Corollary

The group $\mathbf{SL}(n, K)$ is the group of commutators of $\mathbf{GL}(n, K)$, except in the case where $n = 2$ and where $K$ is a field with 2 elements.

As $\mathbf{SL}(n, K)$ is the kernel of the homomorphism det of $\mathbf{GL}(n, K)$ to a commutative group $K^*$, $\mathbf{SL}(n, K)$ contains the commutator group $\Gamma$ of $\mathbf{GL}(n, K)$ (I, § 6, no. 2). To prove that $\mathbf{SL}(n, K) = \Gamma$, it will suffice, by Proposition 17, to show that, for all $\lambda \in K^*$, $B_{ij}(\lambda)$ belongs to $\Gamma$. Now, $B_{ij}(\lambda)$ is a conjugate of $B_{ij}(1)$ in $\mathbf{GL}(n, K)$ since $B_{ij}(\lambda) = Q . B_{ij}(1) . Q^{-1}$, where $Q$ denotes the matrix with respect to the canonical basis $(e_i)$ of the automorphism $v$ of $K^n$ such that $v(e_i) = \lambda e_i, v(e_k) = e_k$ for $k \neq i$. On the other hand, let $u_{ij}$ (for $i \neq j$) be the automorphism of $K^n$ such that $u_{ij}(e_i) = -e_j, \ u_{ij}(e_j) = e_i, \ u_{ij}(e_k) = e_k$ for $k \notin \{i, j\}$, which belongs to $\mathbf{SL}(n, K)$; then $B_{ji}(\lambda) = U_{ij} B_{ij}(-\lambda) U_{ij}^{-1}$, where $U_{ij}$ is the matrix of $u_{ij}$ with respect to the canonical basis. Similarly, if $1 < i < j$, then $B_{1j}(\lambda) = U_{1i} B_{ij}(\lambda) U_{1i}^{-1}$ and finally, for $2 < j$, $B_{12}(\lambda) = U_{2j} B_{1j}(\lambda) U_{2j}^{-1}$. This proves that all the $B_{ij}(\lambda)$ have the same image $s$ in $\mathbf{GL}(n, K)/\Gamma$ and it remains to show that $s$ is the identity element.

Suppose first that $K$ contains an element $\lambda$ distinct from 0 and 1; then $1 = \lambda + (1 - \lambda)$, the two terms on the right hand side being $\neq 0$; the relation $B_{12}(1) = B_{12}(\lambda) B_{12}(1 - \lambda)$ shows that $s^2 = s$ and hence $s$ is the identity element.

Suppose now that $n \geq 3$. The product $B_{21}(1) B_{31}(1)$ is the matrix of an automorphism $u$ of $K^n$ such that $u(e_1) = e_1 + e_2 + e_3, u(e_i) = e_i$ for $i \neq 1$. If $S$ is the matrix of the automorphism $u'$ of $K^n$ such that $u'(e_2) = e_2 + e_3, u'(e_i) = e_i$ for $i \neq 2$, then $S . B_{21}(1) B_{31}(1) . S^{-1} = B_{21}(1)$; we also deduce that $s^2 = s$, which completes the proof.