Every elementary matrix is the commutator of an invertible and an elementary matrix

Statement
Suppose $$k$$ is a field and $$n$$ is a natural number. For $$i \ne j$$ in $$\{ 1,2,3, \dots, n \}$$ and $$\lambda \in k$$, define $$E_{ij}(\lambda)$$ as the matrix with $$1$$s on the diagonal, a $$\lambda$$ in the $$(ij)^{th}$$ position, and $$0$$s elsewhere.

A matrix that is of the form $$E_{ij}(\lambda)$$ is termed an elementary matrix.

Every elementary matrix can be expressed as the commutator of an invertible matrix and an invertible matrix of determinant one (i.e., a unimodular matrix) in either of these cases:


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

In fact, we can further ensure that the unimodular matrix (the matrix of determinant one) is itself an elementary matrix.

Related facts

 * 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 a commutator of invertible matrices: A weaker version of the same statement.

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 $$E_{12}(\lambda)$$.