Two locally linearly dependent operators are linearly dependent Let $S$ and $T$ be operators on a complex vector space $X$. Suppose they are locally linearly dependent, i.e., $Tx=\alpha_{x}Sx \quad \forall x\in X, \quad \alpha_{x}\in \mathbb{C}$. Then we must show that $T=\alpha S$ for some $\alpha \in \mathbb{C}$. 
One way of proving this is by using the following theorem: Let $U$ and $V$ be vector spaces over an infinite field $\mathbb{F}$ and let $V_{0}$ be a finite dimensional subspace of $V$. Let $T_{1}, \dots T_{n}$ be linear operators. If $T_{1}u, \dots T_{n}u$ are linearly dependent modulo $V_{0}$ for every $u \in U$, there exist $\alpha_{1}, \dots, \alpha_{n} \in \mathbb{F}$, not all zero, such that $S=\alpha_{1}T_{1}+\dots \alpha_{n}T_{n}$ satisfies rank $S \leq \text{dim } V_{0}+n-1$. The proof can be found here.
Is there a more elementary way of proving this (without using the above theorem)? I would be grateful for a hint.
 A: Let's first handle the case of an injective $S$, i.e. $\ker S = \emptyset$.
Using the  axiom of choice, $X$ has a (possible infinite) basis, say $B = (b_i)_{i \in I}$. Let $(s_i)$ be the images of the $b_i$ under $S$, and $t_i$ be the images of the $b_i$ under $T$, and $\alpha_i$ the local linear dependence constants for the $b_i$, i.e.  $$
   t_i = \alpha_i s_i \text{.}
$$
Since $S$ is injective, the  $s_i$ are linearly independent, because if there's a non-trivial linear combination $\sum_{k=1}^n \lambda_k s_{i_k} = 0$ then $S\left(\sum_{k=1}^n \lambda_k s_{i_k}\right) = 0$, contradicting the injectivity of $S$.
Now pick arbitrary $j,k \in I$ and let $u = b_j + b_k$. It follows that $$
  0 = T(u) - \alpha_u S(u)
  = \underbrace{\alpha_j s_j+ \alpha_k s_k}_{=T(u)} - (\underbrace{\alpha_u s_j + \alpha_u s_k}_{=\alpha_u S(u)})
  = (\underbrace{\alpha_j - \alpha_u}_{=:\lambda_j}) s_j + (\underbrace{\alpha_k - \alpha_u}_{=:\lambda_k}) s_k \text{.}
$$
Since the $s_i$ are linearly independent (see above), it follows that $\lambda_j = \lambda_k = 0$, i.e. that $\alpha_j = \alpha_u = \alpha_k = \alpha_u$. Thus, all the $\alpha_i$ are in fact equal, and we call them just $\alpha$.
We thus have for all basis vectors $(b_i)_{i \in I}$ that $Tb_i = \alpha Sb_i$, and it follows that indeed $$
  T = \alpha S \text{.}
$$

If $S$ is not injective, i.e. if $U = \ker S \neq \emptyset$, then we can write $X = V \oplus U$ for some subspace $V \subset X$ where $S|_V$ is injective. You can find such a $V$ by starting with a basis $(u_i)_{i \in I_U}$ of $U$ and extending to a basis $(b_i)_{i \in I}$ of $V$ (where $I_U \subset I$). Then, with $I_V := I \setminus I_U$, $(b_i)_{i \in I_V}$ is a basis of some subspace $V$, and since $$
  \iota \,:\, U \times V \to X \,:\, \left( (a_i)_{i \in I_U}, (b_i)_{i \in I_V}\right) \mapsto (c_i)_{i \in I}
  \quad \text{ where } c_i = \begin{cases}
    a_i &\text{if $i \in I_U$} \\
    b_i &\text{if $i \in I_V$}
  \end{cases}
$$
is a bijection between coordinatizations of $U \times V$ and $X$, indeed $X = U \oplus V$.
Since $S|_V$ is injective, it follows from the first part that there is an $\alpha$ such that $$
  t_i = \alpha\, s_i \quad\text{ if $i \in I_V$.}
$$
That leaves $i \in I_U$ to deal with, in which case $s_i = 0$ (because $b_i \in \ker S$, per the construction of our basis). But if $t_i = \alpha_i s_i$ for any $\alpha_i$, it follows that $t_i = 0$, and then $t_i = \alpha_i s_i$ for every $\alpha_i$, and in particular for $\alpha_i = \alpha$. So also have $$
  t_i = \alpha\, s_i \quad\text{ if $i \in I_U$.}
$$
Taken together, we again have $Tb_i = \alpha Sb_i$ for all basis vectors $(b_i)_{i \in I}$ (remember that $I = I_U \cup I_V$), and therefore that $$
  T = \alpha S \text{.}
$$
A: Here is something, although perhaps not what you are looking for :
When $X$ and $Y$ are finite dimensional, define a bilinear map $B:X\times Y^{\ast} \to B(X,Y)^{\ast}$ given by
$$
B(x,f)(T) := f(Tx)
$$
This gives a linear map
$$
X\otimes Y^{\ast} \to B(X,Y)^{\ast}
$$
One can topologize the left hand side so that this is an isomorphism (although this is not trivial). The thing on the LHS is called the projective tensor product and is denoted by $X\hat{\otimes}Y{^\ast}$. (See this)
In particular, when $X$ and $Y$ are finite dimensional, one finds a proof of what you are looking for : 
Suppose $T\notin \operatorname{span}(S) \subset B(X)$, then by Hahn-Banach $\exists h \in B(X)^{\ast}$ such that $\|h\| \leq 1, h(S) = 0$, and $h(T) = 1$.
Consider
$$
\{h \in B(X)^{\ast} : \|h\| \leq 1, h(S) = 0\}
$$
This is compact (in the weak-star topology), convex and non-empty, so is the closed convex hull of its extreme points (by Krein-Milman). Hence there must be such an extreme point such that $h(T) \neq 0$. 
The extreme points are elementary tensors (See this). Hence, $\exists x\in X, f\in X^{\ast}$ such that
$$
(x\otimes f)(S) = 0, \text{ and } (x\otimes f)(T) \neq 0
$$
But
$$
(x\otimes f)(T) = \alpha_x (x\otimes f)(S)
$$
which is a contradiction.
This is the most naive approach I could think of to settle this question. I don't think it extends to the infinite dimensional setting, but it might be worth thinking about.
A: Two locally linearly dependent operators $S$ and $T$ need not be linearly dependent. Indeed, assume that $H$ is a complex Hilbert space and let $u,v\in H$ be two nonzero vectors. Let $u\otimes v$ be the rank one operator defined by $$(u\otimes v)(x):=\langle x,v\rangle u,~(x\in H).$$
Now, assume that $u$ and $v$ are two orthogonal nonzero vectors in $H$, and let $$T:=u\otimes v \text{ and }S:=u\otimes (u+v)=u\otimes u+u\otimes v.$$
For any $x\in H$, we have $$Tx=\langle x,v\rangle u\text{ and }Sx=\langle x,u+v\rangle u.$$
Then $T$ and $S$ are loccaly linealry dependent. But $T=\alpha S$ would imply that $$(1-\alpha)(u\otimes v)=\alpha(u\otimes u), $$
and then $$(1-\alpha)(u\otimes v)(u)=\alpha(u\otimes u)(u).$$
Because $u$ and $v$ are assumed to be two orthogonal, this entails that $$0=(1-\alpha)\langle u, v\rangle u=\alpha\|u\|^2u.$$
Clearly, $\alpha$ must be $0$ and $u\otimes v=T=\alpha S=0$. This arises a contradiction, and $T$ and $S$ are NOT linearly dependent. In fact, if $T$ and $S$ are locally linearly dependent operators, then there are $\alpha,\beta\in\mathbb{C}$ (not both zero) and an operator $R$ of rank at most 1 such that $$\alpha T+\beta S=R.$$
This is known as Kaplansky's theorem or Aupetit's result. It applies to more that two operators but in this case $R$ must be a finite rank operatr.
