Matrix multiplication as a matrix I have the following problem to answer:

Let $V$ be the 4-dimensionnal vector space of $2 \times 2$ matrices (with the elementary matrices $\mathcal{E}$ as a basis). Fix $A \in V$ and let $X$ be any element of $V$. 
1) Write down the $4 \times 4$ matrices (in the basis $\mathcal{E}$) $L_A$ and $R_A$ in $\operatorname{End}(V)$ such that $L_A(X) = AX$ and $R_A(X) = XA$. 
2)Deduce the $4 \times 4$ matrix (in the same basis) of the linear map $\operatorname{ad}(A)(X) = AX - XA$. What is the largest possible rank of this linear map?

In fact, I answered those questions, but I have questions about the answers (as often in maths!). I obtained that $L_A = I_2 \otimes A$, $R_A = A \otimes I_2$, $\operatorname{ad}(A) = I_2 \otimes A - A \otimes I_2$, and the maximal rank of $\operatorname{ad}(A)$ is 2.
But what does it imply that the maximal rank of $\operatorname{ad}(A)$ is $2$? What properties can we deduce about this linear map?
Also, trying to generalize, I obtained that for $n \times n$ matrices, we have $\operatorname{ad}(A) = I_n \otimes A - A \otimes I_n$, of maximal rank $n$. Again, what does it say about this linear operator?
Finally, can we say something about matrices of the form $A \otimes B - B \otimes A$ in general? Are they always, like, an operator with a particular property?
 A: As you probably know, matrix multiplication isn't commutative in general. The kernel of $\operatorname{ad}(A)$ consists of matrices $B$ which commute with $A$ and its dimension measures in some sense how "many" matrices commute with $A$. On one extreme, if $A = I$, then $A$ commutes with all other matrices so $\dim \ker \operatorname{ad}(A) = n^2$. On the other extreme, if you take $A$ to be a diagonal matrix whose entries are all distinct, the only matrices which commute with $A$ must be diagonal so $\dim \ker \operatorname{ad}(A) = n$. It turns out that for other matrices, we will have $n \leq \dim \ker \operatorname{ad}(A) \leq n^2$ and a precise formula for the dimension of the kernel can be given over an algebraically closed field using the information from the Jordan form of $A$. For a "random" matrix over $\mathbb{C}$, $\dim \ker \operatorname{ad}(A) = n$ because $A$ is diagonalizable with distinct eigenvalues so if $\dim \ker \operatorname{ad}(A) > n$, the matrix $A$ will be "special" in some sense (it will have "more symmetries").
Since the largest possible rank of $\operatorname{ad}(A)$ corresponds to the smallest possible dimension of $\ker \operatorname{ad}(A)$, by asking you to find out the largest rank, you are equivalently asked to find the minimal dimension of the subspace of matrices which commute with $A$. For $n = 2$, given a $2 \times 2$ matrix $A$, you are guaranteed to have at least a two-dimensional subspace of matrices which commute with $A$. 
Note that your generalization from $n = 2$ to arbitrary $n$ is false and the maximal possible rank of $\operatorname{ad} A$ is $n^2 - n$. Coincidentally, this coincides with your calculation for $n = 2$.
