Understading wikipedia explanation of tranformation matrix I been staring at this for 2 hours trying to understand the last step, I can't figure out what they mean when they put the $\vec e_i$ to the left of the matrix.
Vector v can be represented in basis vectors, $ E = [\vec e_1 \vec e_2 \ldots \vec e_n]$ with coordinates  $[v]_E = [v_1 v_2 \ldots v_n]$ : 
$$\vec v = v_1 \vec e_1 + v_2 \vec e_2 + \cdots + v_n \vec e_n = \sum v_i \vec e_i = E [v]_E$$
$$A(\vec v) = A \left( \sum {v_i \vec e_i} \right) = \sum {v_i A(\vec e_i)} = [A(\vec e_1) A(\vec e_2) \ldots A(\vec e_n)] [v]_E =\; A \cdot [v]_E = [\vec e_1 \vec e_2 \ldots \vec e_n]
 \begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\
\vdots &  \vdots &  \ddots &  \vdots \\
a_{n,1} & a_{n,2} & \ldots & a_{n,n} 
\end{bmatrix}
\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}  $$
As far as I can tell this $ E = [\vec e_1 \vec e_2 \ldots \vec e_n]$ is just the matrix with basis vectors as columns. But that does not make sense to me. They finish everthing with:

The $a_{i,j}$ elements of matrix A are determined for a given basis E by applying A to every $\vec e_j = [0 0 \ldots (v_j=1) \ldots 0]^T$.

( what is $v_j$ doing inside this "thing"?) and then ending with: 

And observing the response vector A $\vec e_j = a_{1,j} \vec e_1 + a_{2,j} \vec e_2 + \cdots + a_{n,j} \vec e_n = \sum a_{i,j} \vec e_i$. This equation defines the wanted elements, $a_{i,j}$, of j-th column of the matrix A.

For more info, https://en.wikipedia.org/wiki/Transformation_matrix#Finding_the_matrix_of_a_transformation
 A: Am I right in assuming that this is the step you're having trouble with? $$A \cdot [v]_E = [\vec e_1 \vec e_2 \ldots \vec e_n]
 \begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\
\vdots &  \vdots &  \ddots &  \vdots \\
a_{n,1} & a_{n,2} & \ldots & a_{n,n} 
\end{bmatrix}
\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}$$
Obviously $[v]_E = \begin{bmatrix}\vec e_1 & \vec e_2 & \ldots & \vec e_n\end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}$, right?  Try multiplying it out if you don't understand this part.
That means that we just need to prove that $$\begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\
\vdots &  \vdots &  \ddots &  \vdots \\
a_{n,1} & a_{n,2} & \ldots & a_{n,n} 
\end{bmatrix}\left(\begin{bmatrix}\vec e_1 & \vec e_2 & \ldots & \vec e_n\end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}\right) = [\vec e_1 \vec e_2 \ldots \vec e_n]
 \begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\
\vdots &  \vdots &  \ddots &  \vdots \\
a_{n,1} & a_{n,2} & \ldots & a_{n,n} 
\end{bmatrix}
\begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n\end{bmatrix}$$  I.e. that the block matrix of basis vectors commutes with the matrix $A$.

The line 

The $a_{i,j}$ elements of matrix A are determined for a given basis E by applying A to every $\vec e_j = [0 0 \ldots (v_j=1) \ldots 0]^T$.

tells us that the vectors $e_i$ are also expressed with respect to the basis $E$ (I'm not at all sure why they labeled the $j$th component of $e_j$ as $v_j$ -- that seems to be a typo, but otherwise it's pretty clear what they meant).  So because $e_1 = 1e_1 + 0e_2 + \cdots + 0e_n$, we see that $[e_1]_E = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0\end{bmatrix}$ and likewise $[e_2]_E = \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0\end{bmatrix}$, etc.
But then the matrix $\begin{bmatrix} e_1 & e_2 & \cdots & e_n\end{bmatrix}$ is just the identity matrix.  And the identity matrix commutes with every other $n\times n$ matrix.  So of course $A(Iv) = IAv$.
A: Interpret the $[\vec e_1 \vec e_2 \ldots \vec e_n]$ not as a matrix, but as a row of "arrows". Each vector $\vec e_i$ is just an arrow in space.
Now when you multiply this "line" of "arrows" to a "column" of numbers you get:
$x_1 \vec e_1  + x_2 \vec e_2 + \cdots + x_n \vec e_n$ - some big "arrow", which is the result vector.
