Suppose $\mathbf{A}\in\mathbb{C}^{M \times M}$, and diagonal matrix $\mathbf{B}\in\mathbb{C}^{M \times M}$. $\mathbf{A}\succeq \mathbf{0}$, and $\mathbf{0}\preceq\mathbf{B}\preceq\mathbf{I}$.

How to prove $$ (\mathbf{I}+\mathbf{A})^{-1} \succeq \mathbf{B}(\mathbf{I}+\mathbf{A}\mathbf{B})^{-1} $$ When $\mathbf{0}\prec\mathbf{B}\preceq\mathbf{I}$, this equation is easy to prove, but how to prove when $\mathbf{0}\preceq\mathbf{B}\preceq\mathbf{I}$? Thanks.

$\mathbf{B}$ can be expressed as $\mathbf{B}=diag\{b_1,b_2,...,b_M\}$, where $0\leq b_m \leq 1$ is the diagonal elements of $\mathbf{B}$. If we can construct $\mathbf{B}_t =\mathbf{B}+t\mathbf{K}$ for $0 < t \leq 1$ and diagonal matrix $\mathbf{K} \succeq \mathbf{0}$ satisfying $\mathbf{0}\prec\mathbf{B}_t\preceq\mathbf{I}$, then we can proof $$ (\mathbf{I}+\mathbf{A})^{-1} \succeq \mathbf{B}_t(\mathbf{I}+\mathbf{A}\mathbf{B}_t)^{-1} $$ When $t\to0^{+}$, we can have $(\mathbf{I}+\mathbf{A})^{-1} \succeq \mathbf{B}(\mathbf{I}+\mathbf{A}\mathbf{B})^{-1}$. But how to design the diagonal matrix $\mathbf{K}$?

  • 2
    $\begingroup$ Would be great if you define the symboles you used and also show some or all of your work [generally people do not like when you deviate from the latter, on this platform]. $\endgroup$
    – Math-fun
    Apr 16, 2022 at 13:00
  • $\begingroup$ @Math-fun Thanks for your advice. I have edited it again. $\endgroup$
    – ZYX
    Apr 16, 2022 at 13:19
  • $\begingroup$ You can set $K = I-B$. Then $K \ge 0$ and $0< tI\le B+tK \le B+K = I.$ $\endgroup$ Apr 16, 2022 at 13:24
  • $\begingroup$ @mechanodroid Thanks. $\endgroup$
    – ZYX
    Apr 16, 2022 at 13:39
  • $\begingroup$ I have proposed a change in the title in order that the content of the question is a little more explained, and hopefully attract the answer of other people. Do you agree ? $\endgroup$
    – Jean Marie
    Apr 16, 2022 at 19:54


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