1

I need to show if the following inequality is true

$$ (A + B)^{-1}M (A + B)^{-1} - A^{-1} M A^{-1} \preceq 0$$

given that $(A,B)=(A^T,B^T) \succ 0$ and $M = M^T \succeq 0$ also we have that $A + B \succeq A$. If $M = I$, then, it is pretty clear that the above inequality holds, but can we find conditions on $A$ such that the above holds for any PSD $M$?

user2457324
  • 311
  • Feasibility means that a condition or set of conditions can be satisfied. Are you asking about that, or are you asking if the inequality is necessarily satisfied in all cases? You've also used the word "optimization" without specifying an objective function to maximize or minimize. – hardmath Jul 25 '16 at 04:11
  • What is your definition of PSD matrices? If your require a PSD matrix to be Hermitian, then $(A+B)^{-1}M(A+B)^{-1}M$ is not Hermitian in general. – user1551 Jul 25 '16 at 07:39
  • @user1551 I am sorry, but there should not be an $M$ at the end. had a silly typo...sorry about that....these matrices are all real so ok with symmetric...are you saying that the inequality would never be satisfied for any $A$? – user2457324 Jul 25 '16 at 08:06

1 Answers1

0

It's not true. When $M=I$, it reduces to $(A+B)^{-2}\preceq A^{-2}$ or equivalently, $A^2\preceq (A+B)^2$. So it's essentially saying that $0\preceq X\preceq Y$ implies $X^2\preceq Y^2$, which is false in general (see here for a counterexample).

Community
  • 1
user1551
  • 139,064
  • thank you for your reply. But, since I have the condition that $A + B \succeq A$, then, for the case $M=I$, I do have $(A+B)^2 \succeq B^2$ and so the inequality holds? is that correct? I believe that the $M$ is causing problems in my case so I wanted to find conditions on $A$ such that the inequality would hold for any $M = M^T \succeq 0$ – user2457324 Jul 26 '16 at 16:29
  • @user2457324 No. Call $A+B$ as $X$ and $B$ as $Y$. You are saying that $X\succeq Y$ implies that $X^2\succeq Y^2$. As the linked answer shows, this is wrong. – user1551 Jul 26 '16 at 16:34
  • thank you so much for this. Would it be possible for you to provide any insights on the structure of $X,Y$ for why the implication does not hold for higher powers? e.g., something regarding the kernel of some combination of $X,Y$ -- though these are non-zero since it is PD...it would really help me if I could get a mathematical reason why this doesn't hold. hopefully this kind of makes sense. thanks again. – user2457324 Jul 26 '16 at 18:15
  • Also, would you be able to give me any citation of the above in a text, for example? I could not find this in Bernestein's book....your counter is great though :) thanks again and any thoughts (if a precise answer is not clear -- e.g., what you think in the structure may be the culprit) on my previous comment would be much appreciated. – user2457324 Jul 27 '16 at 04:27
  • @user2457324 The keyword here is "operator monotone". It is actually well-known that among all nonnegative exponent $p$, the power function $t\mapsto t^p$ is operator monotone if and only if $0\le p\le1$. In particular, every positive integer power (except the identity map) will not preserve the positive semidefinite partial ordering. E.g. see prop. 2.4 of this paper. – user1551 Jul 27 '16 at 08:07
  • @user2457324 I haven't any references at hand. I seem to recall that some papers of Horn/Johnson has an even more refined result, but I couldn't find it right now. You may also consult Horn and Johnson's Topics in Matrix Analysis or Bhatia's Matrix Analysis to see if they have documented this result or not. – user1551 Jul 27 '16 at 08:07
  • Hi -- I was wondering if you would be able to help me with the Q here https://math.stackexchange.com/questions/2279926/understanding-the-kernel-space-of-the-differences-of-transition-matrices thank you! – user2457324 May 14 '17 at 00:44