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.
 A: First of all, observe that, setting 
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \tag{1}$
we have 
$A = I + J \tag{2}$
which simplifies things somewhat since now we may write
$T(X) = AX - XA = (I + J)X - X(I + J)$
$= X + JX - X - XJ = JX - XJ; \tag{3}$
taking
$X = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} \tag{4}$
it is easily seen that
$JX = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} -x_3 & -x_4 \\ x_1 & x_2 \end{bmatrix} \tag{5}$
and
$XJ = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} x_2 & -x_1 \\ x_4 & -x_3 \end{bmatrix}, \tag{6}$
whence
$T(X) = JX - XJ = \begin{bmatrix} -(x_2 + x_3) & x_1 - x_4 \\ x_1 - x_4 & x_2 + x_3 \end{bmatrix}, \tag{7}$
from which we see that $T(X) = 0$ if and only if
$x_4 = x_1; \; x_3 = -x_2 \tag{8}$
showing that $X$ takes the form
$X =\begin{bmatrix} x_1 & x_2 \\ -x_2 & x_1 \end{bmatrix} = x_1 I - x_2 J; \tag{9}$
furthermore any $X$ as in (9) is easily seen to satisfy $T(X) = JX - XJ = 0$, so the matrices $I$ and $J$ span $\ker T$, and they are easily seen to be linearly independent; thus $\dim(\ker T) = 2$.  Inspection of (7) shows that every $T(X)$ may be written
$T(X) = \begin{bmatrix} -z & y \\ y & z \end{bmatrix} = zD + yP, \tag{10}$
where
$D = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \tag{11}$
and
$P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}; \tag{12}$
$P$ and $D$ are manifestly linearly independent over $\mathbf F$, showing that $\dim \text{Im}T = 2$ as well.  In fact if $\text{char}(\mathbf F) \ne 2$, the matrices $I$, $J$, $P$, $D$ span $M_2(\mathbf F)$, as may seen from
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} =\dfrac{1}{2}(a(I - D) + b(P - J) + c(P + J) + d(I + D)); \tag{13}$
when $\text{char}(\mathbf F) = 2$, a basis for $M_n(\mathbf F)$ is comprised the four matrices 
$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}. \tag{14}$
When $\text{char}(\mathbf F) \ne 2$, we see that the matrices $I$, $J$, $P$, $D$ are linearly independent; if
$aI + bJ + cD + dP = 0, \tag{15}$ 
then we have
$a - c = a + c = 0; \; b - d = b + d = 0, \tag{16}$
which forces $a = b = c = d = 0$ in this case.  The matrices (14) are linearly independent whether $\text{char}(\mathbf F) = 2$ or not.
We have seen in the above that $\dim(\ker(T)) = \dim(\text{Im}(T)) = 2$, and why, which I glean is what our OP user144464 needed.  My night job beckons, so I can say no more at this time; but a usual,
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: 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.
