$\newcommand{\tr}{\operatorname{tr}}$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}


1 Answer 1


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 $p\in (0,1)$.

  • $\begingroup$ I guess the last one is $\text{trace}f(A+C)$; right? $\endgroup$
    – Arash
    Sep 1, 2017 at 14:09
  • $\begingroup$ You are welcome! BTW do you have a reference for this inequality? $\endgroup$
    – Arash
    Sep 2, 2017 at 10:19

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