<s>[INST] 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.
 [/INST] 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 $T(e_i)$.
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