Let $F$ be a field, show that $\phi: M_n(F)\ni X\mapsto AX-XB$ is diagonalizable iff $A,B$ are both diagonalizable.
$\Leftarrow$: Let $\xi_i,\eta_j$ be eigenvectors of $A,B$ respectively, with eigenvalues $\lambda,\mu$, then it is easy to show under the base $\xi_i\eta_j'$, $\phi$ has matrix $\lambda_i-\mu_j, 1\leq i,j\leq n$.
$\Rightarrow$: Any ideas? Thanks.
The matrix of $\phi$ in the canonical base of $M_n(F)$ is:
$$\begin{bmatrix}A_{k,i}\delta_{j,l}-\delta_{i,k}B_{l,j}\end{bmatrix}_{(i,j),(k,l)}$$
If this matrix is diagonal then $A_{k,i} = 0$ if $k\neq i$ and $B_{l,j}=0$ if $l\neq j$. Which shows what you are looking for.