Definition of operation of matrix representation 
Let V and w be finite-dimensional vector spaces with ordered bases $\beta={v_1,v_2,...,v_n}$ and $\gamma={w_1,w_2,...,w_m}$. Let $T:V \rightarrow$ be linear. Then for each j, $1 \leq j \leq n$, there exist unique scalars $a_{ij} \in F$, $1 \leq I \leq m$, such that $T(v_j)= \sum_{i=1}^m a_{ij}w_i$ for $1 \leq j \leq n$.

I don't get the operation of this one. $T(v_j)= \sum_{i=1}^m a_{ij}w_i$ Are we looking for the $j^{th}$ column of $w_j$? 
Expanding, get $T(v_j)$=$a_{1j}w_1 + a_{2j}w_2+...+a_{mj}w_m$. I specifically don't understand the matrix operation here.
 A: I think I know where you're confused. However...
When a vector space is defined in an abstract way as this, you cannot say $j$th column without constructing specific coordinate system.
So since $T:V\to W$, $T(v_j)$ refers to some vector in $W$, i.e., $T(v_j)\in W$. And $\sum_{i=1}^m a_{ij} w_i \in W$ is defined by the vector addition and scalar-multiplication.
However, if it helps you understand the abstract vector space, you can definitely think of $V$ and $W$ as subspaces of $\newcommand{\reals}{{\mathbf R}}\reals^p$ for some $p$.
A: I think the problem is not properly stated. Since for any transformation $T:V\to W$ (whether or not it's linear), there are uniquely determined coefficients $a_{i,j}\in\mathcal{F}$ such that
\begin{equation}
T(v_j) = \sum_{i=1}^m a_{ij} w_i
\end{equation}
by the definition of basis.
I think a right problem can be the one below.

Let $V$ and $W$ be a finite-dimensional vector spaces with bases $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$ respectively. Now let $T:V\to W$ be a linear transformation. Then prove that there exists $\newcommand{\reals}{{\mathbf R}}A\in\mathcal{F}^{m \times n}$ such that for any $x_1,\ldots,x_n\in\mathcal{F}$ and $y_1,\ldots,y_m\in\mathcal{F}$ that satisfies
  \begin{equation}
T(x_1 v_1 + \cdots + x_n v_n) = y_1 w_1 + \cdots + y_m w_m,
\end{equation}
  it is satisfied that $y = Ax$.
  Prove also that this matrix $A\in\mathcal{F}^{m\times n}$ is uniquely determined!

Note that if $V\in\reals^n$, $W\in\reals^m$, and $v_j$s and $w_i$s are the standard unit vectors, $A\in\reals^{m\times n}$ is nothing but the matrix which defines the usual linear transformation!
