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I have three matrices $A \in \mathbb R^{n \times n}$, $B \in \mathbb R^{n \times n}$, and $X \in \mathbb R^{n \times n}$.

Suppose that $A$ is singular, $B = B^\top > 0$ and $X = X^\top > 0$.

Then, does the following inequality true?

$A^\top (A X^{-1} A^\top + B)^{-1} A \leq X$

My approach was decomposing $A$ into singular and non-singular part but, it was still unsuccessful...

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  • $\begingroup$ By $<$ do you mean element-wise smaller or that the matrix difference is positive definite? $\endgroup$
    – Hans
    Aug 4, 2020 at 17:30
  • $\begingroup$ Both $B$ and $X$ are symmetric positive definite. $\endgroup$
    – livehhh
    Aug 5, 2020 at 3:08

1 Answer 1

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Using (say) the Jordan form, approximate $A$ by a nonsingular matrix $C_t$ so that $C_t\to A$ as $t\to0$. Then $$C_t^\top (C_t X^{-1} C_t^\top + B)^{-1} C_t\le C_t^\top (C_t X^{-1} C_t^\top)^{-1} C_t=X.$$ Letting now $t\to0$, we get $$A^\top (A X^{-1} A^\top + B)^{-1} A \le X,$$ as desired.

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