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The following question kept me wondering for some time:

Given the symmetric matrices $A,B,C\in\mathbb{R}^{n×n}$ where $A$ and $C$ are positive definite (hence invertible), and $B$ is positive semidefinite (hence not necessarily invertible) with $\text{trace}(B)\neq0$, prove that

$\text{trace}\big\{C^{-1/2}BC^{−1/2}(A^{−1}+C^{−1/2}BC^{−1/2})^{−1}\big(A+\frac{n}{\text{trace}(B)}C\big)^{-1}\big\}\geq \text{trace}\big\{\big(A+\frac{n}{\text{trace}(B)}C\big)^{−2}A\big\}$

If it would help, one can also consider the simpler version with $C=I$: prove that

$\text{trace}\big\{B(A^{−1}+B)^{−1}\big(A+\frac{n}{\text{trace}(B)}I\big)^{-1}\big\}\geq \text{trace}\big\{\big(A+\frac{n}{\text{trace}(B)}I\big)^{−2}A\big\}$.

Please note that the matrix inversion lemma is not applicable at first to $C^{-1/2}BC^{−1/2}(A^{−1}+C^{−1/2}BC^{−1/2})^{−1}$ since $B$ is positive semidefinite. Although I'm not sure, it seems like $\text{trace}(B)=\text{trace}{(\text{trace}(B)/n)I}$ should be utilized first in some way to arrive at some inequality to which the matrix inversion lemma can later be applied. I would highly appreciate if anyone can provide some help or suggestions on this.

Based on the suggestion of user2097, an invertible (positive definite) $B$ modifies the inequality to

$\text{trace}\big\{(A+C^{1/2}B^{-1}C^{1/2})^{−1}\big(A+\frac{n}{\text{trace}(B)}C\big)^{-1}A\big\}\geq \text{trace}\big\{\big(A+\frac{n}{\text{trace}(B)}C\big)^{−2}A\big\}$

which looks easier, but I was still unable to prove it.

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  • $\begingroup$ Crossposting: math.stackexchange.com/questions/890062/…. $\endgroup$
    – Dietrich Burde
    Commented Aug 8, 2014 at 11:55
  • $\begingroup$ I believe we can assume $B$ is positive definite, without a loss of generality. In fact, if we set $B(\varepsilon)=B+\varepsilon I$ with small $\varepsilon>0$, then both sides of the inequality become continous functions of $\varepsilon$ at point $\varepsilon=0$ (unless $\operatorname{trace}(B)=0$ which is equivalent to $B=0$). In other words, it is sufficient to prove the inequality with $B(\varepsilon)$ instead of $B$, that is, in the case when $B$ is invertible. $\endgroup$
    – user56203
    Commented Aug 8, 2014 at 12:09
  • $\begingroup$ I don't know if it helps -- but this looks like out of a quantum information theory book. $\endgroup$
    – john mangual
    Commented Aug 8, 2014 at 12:21

1 Answer 1

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The last inequality is not true. For example, take $n=2$, $A, C$ to be identity matrix. It suffices to compare $\text{trace}\big\{(I+B^{-1})^{−1}\big\}$ and $\text{trace}\big\{\big(I+\frac{2}{\text{trace}(B)}I\big)^{−1}\big\}$. Now take $B=diag(1,2)$. You find that $\text{trace}\big\{(I+B^{-1})^{−1}\big\}=7/6<6/5=\text{trace}\big\{\big(I+\frac{2}{\text{trace}(B)}I\big)^{−1}\big\}$.

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  • $\begingroup$ Many thanks for the counterexample and showing that what I thought to be true for a long time is actually incorrect. $\endgroup$
    – borntotry83
    Commented Aug 16, 2014 at 22:23
  • $\begingroup$ You are welcome. It happens that people keep a "wrong" belief for some time. $\endgroup$
    – M. Lin
    Commented Aug 17, 2014 at 3:39

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