$1$$s on the diagonal, $$\lambda$$ in the $$(ij)^{th}$$ position, and $$0$$s elsewhere. A matrix that can be written as $$E_{ij}(\lambda)$$ for some $$i,j,\lambda$$ is termed an elementary matrix.

Every elementary matrix can be written as a commutator of two invertible matrices in either of these cases:


 * $$n \ge 3$$.
 * $$k$$ has at least three elements.

Facts about elementary matrices

 * Every elementary matrix is a commutator of elementary matrices: This holds over any unital ring, but requires $$n \ge 3$$.
 * Every elementary matrix is a commutator of unimodular matrices: This requires a somewhat more stringent condition on $$k$$: $$k$$ should have more than three elements.
 * Every elementary matrix is the commutator of an invertible and an elementary matrix: This is a stronger version of the same statement, and has essentially the same proof.

Facts about the general and special linear groups

 * Special linear group is perfect: Holds when $$n \ge 3$$ and $$k$$ has at least four elements.
 * Commutator subgroup of general linear group is special linear group: Holds when $$n \ge 3$$ and $$k$$ has at least three elements.
 * Lower central series of general linear group stabilizes at special linear group: Holds when $$n \ge 3$$ and $$k$$ has at least three elements.

Facts used

 * 1) uses::Every elementary matrix is a commutator of elementary matrices

The case $$n \ge 3$$
This follows directly from fact (1).

The case $$n = 2$$ and $$k$$ has more than two elements
We need to show that the matrices $$E_{12}(\lambda)$$ and $$E_{21}(\lambda)$$ can be expressed as commutators of invertible mtarices. We show this for $$E_{12}(\lambda)$$. A similar argument works for $$E_{21}(\lambda)$$.

Pick $$\mu \in k$$ such that $$\mu \ne 0,1$$. Consider:

$$g = \begin{pmatrix} \mu & 0 \\ 0 & 1 \end{pmatrix}, \qquad h = \begin{pmatrix} 1 & \lambda/(\mu - 1) \\ 0 & 1 \end{pmatrix}$$.

A computation shows that the commutator of $$g$$ and $$h$$ is $$E_{12}(\lambda)$$.