finding a basis for kernal and image when dealing with a matrix vectorspace 
I've calculated the dimension of $M_2(\mathbb{F})$ as $4$, and calculated $T(X) = \begin{pmatrix} -(x_2+x_3) & x_1 - x_4 \\ x_1 - x_4 & x_3 + x_2 \end{pmatrix}$ and noted that $T(X) = 0$ iff $x_2 = -x_3$ and $x_1 = x_4$ but I'm not sure how I can find a basis for it, or for the image - any help please.

Consider arbitrary element 
$$
X =\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix}\in \ker T
$$
As you have calculated, $x_2 = -x_3$, $x_1=x_4$, thus 
$$
X =\begin{pmatrix} x_1 & -x_3 \\ x_3 & x_1 \end{pmatrix}=
\begin{pmatrix} x_1 & 0 \\ 0 & x_1 \end{pmatrix}+
\begin{pmatrix} 0 & -x_3 \\ x_3 & 0 \end{pmatrix}=
$$
$$
=x_1\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+
x_3\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
$$
hence $\ker T$ is spanned by these two matrices. They are linearly independent, hence they form the basis of $\ker T$.
As for the image, we proceed in similar manner.
$$
T(X)=\begin{pmatrix} -x_2-x_3 & x_1 - x_4 \\ x_1 - x_4 & x_3 + x_2 \end{pmatrix}=
$$
$$
=x_1\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+
x_2\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}+
x_3\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}+
x_4\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}
$$
Now, by definition $\mathrm{Im} T$ is spanned by these 4 matrices, but they are not linearly independent. Find maximal linearly independent subset and this will be our basis.
