= Solution
Replace $F^n$ by a <symplectic vector space> $(V,B)$ of dimension $2m$, and elementary row operations by <symplectic transvections>
$$
T_{v,c}(x)=x+cB(x,v)v.
$$
The cross terms in $B(Tx,Ty)$ cancel, and $B(v,v)=0$, so these maps preserve $B$. Their inverses are $T_{v,-c}$.
They generate $Sp(V)$ by the following <symplectic basis> elimination. If $B(u,v)\ne0$, a single <transvection> with direction $v-u$ and parameter $1/B(u,v)$ sends $u$ to $v$. If $B(u,v)=0$ and $u\ne v$, choose $w$ with both $B(u,w)$ and $B(w,v)$ nonzero: the union of two proper <hyperplanes> does not exhaust a <vector space>, even over $\mathbb F_2$. Two <transvections> then suffice. Thus one can make a form-preserving <linear map> fix the first basis <vector> $e$. Next align its partner $f$, while keeping $e$ fixed. Two partners have difference in $e^\perp$, so the preceding single-transvection construction uses a direction in $e^\perp$ whenever their mutual pairing is nonzero. If it is zero, the intermediate partner $f+ae$, $a\ne0$, has nonzero pairing with both. All these <transvections> fix $e$. Once both $e,f$ are fixed, recurse on their <nondegenerate> <symplectic orthogonal complement>. This proves generation over every <field>.
Use the action on the points of <projective space>, whose <kernel> is the scalar center $Z=\{\lambda I:\lambda^2=1\}$. For $m=1$ this is the $SL_2$ case. For $m\ge2$ the action is primitive, although it is generally not two-transitive. The stabilizer of a line $L$ has two orbits on other lines: those orthogonal to $L$ and those not orthogonal to $L$. This follows by extending a <hyperbolic pair>, or an independent isotropic pair, to a <symplectic basis>. A block containing $L$ and a second line $M$ contains the entire corresponding orbit of the stabilizer of $L$. The stabilizer of $M$ then supplies a line of the other type: choose a <vector> orthogonal to one of $L,M$ and not to the other. Nondegeneracy and independence guarantee such a <vector>, also in dimension four. Hence the block contains both orbits and is the whole set.
The <transvections> along $L$ form an abelian normal <subgroup> of its stabilizer; their conjugates generate the <group>. Thus the <Iwasawa simplicity lemma> applies as soon as perfectness is established. For $|F|>3$, every <transvection> lies in an embedded $SL_2(F)$ acting on a hyperbolic plane, and 4(c) proves that embedded <group> perfect.
For the remaining fields, the <perfectness of finite symplectic groups> has a short direct proof. Over $\mathbb F_3$ with $m\ge2$, all parameter-one <transvections> have the same class $t$ in the <abelianization>, and $3t=0$. In an isotropic plane with basis $u,v$, the four directions $u,v,u+v,u-v$ give commuting <transvections> whose product is $1$: their rank-one terms add to zero, since
$$
uu^{\mathsf T}+vv^{\mathsf T}+(u+v)(u+v)^{\mathsf T}+(u-v)(u-v)^{\mathsf T}=0
$$
over $\mathbb F_3$. Thus $4t=0$, giving $t=0$. Parameter-two <transvections> are inverses, so all generators vanish in the <abelianization>. Over $\mathbb F_2$ with $m\ge3$, use the seven nonzero <vectors> of an isotropic three-space. Their commuting <transvections> multiply to $1$, since each diagonal coordinate of $\sum_{0\ne v}vv^{\mathsf T}$ occurs four times and each off-diagonal coordinate twice. Their common abelianized class satisfies $2t=7t=0$, and again the <group> is perfect.
Consequently \b[the projective <symplectic group> is simple], with the necessary exceptions
$$
\boxed{PSp_{2m}(F)\text{ simple, except }(m,F)=(1,\mathbb F_2),(1,\mathbb F_3),(2,\mathbb F_2).}
$$
The first two are the linear exceptions; the last is $Sp_4(2)\cong S_6$, proved in 6(c). The word “projective” matters: the unquotiented <symplectic group> may have the nontrivial scalar center $\{I,-I\}$.
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