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

Statement

Suppose $k$ is a field and $n\ge 2$ is a natural number. For $i\ne j$ in the set $\{1,2,\dots ,n\}$, and $\lambda \in k$, let $e_{ij}(\lambda )$ be the matrix with $\lambda$ in the $(ij{)}^{th}$ position and $0$ elsewhere, and let $E_{ij}(\lambda )$ be the matrix with $1$s on the diagonal, $\lambda$ in the $(ij{)}^{th}$ position, and zeroes elsewhere. Matrices of the form $E_{ij}(\lambda )$ are termed elementary matrices.

Let $S{L}_n(k)$ denote the group of all matrices of determinant $1$ under multiplication. Clearly, $E_{ij}(\lambda )\in S{L}_n(k)$ for all $i\ne j,\lambda \in k$. The following are true:

• If $n\ge 3$, every elementary matrix can be expressed as a commutator of two elementary matrices. In particular, it is the commutator of two elements of $S{L}_n(k)$.
• If $n=2$, and $k$ has more than three elements, every elementary matrix can be expressed as a commutator of two elements of $S{L}_n(k)$.

Facts used

1. Every elementary matrix is a commutator of elementary matrices: This statement is valid for $n\ge 3$.

Proof

The case $n\ge 3$

This follows from fact (1), and the observation that every elementary matrix is unimodular.

The case $n=2$

If $k$ has more than three elements, there exists $\mu \in k$ such that $\mu$ is invertible and ${\mu }^2\ne 1$.

We need to prove that for any $\lambda$, the element $E_{12}(\lambda )$ (and analogously, the element $E_{21}(\lambda )$) is expressible as a commutator. Indeed, set:

$g=\left(\begin{array}{cc}\mu & 0\\ 0& {\mu }^{-1}\end{array}\right),h=\left(\begin{array}{cc}1& \lambda /({\mu }^2-1)\\ 0& 1\end{array}\right)$

A computation shows that the commutator of $g$ and $h$ is $E_{12}(\lambda )$. A similar construction gives $E_{21}(\lambda )$ as a commutator.