Find basis of kernel of linear transformation with matrices? Find the basis of the kernel of the linear transformation $f: M_{2×2} (\mathbb R) \rightarrow  M_{2×2}(\mathbb R)$ given $f(A) = AB − BA$ where $B = \begin{bmatrix}1&1\\0&1\end{bmatrix} $ and $M_{2×2} (\mathbb R)$ is the vectorial space of real $2×2$ square matrices.
My problem is probably that I don't understand the question. I know how to find the basis of the kernel of a linear transformation when $f$ is given as a vector with equations, which i can turn into a matrix, put the identity below, and then echelon it and take the vectors under the null columns. It's probably the same, but I'm a little lost as it's given in  a matricial way. 
 A: As commented by amd, the kernel of $f$ is the set of matrices $A$ such that $f(A) = 0.$  So we compute
$$\begin{align}
f(A) & = AB - BA\\\\
& = \begin{bmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} -
    \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix}\\\\
& = \begin{bmatrix}a_{11} & a_{11} + a_{12}\\ a_{21} & a_{21} + a_{22}\end{bmatrix} -
    \begin{bmatrix}a_{11} + a_{21} & a_{12} + a_{22}\\ a_{21} & a_{22}\end{bmatrix}\\\\
& = \begin{bmatrix}-a_{21} & a_{11} - a_{22}\\ 0 & a_{21}\end{bmatrix}\end{align}$$
which we want equal to the zero matrix.  Hence, we see that the kernel of $f$ consists of matrices such that $a_{21} = 0,$ and $a_{22} = a_{11},$ i.e., matrices of the form
$$\begin{bmatrix}a_{11} & a_{12}\\ 0 & a_{11}\end{bmatrix},$$
which can be expressed as
$$a_{11}\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} + a_{12}\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}$$
where $a_{11}$ and $a_{12}$ here are arbitrary real numbers.  Thus, a (not the as in the question) basis for the kernel of $f$ is
$$\left\{\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}, \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix} \right\}.$$
A: If $A\in \ker (f)$ then $$f(A)=0\implies AB = BA$$
So if $$A = \begin{bmatrix}a  & b\\ c & d\end{bmatrix}$$ we get
$$ \begin{bmatrix}a  & b\\ c & d\end{bmatrix}\begin{bmatrix}1  & 1\\ 0 & 1\end{bmatrix}= \begin{bmatrix}1  & 1\\ 0 & 1\end{bmatrix}\begin{bmatrix}a  & b\\ c & d\end{bmatrix}$$
so $$\begin{bmatrix}a  & a+b\\ c & c+d\end{bmatrix} = \begin{bmatrix}a+c  & b+d\\ c & d\end{bmatrix}$$
So $c =0$ and $a=d$. So $$A = \begin{bmatrix}a  & b\\ 0 & a\end{bmatrix}= a\cdot I +b\cdot J$$
where $ J =\begin{bmatrix}0  & 1\\ 0 & 0\end{bmatrix}$. So $$\ker (f) = \{a\cdot I+b \cdot J;\; a,b\in \mathbb{R}\}$$ and since $I$ and $J$ are linear independent we have def$ (f)=2$.
