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!

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

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