Representative matrix of an operator I've got this course assignment which instructs me to find a representative matrix of an operator T with regard to the standard basis. The vector space is the set of $2\times2$ matrices over the field of real numbers. T is given as:
$$T(A) = MA \qquad \text{in which} \qquad M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
What is the meaning of the term representative matrix in this context? Isn't it M?
IDK if that helps, but there is actually a posted answer which is a $4\times4$ matrix, which is a block matrix constructed from the products of M with each of the standard basis vectors. It makes zero sense to me.
Thanks
 A: Do you agree that a 2x2-matrix is the representative matrix of an operator on a vector space of dimention 2 (say, $\mathbb{R}^2$) ?
You operator T acts on the vector space of all 2x2-matrix. Do you agree this space is of dimension 4 ?
Hence, your task is to write the 4x4-matrix representing T in a base of the 4-dimension space of 2x2-matrix. The canonical base is 
$\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}$  ,
$\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}$  ,
$\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}$  ,
$\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}$ ...  
A: For a matrix
$$
A=\begin{bmatrix} x & y \\ z & t \end{bmatrix}
$$
the given transformation $T$ acts as:
$$
T(A)=\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\begin{bmatrix} x & y \\ z & t \end{bmatrix}=
\begin{bmatrix} ax+bz & ay+bt \\ cx+dz & cy+dt \end{bmatrix}
$$
In the space of $\times 2$ real matrices , the standard basis is:
$$
E_1=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \quad E_2=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\quad E_3=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\quad E_4=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}
$$
so the matrix $A$ is represented as the linear combination of the basis elements:
$$
A=xE_1+yE_2+zE_3+tE_4
$$
or, in vector notation, as a vector in $\mathbb{R}^4$, with components:
$$
A=\begin{bmatrix}x\\y\\z\\t
\end{bmatrix}
$$
so the transformation $T$, in this, notation, is:
$$
T(A)=T\left(\begin{bmatrix}x\\y\\z\\t
\end{bmatrix} \right)=
\begin{bmatrix} ax+bz\\ ay+bt \\ cx+dz \\ cy+dt \end{bmatrix}
$$
and, by a simple inspection , you can see that the matrix that represents this transformation is:
$$
\begin{bmatrix} a&0&b&0\\ 0&a&0&b \\ c&0&d&0 \\ 0&c&0&d \end{bmatrix}
$$
A: To compute the representative matrix $[T]$ of a linear operator $T$ with respect to a basis $({\bf E}_a)$, we compute the vectors $T({\bf E}_a)$ and write them as linear combinations of basis elements:
$$T({\bf E}_a) = t_{1a} {\bf E}_1 + \cdots t_{na} {\bf E}_n .$$
Then, $[T]$ is the matrix whose $(a, b)$ entry is $[T]_{ab} = t_{ab}$. Note that if $T$ is a linear operator from an $n$-dimensional vector space to itself, then $[T]$ is an $n \times n$ matrix. The point of this definition is that it is exactly what is required for the matrix representation $[T({\bf x})]$ of the vector $T({\bf x})$ to be given by $$[T({\bf x})] = [T][{\bf x}] ,$$ that is, so that applying the operator $T$ corresponds to multiplication (on the left) by the representative matrix $[T]$.
In our specific case, we are using the standard basis of the (four-dimensional) vector space of $2 \times 2$ matrices, namely:
$$\left(
\pmatrix{1 & 0\\0 & 0},
\pmatrix{0 & 1\\0 & 0},
\pmatrix{0 & 0\\1 & 0},
\pmatrix{0 & 0\\0 & 1}
\right) .
$$
Computing gives, for example, that
$$
\begin{align}
T\pmatrix{1 & 0\\0 & 0}
&= \pmatrix{a & b\\c & d} \pmatrix{1 & 0\\0 & 0} \\
&= \pmatrix{a & 0 \\ c & 0} .
\end{align}
$$
Decomposing this with respect to the standard basis gives
$$T\pmatrix{1 & 0\\0 & 0}
= a \pmatrix{1 & 0\\0 & 0} + 0 \pmatrix{0 & 1\\0 & 0} + c \pmatrix{0 & 0\\1 & 0} + 0 \pmatrix{0 & 0\\0 & 1} .$$
So, the matrix representation $[T]$ of $T$ has the form
$$
[T] = \pmatrix{a & \ast & \ast & \ast \\ 0 & \ast & \ast & \ast \\ c & \ast & \ast & \ast \\ 0 & \ast & \ast & \ast} .
$$
Repeating the process for the three remaining basis vectors determines the three remaining columns.
