$T\colon M_2(\Bbb R) \to M_2(\Bbb R)$ is the linear transformation that sends $A$ to $AB-BA$, and I want to find the dimension of the kernel of $T$. I have a way of doing this, but I'm not sure if it is correct, and I would like to know if there is a faster way.
I know that the kernel of $T$ will be the set of all matrices that satisfy the solution to
$AB-BA=0$ and therefore $AB=BA$
The solution could then be found by substituting variables in for the entries of $A$, and writing down four equations with four unknowns. You could then solve each equation for zero, and solve the homogeneous system through an augmented matrix.
$\begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{bmatrix}=\begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{bmatrix}$
I'll spare the details, but after doing so for the $B$ given below, I got
B=\begin{bmatrix}2 & 3\\4 & 2\end{bmatrix}$\\4b-3c=0\\3a-3d=0\\4d-4a=0\\3c-4b=0$
and the RREF simplified to
\begin{bmatrix}1 & 0 & 0 & -1\\0 & 1 & -3/4 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}
So then my line of thought is that, since there are two free variables, the solution to the system will consist of two linearly independent vectors, and therefore, the dimension of the kernel of $T$ will be $2$. However, this process is time consuming, so I was wondering if there is s simpler technique that I am overlooking.