Prove that there exists a unique linear map that extends the map from a basis

Question

My professor has stated the following lemma:

Let $U$ and $V$ be two vector spaces and $B$ a basis for $U$. If $f:B\rightarrow V$ is a map, then there exists a unique linear map $A: U \rightarrow V$ such that $A(x)=f(x)\ \forall\ x\in B$

How can I prove this lemma? (This is not a homework problem.)

Answer 1

Hint: Using the definition of basis, you know any $u \in U$ can be written as a unique linear combination of elements of $B$. Given that you already have a map defined on $B$, does this give you a way to define some (linear!) function on $V$?

That is: You have $f(b_i)$ defined for $b_i \in B$. How would you define $f(\sum \alpha_ib_i)$?
(Keep in mind that you need linearity. So this would actually be forced.)

Why will this be well-defined?

Once you've defined this function, can you prove that this is indeed linear? Why must it be unique? (What if you extended $f$ in another way? Can two different linear functions agree on all basis elements?)