Does a linear map change the basis of the target vector-space? If no: Is a linear map clearly defined just by how it transforms the basis vectors? I am learning linear algebra on my own (I only have high school experience). I am not able to find an answer to this seemingly easy question - perhaps because I lack the mathematical lingo to correctly express myself.
If a linear map is defined as V -> W between two vector spaces, are the transformed basis vectors of V also basis vectors of W?
If the answer is yes: could we uniquely define a specific linear map V -> W just by how it transforms the basis vectors of V to the basis vectors of W? Because any vector in the W vector-space would just be a linear combination of these (transformed) basis vectors?
 A: Let $V$ and $W$ be vector spaces, let $B$ be a subset of $V$, and let $f\colon B \to W$ be a function (of sets, ignoring any vector space structure). Then:

*

*If $B$ spans $V$, then there is at most one linear map $T\colon V \to W$ such that $T(b) = f(b)$ for all $b \in B$. (However, if $B$ isn't linearly independent, there might be no such linear map, because the elements of $B$ might satisfy a linear relation that the elements $f(b) \in W$ don't satisfy.)

*If $B$ is linearly independent, then there is at least one linear map $T\colon V \to W$ such that $T(b) = f(b)$ for all $b \in B$. (However, if $B$ doesn't span $V$, there is more than one such linear map, because you can extend $B$ to a basis and make many choices for where to send the other basis vectors.)

*Putting the above two facts together: If $B$ is a basis, then there is a unique linear map $T\colon V \to W$ such that $T(b) = f(b)$ for all $b \in B$.

In other words, given a basis of a vector space, there's a one-to-one correspondence between maps from the basis to another vector space and linear maps to another vector space; to define a linear map, it suffices to define it on a basis. (This is essentially what the matrix representation of a linear map is: you're just taking the image of each vector in a chosen basis of the domain, representing it as a list of numbers via a chosen basis of the codomain, and assembling these lists of numbers into a rectangular array.)
However, given a basis $B$ of $V$ and a linear map $T\colon V \to W$, the image $T(B) = \{T(b) : b \in B\}$ does not have to be linearly independent, nor does it have to span $W$. For example, if you take the map $(x, y, z) \mapsto (x + y, x + y) : \mathbb{R}^3 \to \mathbb{R}^2$, the image of the standard basis $\{(1, 0, 0), (0, 1, 0), (0,0,1)\}$ under this map is $\{(2, 2), (2, 2), (0, 0)\}$, which is neither linearly independent nor spans $\mathbb{R}^2$.
The key here is that these are related to the notions of surjective (onto), injective (one-to-one), and bijective (one-to-one and onto) functions:

*

*A linear map $T\colon V \to W$ is surjective if and only if, for some basis $B$ of $V$, the image set $T(B)$ spans $W$.

*$T$ is injective if and only if, for some basis $B$ of $V$, the image set $T(B)$ is linearly independent.

*Putting the above two together: $T$ is bijective if and only if, for some basis $B$ of $V$, the image set $T(B)$ is a basis of $W$.

Also, you can replace "for some basis" with "for every basis" and the above will still be true (the choice of basis doesn't matter here).
I recommend working out examples of all of these (and drawing pictures, say in 2 or 3 dimensions, if possible), this will help a lot with the intuition.
A: I'm not sure exactly what you are asking in the first part of your question, so I'll address the second part first.
It is true that that a linear transformation is completely determined by its action on a basis.  If the transformation is non-singular, then the images of the basis vectors will be linearly independent.  They may or may not form a basis for $W$.  If the dimension of $W$ is equal to the dimension of $V$ they will, but if it is greater, they will not.  (It's not possible to have a non-singular transformation to a space of smaller dimension.)
If the transformation isn't non-singular, then the images of the basis vectors won't be linearly independent, so they certainly won't form a basis.  If this is what you are asking in the first part, then the answer is "No".  In particular, the $0$ transformation maps everything to the $0$ vector, which isn't an element of any basis.
