If $D$ is a diagonal matrix, when is the commutator $DA - AD$ full rank?
One answer for Q3:
Sufficient condition: $n$ is even and $A$ can be partitioned as \begin{equation} A = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \end{equation} where the $n/2 \times n/2$ blocks $A_{12}$ and $A_{21}$ are each full rank.
Proof: Construct $D$ as \begin{equation} D = \begin{bmatrix} I_{n/2} & 0 \\ 0 & 0_{n/2} \end{bmatrix}, \end{equation} and the commutator is \begin{equation} DA - AD = \begin{bmatrix} 0_{n/2} & A_{12} \\ -A_{21} & 0_{n/2} \end{bmatrix}. \end{equation} The nonzero blocks are full rank by assumption.