Let $A$ be an Hurwitz stable matrix (that is, the real part of the eigenvalues of $A$ is strictly negative) and $Q$ be a positive semidefinite matrix ($Q\ge 0$, for short). Let $P$ denote the solution of the following Lyapunov equation $$ AP+PA^\top = -Q. $$ Consider the following function of $Q$ $$ f(Q):=-\mathrm{tr}(Q(I+P)^{-1}), $$ where $\mathrm{tr}(\cdot)$ denotes the matrix trace.
My questions:
- Is $f(Q)$ a convex function of $Q\ge 0$?
- Does there exist a closed-form expression for the derivative $\displaystyle\frac{\partial f(Q)}{\partial Q}$?
Vectorize the Lyapunov equation $$\eqalign{ q &= -(A\otimes I+I\otimes A)\,p = -B^Tp \cr }$$ Use a colon to denote the trace/Frobenius product $$X:Y={\rm tr}(X^TY)$$ and define some new variables $$\eqalign{ M &= (I+P) \,\,\,\implies\, dM=dP \cr v &= {\rm vec}(M^{-T}Q^TM^{-T}) \cr w &= {\rm vec}(M^{-T}) \cr\cr }$$ Now find the differential and gradient of the function. $$\eqalign{ f &= -Q^T:M^{-1} \cr df &= -Q^T:dM^{-1} - dQ^T:M^{-1} \cr &= Q^T:M^{-1}\,dP\,M^{-1} - M^{-T}:dQ \cr &= M^{-T}Q^TM^{-T}:dP - M^{-T}:dQ \cr &= {\rm vec}(M^{-T}Q^TM^{-T}):dp - {\rm vec}(M^{-T}):dq \cr &= v:dp - w:dq \cr &= -v:B^{-T}\,dq - w:dq \cr &= -(B^{-1}v + w):dq \cr \frac{\partial f}{\partial q} &= -(w+B^{-1}v) \cr }$$ De-vectorize (matricize?) this to recover a matrix-valued gradient
$$\eqalign{ \frac{\partial f}{\partial Q} &= -{\rm Mat}\big(w+B^{-1}v\big) \cr &= -(I+P^T)^{-1}-{\rm Mat}\Big(\big(A^T\otimes I+I\otimes A^T\big)^{-1}{\rm vec}\big((I+P^T)^{-1}Q^T(I+P^T)^{-1}\big)\Big) \cr }$$