# How to prove a result dealing with Loewner order when one of the matrices is only positive semidefinite?

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 $$\mathbf{K}$$?

• 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]. Apr 16, 2022 at 13:00
• You can set $K = I-B$. Then $K \ge 0$ and $0< tI\le B+tK \le B+K = I.$ Apr 16, 2022 at 13:24