Matrix-free proof of $Z(GL_n(F)) = \{\lambda I:\lambda \in F^\times\}$? How does one prove that $$Z(GL_n(F)) = \{\lambda I:\lambda \in F^\times\}$$ without resorting to matrices (and bases)?  (BTW, $Z(GL_n(F))$ is the center of $GL_n(F)$, the general linear group of order $n$ over the field $F$, $I$ is the identity element of $GL_n(F)$, and $F^\times = F\backslash\{0\}$.)
(Obviously, $Z(GL_n(F)) \supseteq \{\lambda I:\lambda \in F^\times\}$, so the problem is to prove the opposite inclusion.)
(I have not been able to make much headway with this.  All I've managed to do is to show that if $T \in GL_n(F)$ is such that every non-zero vector of $F^n$ is an eigenvector of $T$, then $T = \lambda I$, for some $\lambda \in F^\times$ (which, I admit it, is not much).  So I still need to show that any $T\in Z(GL_n(F))$ must have this property that every non-zero vector of $F^n$ is its eigenvector.  Actually, I don't even know if this is the right strategy at all.)
 A: Since you want a matrix-free proof, you also need a matrix-free claim :). That is

$Z(\mathrm{GL}(V))=F^* \cdot 1$

for a vector space $V$ over a field $F$, where $\mathrm{GL}(V) := \mathrm{Aut}_F(V)$. We don't need to assume any finite-dimensionality of $V$ here. We will need the fact that $\mathrm{GL}(V)$ acts transitively on the linearly independent tuples (a fact whose proof needs bases; of course we need bases somewhere since the claim is usually false when $V$ is replaced by a module over a ring which is not a field.)
Let $A \in Z(\mathrm{GL}(V))$ and $v \in V \setminus \{0\}$. I claim that $v$ and $A(v)$ are linearly dependent. If not, there is some $P \in \mathrm{GL}(V)$ such that $P(v)=v +A(v)$ and $P(A(v))=A(v)$ . But then
$$A(v)=P(A(v))=A(P(v))=A(v+A(v))=A(v)+A^2(v) \Rightarrow A^2(v)=0,$$
a contradiction. Thus, there is some $\lambda \in F^*$ such that $A(v)=\lambda v$. Now we have to prove that $\lambda$ doesn't depend on $v$. So assume that $A(v)=\lambda v$ and $A(w)=\mu w$. There is some $P \in \mathrm{GL}(V)$ such that $P(v)=w$. Then $$\mu w = A(w)=A(P(v))=P(A(v))=P(\lambda v)=\lambda P(v)=\lambda w,$$ hence $\mu=\lambda$.
In my opinion this proof is even simpler and more natural than the matrix proof.
A: Let $T\in\text{Aut}(V)$ such that there exists a $v\in V$ with $T(v)\neq\lambda v$ for all $\lambda\in F^*$.
Complete $v$ to a basis $\{v=v_1,v_2,...,v_n\}$ of $V$ in such a way that writing $T(v)=a_1v_1+a_2v_2+...+a_nv_n$ one has $a_1\neq a_2$ and 
$$
T(v_2)\neq a_2v_1+a_2v_1+a_3v_3+...+a_nv_n.
$$
Now define $U\in\text{Aut}(V)$ by declaring
$$
U(v_1)=v_2,\qquad
U(v_2)=v_1,\qquad
U(v_k)=v_k
$$
for all $k\geq3$. Then it is quite obvious that $TU(v_1)\neq UT(v_1)$ and thus $T$ and $U$ do not commute. But then $T$ is not in the center of $\text{Aut}(V)$.
A: Consider the module ${}_{GL_n(F)}F^n$, this is simple (having no nontrivial submodule, i.e. subspace simultaneously invariant to all elements of $GL_n(F)$).
Now, if $T$ commutes with all $A\in GL_n(F)$, then $T:F^n\to F^n$ is an endomorphism of this module, as
$$T(A\cdot v)=TAv=ATv=A\cdot Tv\,.$$
So is $T-\lambda I$ for all $\lambda\in F$. Because of simplicity, such an endomorphism must be invertible or $0$.
If $F$ is assumed to be algebraically closed, then $T$ is forced to have an eigenvalue $\lambda$, but then $T-\lambda I$ is not invertible, so
$$T-\lambda I=0\,.$$
