Showing a linear transformation exists Let $V$ and $W$ be two vector spaces over the field of scalars $F$ and assume that $V$ is finite dimensional. Given an ordered list of vectors $B=(b_1, b_2, ..., b_n)\subset V$ show that if $B$ is linearly independent and $C=(c_1, c_2, ..., c_k)\subset W$ is an arbitrary list of vectors in $W$, then there is a linear transformation $$f:W\to W \,\,\text{such that}\,\, f(b_1) = c_1, ..., f(b_k) = c_k.$$

I don't really know how to start this. Are we meant to show that $f$ is a linear transformation, i.e. satisfies $$f(a+b) = f(a)+f(b), \quad \text{ for } a, b\in W \text{ and }$$ $$f(\lambda a) = \lambda f(a), \text{ for } a\in W \text{ and } \lambda\in F.$$
 A: First, you must certainly have that $k\leq n$.
Second, it looks like you mean $f\colon V\rightarrow W$, and not $f\colon W\rightarrow W$.
Third, I suggest starting by defining your linear transformation on the vectors $b_i$ as given in the problem statement. That is, you start out by defining $f$ on the set $B_k=\{b_1,\ldots,b_k\}$ by $f(b_i) = c_i$ for $i=1,\ldots,k$.
Then extend this definition of $f$ to all linear combinations of the vectors in the set $B_k$, that is, if $x=\sum_{i=1}^k x_ib_i$, then define $f(x) = \sum_{i=1}^k x_ic_i$. Notice how $f$ is now linear on the span of $B_k$.
What you are missing is to define $f$ on the rest of $V$: You have not yet agreed on a value of $f$ on $\{b_{k+1},\ldots,b_n\}$, and you have not yet agreed on a value of $f$ on whatever may remain in $V$ that is not in the span of the vectors $\{b_1,\ldots,b_n\}$. I suggest taking the value $0$ on this remaining part.
I know that this post is mostly words, but you can probably formalize it yourself.
A: 
Lets say that $dim(V)=m>n$ Therefore you can extend your list $B=(b_{1},b_{2},..,b_{n})$ to a base of the space V. Obtaining $B'=(b_{1},..,b_{n},b_{n+1},..,b_{m})$. There is a theorem that says that if you have a base $B'$ of $V$ and $C'=(c_{1},c_{2},..,c_{k},..,c_{m})$ Then there is a unique linear transformation $f:V \rightarrow W$ such that $f(b_{i})=f(c_{i}), \forall i$

To demonstrate what I have just said above, we proceed in the following way; given $v \in V$, since $B'=((b_{1},..,b_{n},..,b_{m})$ is a base, there exists unique $ \alpha_{1}, \alpha_{2},...,\alpha_{n}$ such that
$v=\sum_{i=1}^{n}{\alpha_{i}v_{i}}$. 
We define $f(v)=\sum_{i=1}^{n}{\alpha_{i}f(v_{i})}$
Observe that the $f$ is well defined for the uniqueness of the $\alpha_{i}$'s
I leave you as an exercise to prove that $f$ is a linear transformation. You can also prove that this linear transformation is unique, but I dont think you need that information for your problem. 
