Basis of Domain and Existence of Unique Linear Map Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_n\in W$ where $V,W$ are subspaces of a finite-dimensional vector space over a generic field, $F$. Then there exists a unique linear map $T:V\rightarrow W$ such that
$$Tv_j=w_j\quad \text{for each }j=1,...,n.$$
I understood the proof of this claim.  
$\textbf{My question:}$ What is the significance of this claim? Starting with a basis of the domain, $V$, and landing with the existence of a unique linear map sounds nice, but can someone spell out why this may be of importance?
Reference:
Axler, Sheldon J. $\textit{Linear Algebra Done Right}$, New York: Springer, 2015.
 A: If you want to define a linear map $f \colon V \rightarrow W$ you can do that by just specifying the images of a basis of $V$. This means you get the linearity for free (so you do not need to worry about that) and for many constructions it is very useful too have this kind of control over linear maps. For example, one can show that every subspace $U \subset V$ arises as the kernel of a linear map. How does one prove that? Well... choose a basis of that subspace and extend it to a basis of $V$. Define a linear map by sending the basis of $U$ to $0_W$ and make sure that the rest is not in the kernel (for example by mapping the other basis vectors to themselves). 
Another nice thing is that this property often helps with counting linear maps with additional properties. For example you can ask yourself the following question:
Are there none, one or multiple linear maps $f \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$, such that $f((1,1)) = (2,3)$ and $f((1,0)) = (1,0)$? 
Now you can answer that immediately since the vectors whose image we specified are a basis of $\mathbb{R}^2$. 
There is also another way of defining linear maps by matrices. Actually, matrices correspond to linear maps once you have chosen bases for the involved vector spaces. To get this correspondence you also meed your statement to get the uniqueness of the matrix.
