A problem of the form was hinted at on an upcoming assignment:
$minimize$ $tr(P)$
$s.t.$
$A^TP+PA = -Q$,
$P=P^T>0$
It was hinted that the desired solution require the problem be transformed into a form similar to the one shown below and solved via KKT optimality conditions. Where the transformed constraints will be four equalities or inequalities.
$minimize$ $(C^TX)$
$s.t.$
$AX = b$
My question is how are these four transformed constraints found or derived?
---update---
Perhaps I am over complicating this. If $A$, $Q$ and $P$ are $2x2$, that would be four linear equalities of the form $Ax=b$. For $P=\begin{bmatrix}p_1&p_3\\p_3&p_2\end{bmatrix}$, would it be valid specify the problem as $minimize(p_1+p_2)$, since we want to minimize $trace(P)$? Then taking the 1st equality piecewise:
$(1)$ $2A_{11}p_1+2A_{21}p_3=-Q_{11}$
$(2)$ $A_{12}p_1+A_{21}p_2+(A_{11}p_1+A_{22})p_3=0$
$(3)$ $2A_{22}p_2+2A_{12}p_3=-Q_{22}$
And to define P as PSD:
$(4)$ $p_1 \geq 0$
$(5)$ $p_1p_2-p_3^2 \geq 0$
Then apply the KKT conditions to $min(p_1+p_2)$ and constraints 1-5, treating $p_1, p_2$ & $p_3$ as the variables for which to be solved?
If that is the case, why did the instructor hint that the problem should be transformed to $minimize(C^TX)$?