Enforcing Eigenvalue Constraint on Product of Diagonal Matrix and Full Rank Matrix I have a real diagonal matrix $D$ with diagonal elements $d_1, d_2, ... ,d_n$ as well as a full rank, real $n\times{}n$ matrix $A$. Suppose I have prior knowledge of elements of $A$, and thus its eigenvalues. I'm looking for a way to choose the elements of $D$ such that the eigenvalues of the product $DA$ all have negative real parts (or equivalently, all eigenvalues have positive real parts, requiring only a sign flip in $D$).
Is it possible to choose the elements of $D$ to enforce this constraint? If not, can we prove when it is not and add some additional constraint on $A$?
Thanks
Edit 20210710:
At first I thought I could solve this problem using pole placement techniques from controls. There's a few papers I read discussing techniques, but one of the key assumptions I'm making, that $D$ is diagonal, prohibits me from directly using those techniques. I'm still reading into the theory behind those pole placement techniques, and hoping that some combination of matrix decomposition and numerical methods can get me an answer! If anyone has any other suggestions, I'm all ears!
This is a problem in system stability, so while obtaining negative real apart eigenvalues, it would also be nice to be able to roughly choose the magnitude of the real and imaginary parts. In my case, having a highly damped eigenvalues is desirable, so making the imaginary magnitude small with respect to the real magnitude.
I've played around with some of the $A$ matrices I have, and know it is possible to obtain a stable system given my constraints. But I don't have a technique I can generalize to arbitrary $A$ matrices.

For a very simple example where it is not possible, let $A=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}.$  Then $DA$ will have the form $\begin{pmatrix} 0 & a \\ b & 0\end{pmatrix}$ for some $a,b\in\mathbb{R}$, with characteristic polynomial $x^2-ab$.  Such a polynomial can never have both of its roots with negative real part, since the roots are negatives of each other.  (If you don't need the eigenvalues to have strictly negative real parts, you can get a similar counterexample by taking $A$ to be a permutation matrix of an $n$-cycle for $n>2$.  Then the characteristic polynomial of $DA$ will have the form $x^n-a$ for some $a\neq 0$ and there is always an $n$th root of $a$ whose real part is strictly positive.)
More generally, if all the diagonal entries of $A$ are $0$, then it is not possible, since then $DA$ will also have all diagonal entries $0$ and thus trace $0$ so the real parts of its eigenvalues cannot all be negative.
