Let $A \in \mathbb{R}^{n \times n}$ be a Hurwitz matrix, i.e. $A$ satisfies $\mathrm{Re}\,\lambda_i< 0$, where $\{\lambda_i\}_{i=1}^n$ is the set of eigenvalues of $A$. Suppose that the trace of $A$ is normalized to $-1$, that is $\mbox{trace}(A)=-1$. Further, let $\ge$ denote the standard partial order in the set of positive semidefinite matrices.

Conjecture.$$ \min_{\substack{X\in\mathbb{R}^{n\times n},\ X\ge 0\\ AX+XA^\top \le 0 \\ X-\frac{1}{2} I\le 0}} \mathrm{trace}(AX)=-\frac{1}{2}. $$

I numerically verified the above conjecture for $n=2, 3,\dots,10$ in Matlab using the built-in LMI optimization solver. Any hint/comment towards the (dis)proof of this conjecture is very appreciated.

**The optimal $X$ is not full rank, in general.**
Consider the following $2\times 2$ matrix
$$
A = \begin{bmatrix}-1 & \frac{\sqrt{3}+2}{2} \\ \frac{\sqrt{3}-2}{2} & 0 \end{bmatrix}.
$$
Matrix $A$ has two eigenvalues at $-0.5$.

Let us select $$ X = \begin{bmatrix}\frac{1}{2} & 0 \\ 0 & -\frac{\sqrt{3}-2}{2(\sqrt{3}+2)} \end{bmatrix}. $$ It is easy to see that both constraints are satisfied and $\mathrm{tr}(AX)=-\frac{1}{2}$.

Observe also that, since $A+A^\top$ possesses a positive eigenvalue, $X=\frac{1}{2}I$ violates the constraint $AX+XA^\top\le 0$ and it is not an admissible solution.

1more comments