For arbitrary vectors w and u there is a linear operator T such that T(w) = u? I need to know whether in an arbitrary vector space $V$, given arbitrary non-zero vectors $v,u\in V$, there is a linear operator $T:V\to V$ such that $T(v) = u$.
I know that this statement is true if the dimension of $V$ is finite. But I don't know if the result holds for spaces of infinite dimension. So, being more specific: Let $V$ be a vector space of infinite dimension, given arbitrary non-zero vectors $v,u\in V$, is there a LINEAR map $T:V\to V$ such that $T(v) = u$?
If the latest one is true, I ask a more specific question: Let $V$ be a vector space of infinite dimension, given arbitrary non-zero vectors $v,u\in V$, is there a LINEAR CONTINUOUS map $T:V\to V$ such that $T(v) = u$?
 A: Under the axiom of choice, you may extend the linearly independent set $\{v\}$ to a basis $B$ (see here) for details.  Now, you can map $v$ to $u$, and all other basis elements to $0$.
If you do not accept AC, I have no idea.
A: It continuity is involved then $V$ has to be a topological vector space. Assume that it is a normed vector space. Then for each nonzero $v\in V$ there exists a continuous linera functional $\xi$ on $V$ such that $\xi(v)=1$. 
Let $u, v\in V$ be arbitrary, $v\ne 0$, and $\xi$ as above. Then
$$ x\mapsto \xi(x) u $$
defines a continuous linear map on $V$ such that $v$ is mapped to $u$. This is called rank-one operator.
A: Yes, we may always create a linear transformation $T$ so that $Tv = u$ for any $v \neq 0$ and $u$. We achieve this by finding a basis for $V$ that contains $v$. By a Zorn's Lemma argument, every vector space has a basis. Suppose $\{h_{\alpha} \}$ is a basis for $V - \mathrm{Span}\;{v}$. Then $\{h_{\alpha}\} \cup \{v\}$ will be a basis for $V$.
Now, we can simply define $T$ by its effect on this basis. Let $T$ be a function so that
\begin{align*}
Tv &=u\\
& \; \text{and} \\
Tv_{\alpha} &= v_{\alpha}.
\end{align*}
Then, simply extend $T$ linearly, and it will have the desired properties.
Now for your question on whether there exists a continuous transformation we need to assume $V$ has some additional structure. We need to assume it is a topological vector space. In most cases of interest, it will be a normed space, so assume that it is.
I believe that the same map above will be continuous with respect to the norm topology. It is clearly continuous when restricted to the subspaces $\mathrm{Span} \; v$ and $\mathrm{Span} \; \{h_{\alpha} \}$, as on the latter it is the identity operator, and on the former it is a linear transformation between finite dimensional spaces, so it must be continuous on it. Using these facts, it is straightforward to show that the map is continuous at $0$, and so continuous everywhere, as desired.
