# A matrix inequality involving a singular matrix

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...

• By $<$ do you mean element-wise smaller or that the matrix difference is positive definite? – Hans Aug 4 '20 at 17:30
• Both $B$ and $X$ are symmetric positive definite. – livehhh Aug 5 '20 at 3:08

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.