Why do all linear transformations have no restrictions on their natural domains? Some normal functions have restrictions on their natural domains, but linear functions don't. Why?
Related: Is there a linear transformation who domain isn't all of $\mathbb{R}^n$?
Why my question is not a duplicate of the above question: The above question asks if there asks if there is any function whose domain isn't $\mathbb{R}^n$. This function clearly exists by limiting the domain of any linear transformation. My question is asking about the natural domain of a linear transformation, which none of the answers from the question above address.
Natural Domain: The largest domain where the transformation makes sense. In other words, the domain of a linear transformation, without any artificial "restrictions" put on it.
 A: Here's an attempt to understand the question. 
Imagine that the natural domain of the linear transformation $T$ is some proper subset $A$ of ${\bf R}^n$. By linearity, $T$ extends to the span of $A$, and is linear on that span, so we may assume $A$ is a subspace of ${\bf R}^n$. Now you can define $T$ to be zero on the complement of $A$, and you will have a linear transformation that extends your original to all of ${\bf R}^n$. And that's why there is no restriction on the natural domain of a linear transformation.  
EDIT: As Anthony points out in the comments, I got the extension of $T$ wrong. Extend the basis $\{\,v_1,v_2,\dots,v_r\,\}$ of $A$ to a basis $\{\,v_1,v_2,\dots,v_r,v_{r+1},\dots,v_n\,\}$ of ${\bf R}^n$, define $T(v_j)=0$ for $r+1\le j\le n$, and then extend $T$ to all of ${\bf R}^n$ by linearity. 
A: There are frequent cases where one wants to define a linear function only on a subspace of a vector space. Although any linear map on a subspace can be extended to the whole space, there will in general be many ways to do it, and none of the choices might be better than any others. Also if infinite dimensional spaces are involved the existence of the extension might depend on the axiom of choice.
An archetypical example of the situtation is the follwing. Let $f:V\to W$ be an injective but not surjective linear map, then $f$ induces an isomorphism $\newcommand\im{\operatorname{Im}}\tilde f:V\to\im(f)\subset W$. The inverse isomorphism $\tilde f^{-1}:\operatorname{Im}(f)\to V$ is naturally defined only on $\im(f)$; it could be extended to $W$ but not uniquely so, and the extension would no longer be an isomorphism. For a case where the existence of an extension requires the axiom of choice, take this example with $f$ the inclusion $\Bbb Q\to\Bbb R$ as a linear map of $\Bbb Q$-vector spaces.
For an example of a practical use of such partially defined linear maps, look at the proof of the rank-nullity theorem I gave here. There $\varphi$ is not injective, but its section (right-inverse) $g$ is defined only on$~\im(\varphi)$, and it should not be extended to $K^m$ for the purpose of the proof.
