I want to find sufficient conditions for the following matrix inequality to hold: $$DA^T+AD+I<0$$ based on the eigenvalues of $A$ and $D$, where $D$ is diagonal with negative entries and $A$ is diagonalizable with positive real eigenvalues.
I tried to write $A = T \Lambda T^{-1}$, where $\Lambda$ is diagonal so the inequality becomes $$DT^{-T}\Lambda T^T + T \Lambda T^{-1}D + I < 0$$ $$T^{-1}DT^{-T}\Lambda + \Lambda T^{-1}DT^{-T} + T^{-1}T^{-T} < 0$$ $$S \Lambda + \Lambda S + Q < 0$$ where $S:=T^{-1}DT^{-T}$ and $Q := T^{-1}T^{-T}$, which is essentially the same problem with only difference $S$ being symmetric.
We can also write $$ \lambda_\max(S\Lambda+\Lambda S) < -\lambda_\max(Q)$$ using symmetricity of the matrices. But I want to write something like $$ 2 \lambda_\max(S) \lambda_\max(\Lambda) < -\lambda_\max(Q)$$ Is there any way to do this?
Edit. The solution (I think). Let $x$ be such that $$ x^T (DA^T+AD) x = \lambda_\max(DA^T+AD) $$ Now we can write $x = \sum_i \alpha_i v_i$ where $A^T v_i = \mu_i v_i$ with $\mu_i > 0$. So, $$\begin{align} \lambda_\max(DA^T+AD) &= x^T (DA^T+AD) x \\ &= x^T D A^T \sum_i \alpha_i v_i + \sum_i \alpha_i v_i^T A D x \\ &= x^T D \sum_i \alpha_i \mu_i v_i + \sum_i \alpha_i \mu_i v_i^T Dx \\ &\leq 2 \mu_\max x^T D x \\ &\leq 2 \mu_\max d_\max \end{align}$$ Therefore, $2 \mu_\max d_\max < -1$ is a sufficient condition for the matrix inequality to hold. Is everything all right with this proof?
all matrices below have real scalars. Rearranging terms and using $\Sigma = -D$.
You're trying to find sufficient conditions, purely in terms of eigenvalues of $A$ and $\sigma_i$ to prove
$I\prec \Sigma A^T + A\Sigma $
I'll show it's impossible to use these conditions to prove
$0 \prec \Sigma A^T + A\Sigma $
hence one cannot use them to find $I\prec \Sigma A^T + A\Sigma $
Suppose for a contradiction that we may characterize the positive definiteness of $\big(\Sigma A^T + A\Sigma\big) $ by the eigenvalues of $A$ and the $\sigma_i$'s. and you have selected the eigenvalues of $A$ and $\Sigma$ in such a manner. Note that $\big(\Sigma A^T + A\Sigma\big) $ is positive definite iff $A\Sigma $ is. (If the below non-symmetrized argument makes you uncomfortable you can double everything and recover the symmetric version.)
Now consider the 2x2 case (with obvious blocked structure generalization to higher dim) and since this theorem holds for all choices of $A$, I select $A$ to be upper triangular and
$A = \begin{bmatrix}\lambda_1 & z\\ 0 &\lambda_2 \end{bmatrix}$
then
$\mathbf 1^T A\Sigma \mathbf 1 = \sigma_1\lambda_1 + \sigma_2 z + \sigma_2\lambda_2 \leq 0$
for
$ z \leq -\frac{-1}{\sigma_2}\big(\sigma_1\lambda_1 + \sigma_2\lambda_2\big)$
which contractdicts the positive definiteness of $ A\Sigma$.
book-keeping note:
In the trivial case where $A\propto I$, then the above doesn't hold as $z$ is forced to be equal to zero. But for all other cases the desired sufficient conditions in terms of eigenvalues of $A$ and $\Sigma$ don't exist.