Linear maps that are matrices If I have the linear map $A:\Bbb{R}^3\rightarrow \Bbb{R}^3$ where $A$ is a matrix. Is the matrix $A$  (along with the vectors it operates on) in a basis or not? I think it is not, since the vectors it operators on are not coordinate vectors but rather 'normal' column vectors in   $\Bbb{R}^3$. I do, however, note that if I am correct in my suggestion that it is not in a particular basis, then if we where to represent $A$ in the basis $(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)$ then the matrix representing $A$ would have the same components as $A$ but act on coordinate vectors rather then the 'normal' column 
vectors. Am I correct?
EDIT:
I will try to make my point clearer.
If we have a linear map $f:V\rightarrow W$ we can represent this map with a matrix $A$ (sorry I have changed notation here, $A$ now no longer represents the linear map). This can b represented on the following  commutative diagram:

We have coordinate vectors in $F^n$ and $F^m$ (these represent the proper vectors in a certain basis) and proper vectors  are in $V$ and $W$. The maps $g_1$ and $g_2$ are coordinate maps.
$A$ is the representing matrix of $f$ in a particular bias (or two bases, the basis of $F^n$ and $F^m$).
$f$ is not associated with a basis for a normal linear map.
My question is simply if $f$ is itself a matrix, then surly the above comment about not been associated with a still holds?
Example
The linear map $B:\Bbb{R}^2\rightarrow \Bbb{R}^2$ is defined by the matrix
$$  B= \begin{pmatrix}
        1& 0  \\
        0 & -2 \\ 
        \end{pmatrix} $$
We can represent this linear map in terms of the basis $((1,2)^T,(-1,1)^T)$ as follows:
$$  B'= \begin{pmatrix}
        -1& -1  \\
        -2 & 0 \\ 
        \end{pmatrix} $$
So it is clear that $B'$ has an associated basis, but I don't think the matrix $B$ does?
 A: A matrix is just a bucket of numbers. It's not a linear map until you pick a basis for the domain and codomain vector spaces (at which point every linear map can be written as a matrix in that basis, and every matrix represents a linear map). Sometimes we write down $3\times 3$ matrices and call them linear maps from $\mathbb{R}^3$ to itself without explicitly naming the basis; when we do so we are implicitly picking the Cartesian basis.
So yes, if you interpret a matrix to be a linear map, you must also have a basis (or two bases) in mind. That's the only way the matrix can act on elements of the vector space.
EDIT:
If you have vector spaces $V_1$ and $V_2$, you can find a basis $v_1, \ldots, v_n$ for $V_1$ and $w_1, \ldots, w_m$ for $V_2$, and then can represent every element of $V_1$ using $n$ numbers $\alpha_1, \ldots, \alpha_n$ as the sum $\sum_i \alpha_i v_i$, and similarly for $V_2$ and $m$ numbers $\beta_1, \ldots, \beta_m$.
A linear map, written in the bases $\{a_i\}$ and $\{b_i\}$, acts on the coefficients $\alpha_i$ and transforms them into coefficients $\beta_i$. If you pick different bases, you will get different matrices even for the same linear map.
Any good linear algebra textbook should cover this concept in much greater detail and with many examples.
