Generally, given a matrix $A$ the set
$$
Ker(A):=\{v\mid Av=0\}
$$
is a subvector space and it can be represented in the forms of a system
of linear equations naturally.
For example if
$$
A=\begin{pmatrix}1 & 2\\
3 & 4
\end{pmatrix}
$$
then $Ker(A)$ is given by
$$
Av=0
$$
$$
\begin{pmatrix}1 & 2\\
3 & 4
\end{pmatrix}\begin{pmatrix}x_{1}\\
x_{2}
\end{pmatrix}=0
$$
$$
\begin{cases}
x_{1}+2x_{2}=0\\
3x_{1}+4x=0
\end{cases}
$$
This gives the idea of finding $A$ s.t
$$
ImT=Ker(A)
$$
Now, say we find a basis for $Im(T)$, we can complete this basis
for $Im(T)$ to a basis of $\mathbb{R}^{4}$.
So, say for example $\{v_{1},v_{2}\}$ is a basis for $Im(T)$ and
$\{v_{1},v_{2},v_{3},v_{4}\}$ is a basis for $\mathbb{R}^{4}$.
If we define a mapping $L:\mathbb{R}^{4}\to\mathbb{R}^{4}$ s.t
$$
L(v_{3}),L(v_{4})
$$
are linearly independent, and
$$
L(v_{1})=L(v_{2})=0
$$
then
$$
Ker(L)=sp\{v_{1},v_{2}\}=Im(T)
$$
let $A$ be the representative matrix of $L$, so that
$$
Ker(A)=Ker(L)=Im(T)
$$
This was the idea - I leave it up to you to plug in the details of
your case