A (linear) optimization problem subject to (linear) matrix inequality constraints 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.
 A: Obviously, $trace(AX)=-1/2$ for $X=(1/2) I$. Assume $A$ diagonal. Note $A + A^T \le 0$, so $AX+X A^T \le 0$. If $A$ diagonal, then this $X$ is obviously optimizes the minimum, as $X\le (1/2) I$ is your constraint.
If $A'$ is not is diagonal, write $A'=U A U^*$, $trace(UAU^* UXU^*)=trace(AX)$.
And $UXU^*\le U(1/2)IU^*=(1/2)I$. And $UAXU^* + UXA^*U^* \le 0 \Leftrightarrow AX +X A^* \le 0$, so constraints are invariant under transformation $U$. Take optimum $U X U^*$.
A: Up to a unitary matrix $U$, $X$ is diagonal with all eigenvalues between $0$ and $1/2$ the other constraint on $AX$ implies that the diagonal entries  of $U^*AUU^*XU$ are $\le 0$ and so are those of $U^*AU$.  Take $\text{tr}(AX)$ summing the diagonal terms we see that as convex combination the minimum  could be atteigned as $\frac{1}{2}\text{tr}(A)$.  
The $X$ is the diagonal with entries equal $1/2$ except the terms $x_{i,i}$ where $x_{i,i}=0$ if $ a'_{i,i}=0$ (entry of $U^*AU$)  is possible but in general there (is) should be a counter example to your conjecture.
