I am given real, symmetric matrices $X \succ B \succ A \succ 0$ (where '$\succ$' signifies positive definiteness such that if $B \succ A$ then $B-A \succ 0$ is positive definite). Further let the '$\preceq$' operator signify negative semi-definiteness (and $\succeq$ signify positive semi-definiteness). Does the following hold? \begin{align} (X + B)^{-1} - (X + B + A)^{-1} + (2B + 2A)^{-1} - (2B)^{-1} \preceq 0 \end{align} A few observations:

  1. As the inverse is operator decreasing I know that: \begin{align} (2B + 2A)^{-1} - (2B)^{-1} \preceq 0 \;\;\;\;\;\;\mbox{and}\;\;\;\;\;\; (X + B)^{-1} - (X + B + A)^{-1} \succeq 0. \end{align}

  2. If $A = I$, $B = 2I$, $X = 3I$, where $I$ is the identity matrix, the above inequality holds.

  3. I believe the derivative of the matrix inverse should play a role here. Essentially, how does the addition of $A$ (or $2A$) affect the negative semi-definiteness of the above expression when (i) adding $A$ to $X+B$ (in the inverse) makes the expression more positive semi-definite and (ii) adding $2A$ to $2B$ (in the inverse) makes the expression more negative semi-definite. If the matrix inverse is somehow 'negative', then adding $2A$ should make the expression more negative semi-definite than adding $A$.

Thank you very much for any insights.


It's very easy to generate a random counterexample by computer, e.g. $$ X=I,\ B=\pmatrix{0.9&0\\ 0&0.1},\ A=\pmatrix{0.7&0.07\\ 0.07&0.01}. $$ Numerical computation shows that the eigenvalues of $C=(X + B)^{-1} - (X + B + A)^{-1} + (2B + 2A)^{-1} - (2B)^{-1}$ are approximately $0.007$ and $-0.418$. Therefore $C$ is indefinite.

  • $\begingroup$ But A is not positive definite, the eigenvalues are roughly -0.065 and 0.6165. $\endgroup$ – OttoVonBismarck Feb 11 '15 at 16:51
  • $\begingroup$ @OttoVonBismarck I didn't notice the requirement that $A\succ0$ before. It's fixed now. $\endgroup$ – user1551 Feb 11 '15 at 18:01
  • $\begingroup$ Alas, this does break the inequality. I still wish I had more insight! Do you know if there's an example where the inequality also doesn't hold when $X \succ B \succ A \succ I$, where $I$ is the identity matrix? $\endgroup$ – OttoVonBismarck Feb 11 '15 at 21:10
  • $\begingroup$ @OttoVonBismarck If you scale $X,B$ and $A$ by the same factor $\alpha>0$, then when $\alpha$ is sufficiently large, $A$ will eventually be $\succ I$. Since the new $C$ is just $\frac1\alpha$ times the old $C$, it is still indefinite and your inequality still doesn't hold. $\endgroup$ – user1551 Feb 12 '15 at 1:18
  • $\begingroup$ Ah yes yes, good. Thank you very much! $\endgroup$ – OttoVonBismarck Feb 12 '15 at 22:00

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