TITLE: If $D$ is a diagonal matrix, when is the commutator $DA - AD$ full rank?
QUESTION [12 upvotes]: Suppose $A \in \mathbb{C}^{n \times n}$ is full rank.   I'm looking for a sufficient condition on $A$ such that for some diagonal matrix $D \in \mathbb{C}^{n \times n}$, the commutator $D A - A D$ is full rank.
I've worked out some necessary conditions- $A$ and $D$ cannot share an eigenvector, so no column of $D$ can have $n-1$ zero indices (ie,  no column of $A$ is a scaled column of the identity matrix).
For $n=2$, 
\begin{align}
DA - A D &= 
\begin{bmatrix} 
0 & (d_1 - d_2) a_{1,2} \\
(d_2 - d_1) a_{2, 1} & 0
\end{bmatrix}\\ 
\det(DA - AD) &= (d_1-d_2)^2 a_{1,2} a_{2,1},
\end{align}
so a $a_{1,2}, a_{2,1} \neq 0$ is both necessary and sufficient.
But for $n=3$, 
\begin{align}
DA - A D &= 
\begin{bmatrix} 
0 & (d_1 - d_2) a_{1,2} & (d_1 - d_3) a_{1, 3} \\
(d_2 - d_1) a_{2, 1} & 0 & (d_2 - d_3) a_{2, 3} \\
(d_3 - d_1) a_{3, 1} & (d_3 - d_2) a_{3, 2} & 0
\end{bmatrix}\\ 
\det(DA - AD) &= (d_1-d_2)(d_1-d_3)(d_2-d_3) (a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}).
\end{align}
So if $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, $DA - AD$ is rank deficient regardless of the choice of $D$.
Q1: Is there a geometric interpretation for the constraint $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}) \neq 0$?    
Q2:  I don't get a factored form of the determinant (one term depending on $D$, one term on $A$) for $n > 3$.  Is $n=3$ a special case?  
Q3: Is there a sufficient condition on $A$ such that $DA - AD$ is full rank for $n>3$?  
I'd prefer to not assume $A$ is positive definite.
Edit:  If $A$ is symmetric, $DA - AD$ is skew-symmetric and thus rank deficient if $n$ is odd.  
Q4:  $A$ symmetric is one way to make $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, but clearly other choices result in a rank-deficient commutator.  Is there a clean way to express this for $n > 3$?

REPLY [4 votes]: 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.