= Solution
Let $S=SL_n(F)$ and let $Z$ be its scalar center. Its action on the lines of $F^n$ has <kernel> exactly $Z$: a <linear map> fixing every line is diagonal in a basis, and fixing the lines spanned by $e_i+e_j$ makes all its diagonal entries equal. The action is two-transitive. Indeed an ordered pair of distinct lines can be represented by two independent <vectors> and extended to a basis. A <linear map> between two such bases can be adjusted to <determinant> one by scaling one target basis <vector>; this preserves both target lines. Therefore $S/Z$ has a faithful <primitive group action>.
Fix $L=Fv$. The maps
$$
A_L=\{I+vf:f\in(F^n)^*,\ f(v)=0\}
$$
form an abelian <subgroup>: all products of the rank-one terms vanish. It is normal in the line stabilizer, since <conjugation> replaces $v$ by a <scalar multiple> and transports $f$. Its conjugates contain every <transvection>, and hence generate $S$ by 4(b). The same statements hold for its <image> in $S/Z$.
It remains to prove perfectness rather than assume it. Use the <group commutator> convention $[x,y]=xyx^{-1}y^{-1}$. If $n\ge3$, distinct $i,j,k$ give
$$
[x_{ik}(a),x_{kj}(b)]=x_{ij}(ab).
$$
Every elementary generator is consequently in $S'$. If $n=2$ and $|F|>3$, choose $a\in F^*$ with $a^2\ne1$. There are at most two roots of $a^2=1$, so such a choice exists. For $h=\operatorname{diag}(a,a^{-1})$,
$$
[h,x_{12}(b)]=x_{12}((a^2-1)b).
$$
Varying $b$ gives every upper <transvection>; conjugating by $w(1)$ gives every lower one. Again $S'=S$, and its projective quotient is a <perfect group>.
Apply 4(a) to this faithful primitive projective action. Every nontrivial <normal subgroup> contains the <commutator subgroup>, which is the whole quotient. Thus
$$
\boxed{PSL_n(F)\text{ is simple if }n\ge3\text{ or if }n=2,\ |F|>3.}
$$
The restriction is real: $PSL_2(2)\cong S_3$ and $PSL_2(3)\cong A_4$ are not simple.
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