# Convexity of problem with inverse matrix

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.

• $\log(\det(P))$ is convex on the set of $P$ which are positive-semi-definite and $\det(P)\neq 0$. – Alex R. Jan 31 '15 at 22:57
• No, log(det(P)) is concave on the positive definite cone – Johan Löfberg Feb 1 '15 at 10:17

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)

• Can't we multiply the constraints on the left and right by P? – Michael Grant Feb 1 '15 at 16:52
• Doh! Of course. Read the question too fast and thought this was what had been tried with $Q$ – Johan Löfberg Feb 1 '15 at 18:54
• 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. – sarmnv Feb 2 '15 at 7:33
• 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. – Michael Grant Feb 2 '15 at 14:38
• 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? – Johan Löfberg Feb 2 '15 at 14:56