Matrices representing the same linear transformation Suppose $T:V\rightarrow V$ is a linear operator. Let a basis for $V$ be $B_1=\{e_1,e_2,\cdots,e_n\}$ . 

Let $A$ be the matrix of $T$ relative to this basis.

This means : $T(e_k) = \sum_{i=1}^n a_{ik}~e_i$.
Now, my textbook says : Let us define two more matrices :

$E = [e_1,\cdots,e_n]$ and $E~' = [T(e_1), \cdots,T(e_n)]$

and then concludes that $E~'=EA$.
So, my question is : Aren't $A$ and $E'$ actually the same matrices? $A$ is obtained in the same manner : by taking the linear transformation of each column and writing it down in terms of the basis.
Could anyone please clear this confusion of mine.
Thank you very much for your help in this regard.
 A: Maybe an example would help sort things out. Consider the linear transformation $T(f(x)) = f(x+1)$ on the space of real polynomials of degree at most 3. In the basis $(1,x,x^2,x^3)$ we have
$$ T(1) = 1, $$
$$ T(x) = x+1, $$
$$ T(x^2) = (x+1)^2 = x^2 + 2x + 1, $$
$$ T(x^3) = (x+1)^3 = x^3 + 3x^2 + 3x + 1. $$
Hence, in this basis, the matrix of $T$ is
$$ A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$
so that, writing $(e_1,e_2,e_3,e_4) = (1,x,x^2,x^3)$ to match your notation, we have $E = \begin{pmatrix} 1 & x & x^2 & x^3 \end{pmatrix}$ and
\begin{align}
 EA &= \begin{pmatrix} 1 & x & x^2 & x^3 \end{pmatrix}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{pmatrix} \\
 &= \begin{pmatrix} 1 && 1 + x && 1 + 2x + x^2 && 1 + 3x + 3x^2 + x^3 \end{pmatrix} \\
 &= \begin{pmatrix} 1 && (x+1) && (x+1)^2 && (x+1)^3 \end{pmatrix} \\
 &= \begin{pmatrix} T(1) & T(x) & T(x^2) & T(x^3) \end{pmatrix} = E'
\end{align}
The difference is that $E'$ is not an $n\times n$ matrix where you make the columns by picking off the coefficients of each of the $T(e_i)$, but instead each entry really is just $T(e_i)$.
A: It is  convenient to have operators act on vectors from the left, while scalars act on the vectors from the right. Take a linear operator $T$, $(e_1, \ldots, e_n)$ a basis. You express the images $Te_i$ in terms of this basis with the help of a matrix $A$, the matrix of the operator $T$ corresponding to base $(e_1, \ldots, e_n)$. The defining equation for $A$ is
$$(Te_1, \ldots, Te_n) = (e_1, \ldots, e_n) \cdot A$$
Let's see what happens to a vector $v$. Write $v$ in the basis $(e_1, \ldots, e_n)$. But write this in an efficient way:
$$v = (e_1, \ldots, e_n) \cdot (x_1, \ldots x_n)^{t}$$
Apply $T$ and use that $T$ is linear ( yea, $T$ works from the left). We get
$$Tv = (Te_1, \ldots, Te_n)\cdot (x_1, \ldots x_n)^{t}$$
and using the previous equality:
$$Tv =((e_1, \ldots, e_n)\cdot A )\cdot(x_1, \ldots x_n)^{t}$$
or
$$Tv =(e_1, \ldots, e_n)\cdot (A \cdot(x_1, \ldots x_n)^{t})$$
Therefore, the coordinates for the image $v' = Tv$ are 
$$(x_1', \ldots x_n')^{t} = A (x_1, \ldots x_n)^{t}$$
that is
$$\left (\begin{array}{c} x'_1 \\x'_2\\ \ldots\\ x'_n\end{array} \right) = A \left (\begin{array}{c} x_1 \\x_2\\ \ldots\\ x_n\end{array} \right)$$
