Assume that $D\in{\bf R}^{n\times n}$ is a symmetric positive definite matrix, $B\in{\bf R}^{m\times n}$ is an arbitrary matrix, and $C\in{\bf R}^{m\times m}$ is a symmetric positive semidefinite matrix, such that the matrix $S=(C+BD^{-1}B^T)$ is nonsingular. Let $X=D^{-1}-D^{-1}B^TS^{-1}BD^{-1}$. Prove that all the eigenvalues of the matrix $D^{1/2}XD^{1/2}$ satisfy $0\leq\lambda\leq1$？
I try to utilize Woodbury formula to prove the above assertion, but it requires $C$ to be nonsingular which is not guaranteed. How can I prove the above assertion? Is there any hint related?
As usual, write $A\le B$ to mean that the spectrum of $B-A$ is positive.
Then $S=C+BD^{-1}B^T\ge C$ since $D^{-1}>0$.
Hence $$0\le S^{-1}C\le 1\\ I=S^{-1}(C+BD^{-1}B^T)\\ \therefore\ 0\le I-S^{-1}BD^{-1}B^T\le 1\\ \therefore\ 0\le S^{-1}BD^{-1}B^T\le 1$$ But in general $AB$ and $BA$ have the same eigenvalues, except possibly for $0$. In this case, this does not matter, so $$0\le B^TS^{-1}BD^{-1}\le1\\ \therefore\ 0\le I-B^TS^{-1}BD^{-1}\le 1\\ \therefore\ 0\le DX\le 1\\ \therefore\ 0\le D^{1/2}XD^{1/2}\le 1$$ by conjugating with $D^{1/2}$.