Proving linear transformation For $A =\begin{bmatrix}1&2\\0&1\end{bmatrix}$, show that the mapping $\operatorname{ad}_AM_{2\times2} (R) → M_{2\times2} (\mathbb R)$ defined by $\operatorname{ad}_A (B) =
AB − BA$ is a linear transformation. Find bases for the kernel and the range of $\operatorname{ad}A$.
I am aware that to prove a linear transformation one needs to prove that $T(v)+T(w)=T(v+w)$ and $T(cv)=cT(v)$
I am not sure what the "$\operatorname{ad}$" is referring to though? 
What part am I using to prove that it is a linear transformation?
 A: I will use the typical notation
$AB - BA = [A, B]; \tag 1$
thus
$\text{ad}_A(B) = [A, B]; \tag 2$
it is easy to verify that $\text{ad}_A$ is linear.  Indeed
$\text{ad}_A(B_1 + B_2) = [A, B_1 + B_2] = A(B_1 + B_2) - (B_1 + B_2)A$
$= AB_1 + AB_2 - B_1A - B_2A = (AB_1 -B_1A) + (AB_2 - B_2A)$
$= [A, B_1] + [A, B_2] = \text{ad}_A(B_1) + \text{ad}_A(B_2); \tag 3$
also
$\text{ad}_A(cB) = [A, cB] = A(cB) - cBA = c[A, B] = c \; \text{ad}_A(B). \tag 4$
(3) and (4) together show that $\text{ad}_A$ is linear.
We compute $\ker \text{ad}_A$; to this end, we observe that
$A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = I + 2N, \tag 5$
where
$N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}; \tag 6$
then
$\text{ad}_A(B) = [I + 2N, B] = -[B, I + 2N] = -\text{ad}_B(I + 2N) = -\text{ad}_B(I) - 2 \; \text{ad}_B(N) \tag 7$
by the linearity of $\text{ad}_B$; continuing,
$\text{ad}_B(I) = [I, B] = IB - BI = 0; \tag 8$
thus
$\text{ad}_A(B) = -2 \; \text{ad}_B(N) = -2 [B, N] = 2[N, B]; \tag 9$
we set
$B = \begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \end{bmatrix}, \tag{10}$
and find
$NB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \end{bmatrix} = \begin{bmatrix} b_3 & b_4 \\ 0 & 0 \end{bmatrix}, \tag{11}$
$BN = \begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}  = \begin{bmatrix} 0 & b_1 \\ 0 & b_3 \end{bmatrix}, \tag{12}$
$[N, B] = NB - BN = \begin{bmatrix} b_3 & b_4 - b_1 \\ 0 & -b_3 \end{bmatrix}; \tag{13}$
thus
$\text{ad}_A(B) = 0 \Longleftrightarrow [N, B] = 0 \Longleftrightarrow b_3 = 0, \; b_4 = b_1; \tag{14}$
a matrix $B \in \ker \text{ad}_A$ thus takes the form 
$B = \begin{bmatrix} b_1 & b_2 \\ 0 & b_1 \end{bmatrix} = b_1 I + b_2 N; \tag{15}$
it follows that $\{I, N \}$ is a basis for $\ker \text{ad}_A$, since $I$ and $N$ are clearly linearly independent.
Finally, we find the range of $\text{ad}_A$; from (9) and (13) it is easy to see that a matrix is in $\text{Range}(\text{ad}_A)$ if and only it is of the form
$\begin{bmatrix} \alpha & \beta \\ 0 & -\alpha \end{bmatrix}; \tag{16}$
it is easy to see that such a matrix is given by $2[N, B]$ for
$B = \dfrac{1}{2} \begin{bmatrix} b_1 & b_2 \\ \alpha & b_1 + \beta \end{bmatrix}; \tag{17}$
we see that
$\begin{bmatrix} \alpha & \beta \\ 0 & -\alpha \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} + \beta N; \tag{18}$
it is then evident from (18) that the range of $\text{ad}_A$ is spanned by the two matrices
$D = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \tag{19}$ 
and $N$; since they are linearly independent, $\{D, N \}$ forms a basis for $\text{Range}(\text{ad}_A)$.
Since $\{I, N \}$ is a basis for $\ker \text{ad}_A$, and $\{D, N \}$ is a basis for $\text{Range}(\text{ad}_A)$, we see that both the kernel and range of $\text{ad}_A$ are two-dimensional over $\Bbb R$.
