Linear maps for $\Bbb{R}^n$ to $\Bbb{R}^m$? This question is related to:


*

*What is $\Bbb{R}^n$?

*The basis of a matrix representation
I am still confused about the topics in these questions and am going to ask another question that will hopefully clarify this for me.
Consider the following theorems:

Theorem 1
  
  
*
  
*Let $A$ be an $m\times n$ matrix with components in the field $\Bbb{F}$ then the map:
  
  
  $$\Bbb{F}^n\rightarrow \Bbb{F}^m$$
  $$x \mapsto  Ax$$
  is linear.

  
*Conversely, if $f:\Bbb{F}^n \rightarrow \Bbb{F}^m$ is linear , there exists a unique $m \times n$ matrix $A$  with components in the field $\Bbb{F}$  for which $f(x)=Ax$.
  
  
  Hence one can interoperate the $m \times n$ matrices as the linear map from $\Bbb{F}^n$ to $\Bbb{F}^n$ 

This was taken from "Linear Algebra" by Janich, Kalus (with minor changes).
Now consider the similar theorem from the same book. 

Theorem 2
Let $f:V\rightarrow W$ be a linear map between vector spaces over $\Bbb{F}$, and let $(v_1,...,v_n)$ and $(w_1,...,w_n)$ be bases for $V$ and $W$ respectively. Then the $m \times n$ matrix $B$ determined by the commutator diagram:
$$
\require{AMScd}
\begin{CD}
V @>{f}>> W\\
@VVV @VVV \\
\mathbb{F}^n @>{B}>> \mathbb{F}^n
\end{CD}
$$
  is called the matrix associated to $f$ relative to the two chosen bases.

The matrix $B$ is associated with the basis $v= (v_1,...,v_n)$ and $w=(w_1,...,w_n)$, and we could write $B$ in terms of another basis which will change its components and be representing $f$ in the new basis. It is thus not theorem 2 that I have a problem understanding. It is rather theorem 1.

Example
Consider the linear map $$f(
        \begin{bmatrix}
        x \\ y
        \end{bmatrix})=\begin{bmatrix}
        2x+y \\ y
        \end{bmatrix}$$
  Clearly here the vector  $$\begin{bmatrix}
        x \\ y
        \end{bmatrix}$$
is in $\Bbb{R}^n$ (taking $x,y\in \Bbb{R}$) and is not a coordinate vector, but an arbitrary vector in $\Bbb{R}^n$. We could write the linear map $f$ as follows:
  $$f(
        \begin{bmatrix}
        x \\ y
        \end{bmatrix})= \begin{pmatrix}
        2 & 1\\ 0&1
        \end{pmatrix}
\begin{bmatrix}
        x \\ y
        \end{bmatrix}
$$
  In this case the matrix $A$ in theorem 1 is given by:
  $$A=\begin{pmatrix}
        2 & 1\\ 0&1
        \end{pmatrix}$$

The thing that I am confused about is simply which bias the matrix $A$ is in as described above. My view (which goes against the answers in the linked questions) is that $A$ should not be associated with a basis as described above. The reasoning behind this is that $A$ performers exactly the same operation on a vector $x$ as $f$ does with $x$ not been a coordinate vector. If we however say that $A$ is in a basis (by in a basis I mean like the matrix $B$ in theorem 2, so that is represents $f$ with respect to one (or two, if the bases of the domain and codomain are different) bases). Then it follows that $x$ also has to be a coordinate vector in this same basis (the basis of the domain). But the linear map $f$ does not act on coordinate vectors, it is not associated with a basis and therefore $x$ is not a coordiante vector. So my conclusion is that:
The matrix $A$ is not a representation of $f$ in a particular basis but represents $f$ in general and performing $A$ on an arbitrary vector in $\Bbb{R}^n$ is equivalent to performing the linear map $f$ on it. And both $A$ operate on vectors in $\Bbb{R}^n$ which themselves are not coordinate vectors.
So the matrix $A$ is not the same as $B$ i.e. $B$ represents its linear map in with respect to a particular set of basis and acts on coordinate vectors only. Whilst the matrix $A$ is not related to a basis and acts on any vectors, as $f$ does in $\Bbb{R}^n$.
Please could you either confirm that this argument/analysis is right or wrong. I am really confused about this topic, so if you could give sources in you questions, this would be a great help. 
p.s. I note that this question is very similar to my other questions, I have however tried to make it slightly more detailed and asking a different question at the end. I have asked this one since I am still confused about the subject and want a detailed answer, preferably backed up by sources.
 A: The matrix $A$ in Theorem 1 is the matrix corresponding to the orthonormal base
$\{(1,0,\dots,0),(0,1,0,\dots,0),\dots(0,\dots,0,1)\}$.
A: I believe the main misunderstanding is the idea that a vector like $\begin{bmatrix}
        x \\ y
        \end{bmatrix}$ is not a coordinate vector, which in fact it is. Indeed we have the standard basis $\left\{{\begin{bmatrix} 
        1 \\ 0
        \end{bmatrix}}, \begin{bmatrix}
        0 \\ 1
        \end{bmatrix} \right\}$
of $\mathbb{R}^2$ and thus 
$$ \begin{bmatrix} x \\ y \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \end{bmatrix} + y\begin{bmatrix} 0 \\ 1 \end{bmatrix}. $$
Now any linear map on $\mathbb{R}^2$ like the map $f$ you defined is represented by a unique matrix when considering the standard basis.
More general, when we talk about $\mathbb{F}^n$ as a vector space we have already fixed a certain basis, the standard basis (the columns of the identity matrix). Through this choice the matrices representing linear maps become unique. It could be helpful to look at the special case of theorem 2 where we take $V = \mathbb{F}^n$ and $W = \mathbb{F}^m$ and the respective standard bases.
