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.

  • $\begingroup$ Crossposting: math.stackexchange.com/questions/890062/…. $\endgroup$ 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
    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. $\endgroup$ Aug 8, 2014 at 12:21

1 Answer 1


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 $\text{trace}\big\{(I+B^{-1})^{−1}\big\}=7/6<6/5=\text{trace}\big\{\big(I+\frac{2}{\text{trace}(B)}I\big)^{−1}\big\}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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