What does this theorem in linear algebra actually mean? I've just began the study of linear transformations, and I'm still trying to grasp the concepts fully.
One theorem in my textbook is as follows:
Let $V$ and $W$ be vector spaces over the field $F$, and suppose that $(v_1, v_2, \ldots, v_n)$ is a basis for $V$. For $w_1, w_2, \ldots, w_n$ in $W$, there exists exactly one linear transformation $T: V \rightarrow W$ such that $T(v_i) = w_i$ for $i=1,2,\ldots,n$.
The author doesn't explain it, but gives the proof right away (which I understand). But I'm trying to figure out what this theorem actually states, and why it is so important? So in words it means: if I have a basis for my domain, and a basis for my codomain, then there exists just one linear transformation that links both of them.
So let's say I have a linear map $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, with $T(1,0) = (1,4)$ and $T(1,1)=(2,5)$. So because $(1,0)$ and $(1,1)$ is a basis for my domain, it is an implication of the theorem that $(1,4)$ and $(2,5)$ is automatically a basis for my codomain?
 A: The theorem says that any map from the finite set $\{v_1,\ldots,v_n\}$ to a vector space $W$ can be uniquely extended to a linear map $V\to W$; this is true if (and only if) $[v_1,\ldots,v_n]$ forms a basis of$~V$. It's importance is that it allows, at least in the case where $V,W$ are finite dimensional, any linear maps to be represented by finite information, namely by a matrix, and that every matrix so represents some linear map. In order to get there, we must also choose a basis in $W$; then by expressing each of the images $f(v_1),\ldots,f(v_n)$ in that basis, we find the columns of the matrix representing $f$ (with respect to $[v_1,\ldots,v_n]$ and the chosen basis of $W$). Note that this information only explicitly describes those $n~$images; the actual linear map is implicitly defined as its unique linear extension to all of$~V$. The existence part of the theorem ensures that we never need to worry whether there is actually a linear transformation that corresponds to a freely chosen matrix: one can always map $v_j$ to the vector represented by column$~j$, for all$~j$ at the same time.
It is only thanks to this theorem that we can work with matrices as if we work with the linear transformation they encode; as long as we fix our bases of $V$ and $W$, we have a bijection between linear transformations $V\to W$ on one hand and $m\times n$ matrices (where $m=\dim W$) on the other. In fact this bijection is itself linear, so an isomorphism of the $F$ vector spaces $\mathcal L(V,W)$ and $\operatorname{Mat}_{m,n}(F)$.
A: Here is a corollary of (the "uniqueness" part) of the theorem that reflects the way that it is often used:

Suppose that $T_1$ and $T_2$ are linear transformations from $V$ to $W$ such that $T_1(v_i) = T_2(v_i)$ for $i = 1,\dots,n$, where $(v_1,\dots,v_n)$.  Then $T_1 = T_2$.

It is common to ask whether two linear transformations are the same, and this theorem gives us a good way to check.
The existence part can be phrased like so:

For any basis $(v_1,\dots,v_n)$ of $V$ and set of vectors $w_1,\dots,w_n \in W$, we can construct a linear transformation such that $T(v_i) = w_i$.

I find that it is helpful to consider the existence and uniqueness aspects separately.
Note that $T(v_1),\dots,T(v_n)$ will generally not be a basis for the codomain.  For example, consider $T$ as given by $T(x,y) = (x,0)$. We know that $((1,0),(0,1))$ is a basis, but $(T(1,0),T(0,1))$ is not a basis.
In fact, we can deduce from this theorem that if there is a way to map the basis of the domain to the basis of the codomain, then the two vector spaces must be isomorphic.  That is, any two spaces with the same dimension are isomorphic.
A: No, because you don't even know the dimension of the codomain. You do get that $w_i$ are a spanning set for the image of $T$, whatever that is. This theorem as stated doesn't imply that, but it would follow from the proof. (Any vector $b$ in the image is mapped to from some vector $x$ in the domain; $x$ is some linear combination of the $v_i$; by linearity of $T$, $Tx$ is a linear combination of the $w_i$; so $b=Tx$ is a linear combination of the $w_i$.) 
Ultimately this theorem is simpler than all that, though. It just says that you can uniquely define a linear map by its action on a basis of the domain. 
A: the way $T$ is defined on a basis $\{v_1,v_2, \cdots, v_n\}$ and the image $\{Tv_1 = w_1, Tv_2=w_2, \cdots, Tv_n=w_n\}$ and made $T$ linear and defined for every vector by $$T(x_1v_1+x_2v_2+\cdots+x_nv_n = x_1w_1+x_2w_2+\cdots+x_nw_n$$ in all of this, nothing is said about the vectors $\{w_1, w_2, \cdots, w_n \}.$ all you got is a linear transformation $T \colon V \rightarrow W.$  for all we know all $w_j$'s could be the zero vector.
further constraints on $T,$ like $T$ is $1-1$ or onto will reflect constraints on $w_j$'s. these also show in terms of the $rank(T).$   
