= Solution
For distinct coordinates $i,j$, let $E_{ij}(t)=I+t e_{ij}$, where $e_{ij}$ is the <matrix unit>. For $t\ne0$ this is a <transvection>: it fixes the <hyperplane> with $j$th coordinate zero pointwise, has <determinant> one and inverse $E_{ij}(-t)$. We show these <elementary transvection matrices> generate $SL(n,F)$ over any <field>.
Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular <matrix> with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a <diagonal matrix> $D=\operatorname{diag}(a_1,\ldots,a_n)$ with $\prod_i a_i=1$. Thus it remains to realize determinant-one <diagonal matrices> using elementary <transvections>.
In a two-coordinate block put
$$
w(a)=E_{12}(a)E_{21}(-a^{-1})E_{12}(a)
=\begin{pmatrix}0&a\\-a^{-1}&0\end{pmatrix}.
$$
Direct multiplication gives
$$
\boxed{w(a)w(-1)=\operatorname{diag}(a,a^{-1}).}
$$
The <diagonal matrix> $D$ is the product, for $i=1,\ldots,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is $\prod_{i<n}a_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary <transvections>. All row operations used are invertible <transvections>, and no division except by a nonzero pivot or $a_i$ has been made. This proves
$$
\boxed{SL(n,F)=\langle E_{ij}(t):i\ne j,\ t\in F\rangle.}
$$
The identity element $E_{ij}(0)$ can be omitted from the generating set. This is <transvections generate the special linear group>; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.
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