Dependency of linear map definition on basis The definition of linear map depends on the basis. Is this a flaw of the "linear map" construct?
 A: It is instructive to reflect on the fact that the linear map is independent of basis, though the matrix which represents a linear map will change according to which basis you use. What allows you to pass from the world of matrices to the world of linear maps is precisely a choice of basis.
A: A map $F:\mathcal{V}\rightarrow\mathcal{W}$ is linear iff $F(\alpha x+\beta y) = \alpha F(x) + \beta F(y)$ for scalars $\alpha,\beta$ and vectors $x,y\in\mathcal{V}$. It does in no way depend on any basis. What depends on the basis (or bases if $\mathcal{V}\neq\mathcal{W}$) is the matrix which represents the linear maping w.r.t. to such a basis/bases.
That is, if $V=[v_1,\ldots,v_n]$ and $W=[w_1,\ldots,w_m]$ are the bases of $\mathcal{V}$ and $\mathcal{W}$, respectively, and $x\in\mathcal{V}$ with coordinates $\xi_1,\ldots,\xi_n$ in the basis $V$ ($x=\sum_{i=1}^n\xi_iv_i$) then $y=F(x)$ can be written as:
$$
y=F(x)=F\left(\sum_{j=1}^n\xi_jv_j\right)=\sum_{j=1}^n\xi_jF(v_j).
$$
Since $F(v_j)\in\mathcal{W}$, it can be uniquely expressed in terms of the basis $W$:
$$
F(v_j) = \sum_{i=1}^m\phi_{ij}w_i.
$$
Putting it together gives
$$
y=\sum_{j=1}^n\xi_jF(v_j)
=\sum_{j=1}^n\xi_j\sum_{i=1}^m\phi_{ij}w_i
=\sum_{i=1}^m\left(\sum_{j=1}^n\phi_{ij}\xi_j\right)w_i
=\sum_{i=1}^m\eta_iw_i,
$$
where
$$
\eta_i=\sum_{j=1}^m\phi_{ij}\xi_j
$$
is the coordinate of $y$ w.r.t. the basis $W$.
The formula for $\eta_i$ is a well-known formula for matrix-vector multiplication.
If $\mathbf{x}=[\xi_1,\ldots,\xi_n]^T$, $\mathbf{y}=[\eta_1,\ldots,\eta_m]^T$,
and the $i,j$  component of a matrix $\mathbf{F}$ is $\phi_{ij}$, then the relation between these three guys is simply
$$
\mathbf{y}=\mathbf{F}\mathbf{x}.
$$
The matrix $\mathbf{F}$ is the representation of the linear map $F$ w.r.t. to bases $V$ and $W$ of $\mathcal{V}$ and $\mathcal{W}$, respectively, which does depend on the bases, the linear map itself does not.
A: A matrix representation (if exists) of any linear map depends on a basis but the definition of linear map is independent of basis. Not every linear map can even be represented with a matrix. As an illustration, for two vector spaces over the field of reals $\mathbb{R}$ consider the linear mapping, $$f:\mathbb{R}^3\rightarrow\mathcal{P}$$ 
where $\mathcal{P}$ is the set of all real quadratic expressions with indeterminate $x$. For example $\left(2, 3, 4\right)$ in $\mathbb{R}^3$ is mapped to $2x^2+3x+4$. This linear map can't be represented as a matrix formed by elements of the underlying field $\mathbb{R}$. So think of matrices as easy representations of certain linear maps but don't think of a general linear map as a matrix. 
