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I am trying to solve the next problem \begin{aligned} & \underset{P}{\text{maximize}} & & \log \det P \\ & \text{subject to} & & A^T P^{-1} + P^{-1} A \preceq 0 \\ &&& P^{-1} \succeq 0 \\ &&& \ldots \end{aligned}

Is this problem convex or can be transfrom to convex?

I try to introduce the new variable Q

\begin{aligned} & \underset{P}{\text{maximize}} & & \log \det P \\ & \text{subject to} & & A^T Q + P Q \preceq 0 \\ &&& Q \succeq 0 \\ &&& P Q = I &&& \ldots \end{aligned} but last constrainst is BMI.

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    $\begingroup$ $\log(\det(P))$ is convex on the set of $P$ which are positive-semi-definite and $\det(P)\neq 0$. $\endgroup$
    – Alex R.
    Jan 31, 2015 at 22:57
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    $\begingroup$ No, log(det(P)) is concave on the positive definite cone $\endgroup$ Feb 1, 2015 at 10:17

1 Answer 1

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EDIT: Removed my incorrect answer.

However, it is trivial/ill-posed due to the homogenous form you use. If there exist a feasible solution ($A$ stable) you can let $P$ tend to infinity and obtain an arbitrarily good solution)

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  • $\begingroup$ Can't we multiply the constraints on the left and right by P? $\endgroup$ Feb 1, 2015 at 16:52
  • $\begingroup$ Doh! Of course. Read the question too fast and thought this was what had been tried with $Q$ $\endgroup$ Feb 1, 2015 at 18:54
  • $\begingroup$ Thanks for answer, but there is other constraints. I try to find an ellipsoid with maximum volume inside the region of attraction. But this problem is likely nonconvex. $\endgroup$
    – sarmnv
    Feb 2, 2015 at 7:33
  • $\begingroup$ Indeed, if you have other constraints on $P$ that cannot be expressed in terms of $P^{-1}$, then it is nonconvex. This suggests that finding the minimum volume ellipsoid is convex, then, and that is of course not going to be useful to you. $\endgroup$ Feb 2, 2015 at 14:38
  • $\begingroup$ You would have to specify what you mean more precisely. This is a linear system so the region of attraction in $R^n$. Do you mean the largest invariant ellipsoid such that linear constraints on the states hold, or something like that? $\endgroup$ Feb 2, 2015 at 14:56

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