What conditions for the diagonal matrix $\mathbf{D}$ satisfying $\operatorname{rank}([DA,A]) = 2n$?

Question

Given a matrix $\mathbf{A} \in \mathbb{R}^{m\times n}(m > n)$ with $\operatorname{rank}(\mathbf{A}) = n$. What conditions the diagonal matrix D satisfies $\operatorname{rank}([\mathbf{DA,A}]) = 2\operatorname{rank}(\mathbf{A})$?

Since \begin{equation} \begin{pmatrix} \mathbf{A}^{-1}_{\text{left}}\mathbf{D}^{-1}\\ \mathbf{A}^{-1}_{\text{left}} \end{pmatrix} [\mathbf{DA,A}]= \begin{pmatrix} \mathbf{I}_{n} & \mathbf{0}\\ \mathbf{0} & \mathbf{I}_{n} \end{pmatrix}, \end{equation} where $\mathbf{A}^{-1}_{\text{left}}$ is Left inverse of $\mathbf{A}$, so the condition for $\mathbf{D}$ which defined by \begin{equation} \begin{cases} \mathbf{A}^{-1}_{\text{left}}\mathbf{D}^{-1} \mathbf{A} = \mathbf{0} \\ \mathbf{A}^{-1}_{\text{left}}\mathbf{DA} = \mathbf{0} \end{cases} \end{equation} Is this all right? If not, what the other condition for $\mathbf{D}$?

Answer 1

No. If $[DA,A]$ has full rank, we do have $$ \pmatrix{XD^{-1}\\ Y}[DA,A]=\pmatrix{I_n\\ &I_n} $$ for some left inverses $X$ and $Y$ of $A$, but $X$ and $Y$ are not necessarily equal. E.g. consider $$ A=\pmatrix{1\\ 1},\ D=\pmatrix{2\\ &1},\ [DA,A]^{-1}=\pmatrix{1&-1\\ -1&2}=\pmatrix{XD^{-1}\\ Y} $$ where $X=(2,-1)\ne(-1,2)=Y$.

Provided that $A$ is a real tall matrix of full column rank, here are two necessary and sufficient conditions for $[DA,A]$ to have full column rank:

1. $\det([DA,A]^T[DA,A])>0$.
2. $\operatorname{rank}((I - AA^+)DA)=n$, or equivalently, $\det(A^TD(I - AA^+)DA)>0$. Its connection to the previous condition is that $A^TD(I - AA^+)DA$ is the Schur complement of $A^TA$ in $[DA,A]^T[DA,A]$.