# On the solution of a matrix optimization problem

Let $$A\in\mathbb{R}^{n\times n}$$ be a matrix with eigenvalues $$\{\lambda_i\}_{i=1}^n$$ such that $$\mathrm{Re}(\lambda_i)<0$$. Moreover, let $$\succeq$$ denote the standard partial order in the cone of positive definite matrices.

Consider the following maximization problem \begin{align}\label{eq:opt}\tag{\ast} M=\sup_{P,X\in\mathbb{R}^{n\times n}} \quad & \mathrm{tr}(XP^{-1})\\ \text{s.t.} \quad & AP+(AP)^\top=-X \\ & X=X^\top \succeq 0 \\ & \mathrm{tr}(X)=1 \\ & P=P^\top \succeq 0. \end{align}

Notice that we can rewrite \eqref{eq:opt} as \begin{align} M&=\sup_{\substack{X\in\mathbb{R}^{n\times n}\\ X=X^\top\succeq 0,\ \mathrm{tr}(X)=1}} -\mathrm{tr}((AP+PA^\top)P^{-1})=-2\mathrm{tr}(A). \end{align}

My question. If the cost function $$\mathrm{tr}(XP^{-1})$$ in \eqref{eq:opt} is replaced by $$\mathrm{tr}(X(\alpha I+ P)^{-1})$$, with $$\alpha>0$$, is it true that $$M= \frac{1}{\alpha-\frac{1}{2\mathrm{tr}(A)}} \ \ \ ?$$

From an extensive amount of numerical simulations it seems that answer to my latter question is in the affirmative. However, after spending a lot of time thinking about this, I couldn't prove it. So, any help is really appreciated! Thank you!

• What does $X>0$ mean? Could you properly add constraints to (*) and add P as an optimization variable? – LinAlg Oct 11 '18 at 12:32
• @LinAlg $X>0$ means that $X$ is positive definite. I’ve explicitly added the constraints to the problem. – Ludwig Oct 11 '18 at 13:54
• can you use the well known result that $(I+P)^{-1} = I + P + P^2 + P^3 + \ldots$? – LinAlg Oct 11 '18 at 15:58
• @LinAlg Yes, I've thought about this fact, but I don't see how this can be helpful... – Ludwig Oct 11 '18 at 15:59