= Solution
For $i\ne j$, put $x_{ij}(a)=I+aE_{ij}$. When $a\ne0$ this is a <transvection>: it fixes the <hyperplane> $v_j=0$ pointwise and has rank-one difference from the identity. Its inverse is $x_{ij}(-a)$ and its <determinant> is one.
Left multiplication performs a row addition. For an invertible <matrix>, bring a nonzero entry into a pivot position by a signed interchange, then clear its column by row additions. A signed interchange on two coordinates is itself a product of row additions, because
$$
w(a)=x_{12}(a)x_{21}(-a^{-1})x_{12}(a)
=\begin{pmatrix}0&a\\-a^{-1}&0\end{pmatrix}\qquad(a\ne0).
$$
Repeating the pivot construction and then clearing above the pivots reduces any determinant-one <matrix> to a diagonal <matrix> of <determinant> one. Finally
$$
w(a)w(-1)=\operatorname{diag}(a,a^{-1}).
$$
A determinant-one diagonal <matrix> is a product of such two-coordinate matrices, using the last coordinate to compensate each of the first $n-1$ diagonal entries. Each displayed <matrix> is a product of <elementary transvection matrices>. Hence reversing the elimination proves
$$
\boxed{SL_n(F)=\langle x_{ij}(a):i\ne j,\ a\in F\rangle.}
$$
This argument works over every <field>, including <characteristic> two.
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