Let $V = M_n(\mathbb{K})$ and $S: V \rightarrow V$ linear such that, given $A \in V$, $S(B) = AB - BA$. If $A$ is diagonalizable, is $S$ also diagonalizable?
I tried constructing the matrix for $S$ for $n = 2$ using the basis $\beta = \{E_{11}, E_{21}, ..., E_{n1}, E_{12}, ..., E_{nn}\}$ with hopes of finding a pattern (it worked for a similar problem), but that didn't seem to help.
Yes. In the basis $\beta$, the matrix representation of $S$ is $S=I\otimes A - A^\top\otimes I$, where $X\otimes Y$ denotes the Kronecker product of two matrices $X$ and $Y$. When $A$ is diagonalisable, we can write $A=PDP^{-1}$ for some invertible matrix $P$ and diagonal matrix $D$. Therefore \begin{align*} S &= I\otimes A - A^\top\otimes I\\ &= (P^{-\top}IP^\top)\otimes(PDP^{-1}) - (P^{-\top}DP^\top)\otimes(PIP^{-1})\\ &= (P^{-\top}\otimes P)(I\otimes D)(P^\top\otimes P^{-1}) - (P^{-\top}\otimes P)(D\otimes I)(P^\top\otimes P^{-1})\\ &= (P^{-\top}\otimes P)(I\otimes D - D\otimes I)(P^\top\otimes P^{-1})\\ &= (P^{-\top}\otimes P)(I\otimes D - D\otimes I)(P^{-\top}\otimes P)^{-1}, \end{align*} which is similar to the diagonal matrix $I\otimes D - D\otimes I$.