Consider the following equation: \begin{equation} \frac{ \mathrm{d} \mathbf{p}(t)}{ \mathrm{d} t}=(\mathbf{D}-\mathbf{A})\mathbf{p}(t) \end{equation}
Where $\mathbf{A}$ is a symmetric $N\times N$ matrix and $\mathbf{D}$ is a diagonal matrix.
The solution is simply: \begin{equation} \mathbf{p}(t)=e^{(\mathbf{D}-\mathbf{A})t}\mathbf{p}(0) \end{equation}
Now consider the following equation: \begin{equation} \frac{ \mathrm{d} \mathbf{q}(t)}{ \mathrm{d} t}=(\mathbf{I}-\mathbf{A}\mathbf{D}^{-1})\mathbf{q}(t) \end{equation}
My question: Can I express $\mathbf{q}(t)$ in terms of $\mathbf{p}(t)$?
$ \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\vec#1{\operatorname{vec}\LR{#1}} \def\Unvec#1{\operatorname{Unvec}\BR{#1}} \def\qiq{\quad\implies\quad} $ Assume that a matrix $X$ exists such that $\,q = Xp.\;$
This ansatz yields $$\eqalign{ \dot q &= X\dot p \\ &= X\LR{D-A}p \\ &= X\LR{D-A}\LR{X^{-1}q} \\ &= \LR{XFX^{-1}}q \\ }$$ You would like the matrix in parentheses to equal the matrix $\,G=(I-AD^{-1})$
Therefore $$\eqalign{ G &= XFX^{-1} \qiq GX &= XF \\ }$$ which is a Sylvester Equation which can sometimes be solved for the unknown $X$ matrix.
One approach is to use vectorization $$\eqalign{ 0 &= \BR{I\otimes G - F^T\otimes I}\,\vec{X} \\ 0 &= M\,\vec{X} \\ }$$ This means that $\vec{X}$ lies in the nullspace of $M$.
Continuing with an arbitrary vector $y$ and the pseudoinverse $M^+$ $$\eqalign{ \vec{X} &= \LR{I-M^+M}y \\ X &= \Unvec{y-M^+My} \\ }$$ If $\,M^+M=I\,$ then this approach fails, and a solution of the assumed form does not exist.