I am reading Peter Cameron's note on Classical Groups and I got confused with Proposition 2.1 on page 14.
I have no problem in proving that the elements in kernel are scalars. However, I don't understand the last paragraph of the proof, which shows that the kernel lies in ${\bf Z}(F)$. It seems that the argument tries to insist that an element $A$ in the kernel should induce a linear map. But I don't see any reason for so. (To be in the kernel, it just needs to stabilize all 1-spaces in $F^n$ but need not to stabilize any particular vectors, right?)
In fact, I believe that there is also a mistake in the paragraph before the proposition statement, which says that every $A\in{\rm GL}(n,F)$ induces a linear map on $F^n$. For a simple case, take $A=\lambda I$, then $$(kv)A=(\lambda k)v=(\lambda k\lambda^{-1})(\lambda v)=k^\lambda(vA)$$ so the induced map of $A$ should be semilinear instead?
I think the result about only central scalars being in the kernel only makes sense if $n > 1$.
We have shown so far that $A = cI$ for some scalar $c$. Now let $a \in F$ and consider $(ae_1+e_2)A = ace_1 + ce_2$. This must equal $d(ae_1+e_2)$ for some scalar $d$. From the $e_2$ coefficient, we get $c=d$, and hence $ac=da=ca$.