# If $D$ is a diagonal matrix, when is the commutator $DA - AD$ full rank?

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 $$n > 3$$?

• 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 $\det([D,A])=0$. Mar 7, 2019 at 19:20

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