2
$\begingroup$

$\newcommand{\tr}{\operatorname{tr}}$Does submodularity property hold for the trace of a positive-definite hermitian matrix?

I.e., does given any real symmetric positive-definite matrices $X,A,B$ $$ \tr X^{-1} + \tr(X+A+B)^{-1} \geq \tr(X+A)^{-1} + \tr(X+B)^{-1} $$ hold?

UPD: I have checked it numerically, it appears that it does not hold for the following matrices: \begin{equation} \begin{split} X &= \begin{pmatrix} 0.08151549 & 0.05234424\\ 0.05234424 & 0.17050588 \end{pmatrix} \\ A &= \begin{pmatrix} 0.29185525 & 0.29699319\\ 0.29699319 & 0.30792421 \end{pmatrix} \\ B &= \begin{pmatrix} 0.65213446 & 0.43711443\\ 0.43711443 & 0.2932183 \end{pmatrix} \end{split} \end{equation}

$\endgroup$
3
$\begingroup$

This submodularity property is known to not hold, as the OP has found out already. However, I'd like to mention the following observation:

Let $f$ be defined on $(0,\infty)$ such that $-f'$ is operator monotone (i.e., for $A\le B \implies f'(A) \ge f'(B)$), then \begin{equation*} \operatorname{Tr} f(A+B+C)+\operatorname{Tr} f(A) \le \operatorname{Tr} f(A+B)+\operatorname{Tr} f(A+C). \end{equation*}

Example: the above inequality holds for $f(t)=t^p$ for $p\in (0,1)$.

$\endgroup$
  • $\begingroup$ I guess the last one is $\text{trace}f(A+C)$; right? $\endgroup$ – Arash Sep 1 '17 at 14:09
  • $\begingroup$ You are welcome! BTW do you have a reference for this inequality? $\endgroup$ – Arash Sep 2 '17 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.