<s>[INST] Finding an explicit isomorphism between $\mathbb R^4 / \ker \ T$ and $\mathbb R^2$ I'm wondering if I have a valid answer to this. 
It is exactly (e) of the following: 

I first state that the two vector spaces are isomorphic because they have equal dimension. I then define a linear transform $S:  \mathbb R^4  / Ker  \ T \to \mathbb R^2$ by 
$$S \begin{pmatrix}
0 \\ 
a \\ 
b \\ 
0 \\ 
\end{pmatrix}  \to 
\begin{pmatrix}
a \\
b \\
\end{pmatrix}: a,b \in \mathbb R$$
Where $\begin{pmatrix}
0 \\ 
a \\ 
b \\ 
0 \\ 
\end{pmatrix}$ is an arbitrary element in the departure space. 
If this mapping translates basis to basis, this is an isomorphism.
$$
S \begin{pmatrix}
0 \\ 
a \\ 
b \\ 
0 \\ 
\end{pmatrix}  \to 
\begin{pmatrix}
a \\
b \\
\end{pmatrix} \in \mathbb R^2 \implies S\left(a \begin{pmatrix}
0 \\ 
1 \\ 
0 \\ 
0 \\ 
\end{pmatrix}  + 
b\begin{pmatrix}
0 \\
0 \\
1 \\
0 \\ 
\end{pmatrix}\right) = a \  \vec e_1 + b  \vec \ e_2 $$
$$
\text{ as
 $S\begin{pmatrix}
0 \\ 
1 \\ 
0 \\ 
0 \\ 
\end{pmatrix} = 
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}$ and 
$
S\begin{pmatrix}
0 \\ 
0 \\ 
1 \\ 
0 \\ 
\end{pmatrix} = 
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}$}$$
Is this a valid answer?
 [/INST] Your answer is not valid, since you don't even mention $F^4/\ker T$ in it. An explicit isomorphism would be$$\begin{array}{rccc}\Psi\colon&F^4/\ker T&\longrightarrow&F^2\\&\begin{pmatrix}a\\b\\c\\d\end{pmatrix}+\ker T&\mapsto&T\begin{pmatrix}a\\b\\c\\d\end{pmatrix}.\end{array}$$
</s>