Universal Matrix Equation

Question

Let $D$ be a fixed invertible $n\times n$ matrix. When (besides the case where $D=I_{n\times n}$) do we have the following. For any matrices $X,Y$ such that the product $DXY$ is well-defined does there exist a matrix $Z$ such that $$ DXY = XZ? $$

Answer 1

Let's assume that that $Y = [1]$ (a $1 \times 1$ matrix). Then your question becomes whether for any column vector $x \in M_{n \times 1}(\mathbb{F})$ there exists $c \in \mathbb{F}$ such that $Dx = cx$. In other words, you ask whether any vector $x$ is an eigenvector for $D$. This can happen only if $D = \lambda I_n$ for some $\lambda \in \mathbb{F}$.

On the other hand, if $D = \lambda I_n$ then indeed for any $X,Y$ you can find $Z$ such that $DXY = \lambda XY = XZ$.

Answer 2

This is even wrong if $X$ is an $n\times n$ matrix and $Y$ is the identity matrix. Let $$ D = \pmatrix{0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0} \quad X = \pmatrix{0&0&1&0\\0&0&0&0\\0&0&0&0\\0&0&0&0} $$ Then the monoid generated by $D$ and $X$ is $\{I, D, X, DX, XD, DXD, 0\}$ and is presented by the following relations: $D^2 = I$, $X^2 = 0$ and $XDX = 0$. Then $$ DX = \pmatrix{0&0&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0} $$ but there is no matrix $Z$ such that $XZ = DX$.