TITLE: Linear Maps and Basis of Domain
QUESTION [6 upvotes]: I don't understand the theorem from Axler's Linear Algebra Done Right:
Suppose $v_1,v_2,...,v_n$ is a basis of $V$ and $w_1,....,w_n \in W$. Then there exists a unique linear map $T: V \to W$ such that $Tv_j = w_j$ for $j = 1,...,n$
What is this trying to say? I don't understand.
In addition, I don't understand the proof of this by construction. How does this show it exists, and how does the uniqueness proof indeed show that it is unique?

REPLY [3 votes]: It says that once you know how $T$ acts on a basis, you know how it acts on ALL vectors $v\in V$. To see this, suppose we have defined $T$ on , $\left \{ v_{1}, v_{2},v_{3}\cdots v_{n}\right \}$, a basis for $V$:
$Tv_{1}=w_{1}, Tv_{2}=w_{2}, \cdots ,Tv_{n}=w_{n}$
We can express $v$ as a linear combnation of the basis vectors, by writing
$v=a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}\cdots a_{n}v_{n}$. 
Now apply $T$:
$Tv=T(a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}\cdots a_{n}v_{n})$. 
But $T$ is linear so we get
$Tv=a_{1}Tv_{1}+a_{2}Tv_{2}+a_{3}Tv_{3}\cdots a_{n}Tv_{n}$. 
So the effect of $T$ on our arbitrary $v$ only depended on how we defined $T$ on the basis $\left \{ v_{1}, v_{2},v_{3}\cdots v_{n}\right \}$