Linear operator from F^n to F^n without reference to any basis Let $T:F^n \to F^m$ be a linear operator defined by $T(v)=Av$ where $A$ is any $m\times n$ matrix and $v$ is an element of $F^n$.
Does this transformation make any reference to some basis of $F^n$? I mean we are just taking any vector in $F^n$, which is an $n$-tuple of scalars in $F$ and then we get an output by left multiplication with $A$. I think there is no basis involved here.
But I read that if you fix a basis for the vector space then corresponding to every $m\times n$ matrix there is a linear transformation defined by taking the columns of the matrix as images of the basis vectors. Conversely, given any linear transformation we can construct an $m\times n$ matrix.
So for the above arbitrary transformation, is there any basis involved? Do we assume that the basis is the standard basis?
 A: The notation $T : F^n \to F^n$ suggests a basis by convention, namely the "standard" basis that comes from the Cartesian product structure on $F^n$: $(1,0,0,…,0)$, $(0,1,0,…,0)$, etc. So for a linear operator on $F^n$ it does make sense to define it with a matrix, using the standard basis by convention. 
If instead your first sentence were replaced by "Let $V$ be a vector space and let $T : V \to V$ be a linear operator" then it would make no sense to define $T$ by a matrix. Not even convention let's you distinguish one basis of $V$ from any other.
A: A linear map $T\colon V\to W$ ($V$ and $W$ are finitely generated $F$-vector spaces) can be represented by a matrix once a basis $\mathscr{B}$ and a basis $\mathscr{D}$ on $W$ are selected. The representing matrix is the unique $m\times n$ matrix $A$ such that
$$
C_{\mathscr{D}}(T(v))=AC_{\mathscr{B}}(v),\qquad\text{for all $v\in V$}
$$
where $C_{\mathscr{B}}\colon V\to F^n$ is the “coordinate map”:
$$
C_{\mathscr{B}}(v)=\begin{bmatrix}\alpha_1\\\alpha_2\\\vdots\\\alpha_n\end{bmatrix}
\quad\text{if and only if}\quad
v=\alpha_1v_1+\alpha_2v_2+\dots+\alpha_nv_n
$$
Notation I assume $n=\dim V$, $m=\dim W$ and $\mathcal{B}=\{v_1;v_2;\dots;v_n\}$
Knowing the representing matrix allows making computations about $T$, which of course may also be possible with just $T$. However, representing matrices allow for standard techniques such as LU decomposition, Gaussian elimination and so on. However, the distinction between $T$ and its representing matrix should always be kept in mind.
In the special case when $V=F^n$ there is a “canonical” or standard basis, which can make ideas a bit confused. But there's a way out. The standard basis $\mathscr{E}_n$ on $F^n$ has the peculiar property that, for all $v\in F^n$,
$$
C_{\mathscr{E}_n}(v)=v.
$$
Therefore, if we have $V=F^n$ and $W=F^m$ we can select the standard bases on the domain and codomain. If $A$ is the representing matrix, the relation
$$
C_{\mathscr{E}_m}(T(v))=AC_{\mathscr{E}_n}(v),\qquad\text{for all $v\in F^n$}
$$
boils down to
$$
T(v)=Av,\qquad\text{for all $v\in F^n$}
$$
because $C_{\mathscr{E}_m}(w)=w$ for all $w\in F^m$ and $C_{\mathscr{E}_n}(v)=v$ for all $v\in F^n$.
The conclusion is that any linear map $T\colon F^n\to F^m$ is of the form $T(v)=Av$ for a unique $m\times n$ matrix $A$. So your starting point could not be really different. The point is that the spaces of the form $F^n$ ($n\ge1$) are special, because they're the only vector spaces where a basis can be selected with the property that the coordinate map is the identity.
A: As soon as you specify a matrix by components, you have to specify the basis for both domain and codomain. A matrix already implies selection of a basis. Even if you say standard basis, you still have to know the meaning of the principal axes (is the first component meant to be north? south? Along polynomial $f(x)=x^2$? Ground state of hydrogen atom? You name it). The effect of the operator itself is unaware of the basis, and you can specify it as such. Typical examples are expressions that involve vectors (pure vectors without being written out in components), integral and derivative expressions for functional operators, or expressions involving commutators.
For instance, you can wrate a projection operator as $T: \vec{x}\mapsto \vec{v}(\vec{v}\cdot \vec{x})$ which is basis-independent. Even if you specify $T=\vec{v}\otimes \vec{v}$ it still doesn't matter which basis you use, until you express $\vec{v}$ in some basis and write out components.
