Confusion about a Linear Transformation question. Let $\beta := [M_1, M_2, M_3, M_4]$ be the ordered basis of $R^{2×2}$ defined by:
$$ M_1 := \begin{pmatrix}
1 & 0\\
0 & 0 \end{pmatrix}, 
M_2 := \begin{pmatrix}
0 & 1\\
0 & 0 \end{pmatrix},
M_3 := \begin{pmatrix}
0 & 0\\
1 & 0 \end{pmatrix},
M_4 := \begin{pmatrix}
0 & 0\\
0 & 1 \end{pmatrix}
$$
Define the linear transformation
$L := A → (A + A^T)$.
a) What is $L \left( \begin{pmatrix}
a & b\\
c & d \end{pmatrix} \right)$?
b) Find the matrix representation of $L$ with respect to the basis $\beta$.
c) Find all solutions to $L(A) = \begin{pmatrix}
4 & 0\\
0 & -6 \end{pmatrix}$.
Does the following look correct?
$L \left( \begin{pmatrix}
a & b\\
c & d \end{pmatrix} \right)$ = $ \begin{pmatrix}
2a & b + c\\
b+ c & 2d \end{pmatrix}$
$Ker(L) = span\{\begin{pmatrix}
0 & c\\
-c & 0 \end{pmatrix}\}$
$range(L) = span\{ \begin{pmatrix}
a & 0\\
0 & 0 \end{pmatrix}, \begin{pmatrix}
0 & s\\
s & 0 \end{pmatrix}, \begin{pmatrix}
0 & 0\\
0 & d \end{pmatrix}\}$ where s denotes the quantity $c + d$
Matrix Representation: $\begin{pmatrix}
2 & 0 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 2\end{pmatrix} \begin{pmatrix}
a\\
b\\
c\\
d\end{pmatrix}=\begin{pmatrix}
2a\\
b + c\\
b + c\\
2d\end{pmatrix}$
Thanks for the help!

Your transformation is from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$; and from the given question, it looks like both the domain and the range spaces are represented using the same basis that is $\left\{b_1=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},b_2=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix},b_3=\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, b_4=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\right\}$. Now, given a vector $x$ in the domain w.r.t. the basis of the domain, the actual vector in $\mathbb{R}^{2 \times 2}$ is $x_1 b_1 + x_2b_2+x_3b_3+x_4b_4$. The transformation $L:A\rightarrow A+A^T$ w.r.t. the given basis for the domain and the range spaces can be represented through a matrix, call it, $M_L$. $M_L$ consists of columns that are essentially $L(b_1), L(b_2), L(b_3),L(b_4)$ represented w.r.t. the basis (of the range space). 
For example, the first column of $M_L$ can be found out as the vector $M_{L_1}$ that is given by the components $[m_{11},m_{21},m_{31},m_{41}]^T$ that are found out by solving the following: $$L(b_1)=m_{11}b_1+m_{21}b_2+m_{31}b_3+m_{41}b_4.$$
Note that in the above I've used $b_1, b_2, b_3,b_4$ on the right hand side since the same basis is used to represent the range space as well. Were a different basis given for the range space then the elements of that basis should be plugged in instead. I hope you can take it from here and find the matrix $M_L$. The other two parts should be harmless; it's usually the matrix representation that's confusing at first looks.
