Given a (not necessarily finite dimensional) vector space $V$ prove that the center of $\operatorname{GL}(V)$ is the set of all scalar transformations (i.e all transformations of the form $a\operatorname{Id}$)?
I know how to prove this for general linear group of degree $n$, please help me solve this for the case of a general linear map.
Let $T$ in the center. For any $L$ we have $T\circ L= L\circ T$, that is $$T(Lx) = L(Tx)$$ for all $L$ and all $x \in V$.
Let $x$ in $V$. There exists $L$ linear map so that the subspace $\{y \ | \ Ly = y\}$ equals $\mathbb{F} \cdot x$ ( use a basis starting from $x$).
We get $L(Tx) = T(Lx) = Tx$ and so $Tx \in \mathbb{F} \cdot x$.
So, for any $x\in V$ we have $T(x)$ proportional to $x$. It's easy now.
Let assume that there exists an element $A$ in the center of a general linear group over an arbitrary vector space $V$ such that $A$ is not a scalar transformation. ($\dim V$ is infinite)
CLAIM: $\exists\, v$ such that $v$ is not an eigenvector of $A$.
Let assume that every vector in $V$ is an eigenvector of $A$. Pick two non-parallel vectors $v_1$, $v_2$ and let $a_1$ and $a_2$ be corresponding eigenvalues. The assumption implies that $v_1+v_2$ should be an eigenvector of $A$ and let $a$ be the eigenvalue of $v_1+v_2$. $$A \cdot v_1 + A \cdot v_2 = a_1v_1 + a_2v_2 = av_1 + av_2$$ $$(a_1 - a)v_1 = (a-a_2)v_2$$ $v_1$ and $v_2$ are not parallel, so $a_1=a_2=a$ for arbitrary vector $v_1$ and $v_2$. So $A$ should be a scalar transformation. This is a contradiction.
So we can assume that there exists a vector $v$ which is not an eigenvector of $A$. Let $w = A \cdot v$. Then $w \notin \left<v\right>$. Let $S$ be the vector subspace generated by $v$ and $w$. Then $\left\{ v, w \right\}$ is a basis of $S$ and there is a basis $T$ of $V$ s.t. $\left\{ v, w \right\} \subset T$. Let $S'$ be the vector subspace of $V$ generated by $T \setminus \left\{ v, w \right\}$.
We can construct two general linear matrices $B_1$ and $B_2$ s.t. $$B_1 \cdot v = w,\ B_1 \cdot w = v$$ $$B_2 \cdot v = \frac{1}{2}w,\ B_2 \cdot w = v$$ $$B_1|_{S'}=B_2|_{S'}=\mathrm{id}|_{S'}.$$
Then we get following results: $$A \cdot w = A \cdot B_1 \cdot v = B_1 \cdot A \cdot v = B_1 \cdot w = v$$ $$A \cdot w = A \cdot B_2 \cdot 2v = B_2 \cdot A \cdot 2v = B_2 \cdot 2w = 2v$$ This is a contradiction. So $A$ should be a scalar transformation.
