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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 $n > 3$?

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  • $\begingroup$ Very interesting problem! The only thing I see for now is a generalization of what you get in Q4: if $n$ is odd and $A$ is symmetric, then the commutator $[D,A]$ is never of full rank--regardless of the choice of $D$--due to $$\det([D,A])=\det([D,A]^T)=\det(A^TD^T-D^TA^T)=\det(-[D,A])=(-1)^n\det([D,A])=-\det([D,A])$$ so $\det([D,A])=0$. $\endgroup$ Mar 7, 2019 at 19:20

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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.

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