Finding Matrix Representation of Linear Operator 
Definition: Using the notation above, we call the $m \times n$ matrix $A$ defined by $A_{ij} = a_{ij}$ the matrix representation of $T$ in the ordered bases $\beta$ and $\gamma$.

The "notation above" to which the author refers is $T(v_j) = \sum_{i=1}^m a_{ij} w_i$. The author then remarks that the $j$-th column of $A$ is simply $[T(v_j)]_{\gamma}$ 
Here is one problem I am having trouble with: 
The question is: Define $T: M_2(\Bbb{R}) \to M_2(\Bbb{R})$ by $T(A) = A^T$. Compute $[T]_{\alpha}$. 
From my understanding, $T(E_{11}) = T \left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \right) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $T(E_{12}) = T \left(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right) = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, etc. And so the first column is $T(E_{11})$, the second column $T(E_{12})$, etc. Hence, the matrix representation should be
$$[T]_{\alpha} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ 
But evidently the answer is
$$[T]_{\alpha} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}$$
Why is this the case? Moreover, how could either answer be an operator on $M_2(\Bbb{R})$ when the dimensions of the matrix are wrong? 
Here is a related problem. Let $T : P_3(\Bbb{R}) \to P_2(\Bbb{R})$ by defined by $T(f(x)) = f'(x)$. According to my book, the matrix representation is 
$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ \end{bmatrix}$$ 
but why wouldn't it be
$$\begin{bmatrix} 0 & 1 & 2x & 3x^2 \end{bmatrix}~~?$$
After all, $T(x^3)$ literally equals $3x^2$ and so should be the fourth column. Again, how could this matrix operate on $P_3(\Bbb{R})$. 
In each problem, is the author implicitly using isomorphisms between $M_2(\Bbb{R})$ and $\Bbb{R}^4$, and $P_3(\Bbb{R})$ and $\Bbb{R}^4$? If so, this would be fallacious since isomorphisms haven't been introduced. 
 A: A matrix representation for a linear map describes how the transformation acts in the coordinate space (what you think as an implicit isomorphism is simply the definition). 
If we fix a basis $u_1,u_2,\ldots,u_n$ of $U$ and a basis $v_1,\ldots,v_m$ of $V$ then a linear map $T\colon U\to V$ can be described as $y=[T]_ax$ where $x$ is the coordinate vector for $u$ and $y$ is the coordinate vector for the image of $u$, i.e.
$$
u=\color{red}{x_1}u_1+\ldots+\color{red}{x_n}u_n,\quad Tu=\color{blue}{y_1}v_1+\ldots+\color{blue}{y_m}v_m.
$$
Coordinate vectors can be organized as we like, for example, as a list $\color{red}{(x_1,x_2,\ldots,x_n)}$, but often it is convenient to set them up as a column vector in order to be able to refer to $[T]_a\color{red}x$ as a matrix multiplication. Thus, a matrix representation will be an $m\times n$ matrix. 
Now in your example $T(A)=A^T$. We need to get the coordinate vector for $A$:
$$
A=\begin{bmatrix}a & b\\c & d\end{bmatrix}=\color{red}a\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}+\color{red}b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}+\color{red}c\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}+\color{red}d\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix},
$$
that is
$$
\color{red}{x=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}},
$$
and similarly the coordinate vector for $A^T$
$$
A^T=\begin{bmatrix}a & c\\b & d\end{bmatrix}=\color{blue}a\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}+\color{blue}c\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}+\color{blue}b\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}+\color{blue}d\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix},
$$
that is
$$
\color{blue}{y=\begin{bmatrix}a\\c\\b\\d\end{bmatrix}}.
$$
Therefore, the matrix representation $[T]_a$ is the $4\times 4$ matrix that $\color{blue}y=[T]_a\color{red}x$. Note that the order of the basic vectors must be fixed. 
The more systematic way to build $[T]_a$ is to map the basic vectors and organize the result column-wise. For example,
$$
T(E_{11})=E_{11}=\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}=\color{blue}1\cdot E_{11}+\color{blue}0\cdot E_{12}+\color{blue}0\cdot E_{21}+\color{blue}0\cdot E_{22},
$$
hence,
$$
[T]_a=\begin{bmatrix}\color{blue}1 & ? & ? & ?\\
\color{blue}0 & ? & ? & ?\\
\color{blue}0 & ? & ? & ?\\
\color{blue}0 & ? & ? & ?
\end{bmatrix},
$$
and so on.
