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

Statement
Let $$R$$ be any unital ring (not necessarily commutative) and $$n$$ be a natural number. Define, for $$i,j \in \{ 1,2,3, \dots n\}$$ with $$i \ne j$$ and $$\lambda \in R$$:


 * $$e_{ij}(\lambda)$$ is the matrix with $$\lambda$$ in the $$(ij)^{th}$$ position and $$0$$s elsewhere.
 * $$E_{ij}(\lambda)$$ is the matrix with $$1$$s on the diagonal, $$\lambda$$ in the $$(ij)^{th}$$ position, and $$0$$s elsewhere.

An elementary matrix is a matrix that can be written as $$E_{ij}(\lambda)$$ for some $$i \ne j$$ and some $$\lambda \in R$$.

If $$n \ge 3$$, every elementary matrix can be expressed as the commutator of two elementary matrices.

Proof
Given: A ring $$R$$, a $$n \times n$$ elementary matrix $$E_{ij}(\lambda)$$ for $$n \ge 3$$.

To prove: $$E_{ij}(\lambda)$$ is the commutator of two elementary matrices.

Proof: Pick $$l \in \{ 1,2,3, \dots, n \}$$ different from $$i,j$$. Consider elements:

$$g = E_{il}(\lambda), \qquad h = E_{lj}(1)$$.

Then, the commutator is given by:

$$ghg^{-1}h^{-1} = E_{il}(\lambda)E_{lj}(1)E_{il}(-\lambda)E_{lj}(-1)$$.

Rewriting in terms of the $$e_{ij}$$s:

$$ghg^{-1}h^{-1} = (I + e_{il}(\lambda))(I + e_{lj}(1))(I - e_{il}(\lambda)))I - e_{lj}(1))$$.

Simplifying this yields:

$$ghg^{-1}h^{-1} = (I + e_{il}(\lambda) + e_{lj}(1) + e_{ij}(\lambda))(I - e_{il}(\lambda) - e_{lj}(1) + e_{ij}(\lambda))$$.

Further simplification yields:

$$ghg^{-1}h^{-1} = I + e_{ij}(\lambda) = E_{ij}(\lambda)$$.

This completes the proof.