Converting a quadratic matrix inequality to a linear matrix inequality

[Universo]

I would like to solve the following problem by converting it to an SDP.

What I can do it to first cast it as a quadratic program and cast the QP as an SDP. But I don't see how I can do it directly. I thought this optimization problem is equivalent to

I am not sure about this equivalency but if it is true I still need to convert the quadratic matrix inequality above to an LMI. I would appreciate any help/tips on this.

[Johan Löfberg]

No images visible

[Universo]

The original problem I would like to solve is in prob1.png. The problem that I think is equivalent to the first one is in prob2.png.

[Universo]

Thanks for pointing it out. I posted a reply to my original post by attaching the images. They are visible now I hope.

[Johan Löfberg]

They are equivalent, and the second model is converted to an LMI by applying a Schur complement (assuming P and R are positive semidefinite)

[Johan Löfberg]

having said that, writing this as a semidefinite program is a waste of resources. Since the objective is a quadratic function, and you have no constraints, the problem can be solved analytically. It's just a matter of matrix algebra and book-keeping. If you plug in the model in YALMIP, YALMIP will solve it as an unconstrained QP. 

[Universo]

There are also linear constraints but even then you are absolutely right. I already use a QP solver for this application but I need to demonstrate that they can be solved using an SDP solver as well. Next I will also cast it as an SOCP and use an SOCP solver on it.

[Universo]

Do I need to separate Q into blocks while applying Schur complement?

[Johan Löfberg]

Well, since it is a QP, it is trivially solvable using SOCP and SDP, since those are levels of generalizations of QP (the SDP formulation would be very ineffcient comparing to solving the original QP)

[Johan Löfberg]

No (but I don't even understand the premise of the question why Q would have blocks). Simply apply a Schur complement twice, or if you want, write as one quadratic term with the inner matrix [P 0;0 R]

[Universo]

Are these the LMIs you had in mind?

[Johan Löfberg]

If you want to work with a second upper-bounding variable, you don't have to (cannot) have equality, but it is sufficient with Q-KRK^T >= S. You middle line does not make sense (it is a non-convex constraint if R is PSD, so it can never be transfered to an LMI, which shows up in the fact that the resulting LMI is nonsensical as -R^-1 never can be positive PSD.)

However, you can just as well apply schur on A-BC^-1B'-DE^-1D^T>=0 to arrive at [A B' D'; B C 0;D 0 E]>=0