Suppose block matrix $D$ in $\begin{pmatrix} A & B \\ C&D\\ \end{pmatrix}$ has Schur complement $S$. Is $S^{-1}-A^{-1}$ positive semi-definite?

Question

Suppose that the positive semi-definite matrix $M$ is partitioned into four submatrices blocks as $$M = \begin{pmatrix} A & B \\ C&D\\ \end{pmatrix}$$ and block matrix $D$ has Schur complement $$S = A-BD^{-1}C$$ I'm interested in the difference between $S^{-1}$ and $A^{-1}$. Is $S^{-1}-A^{-1}$ positive semi-definite? Or negative semi-definite? Or neither?

Answer 1

Edit: Below, I am writing $A \succeq B$ for the Loewner order, meaning that $$ A \succeq B \Leftrightarrow A - B \;\; \text{is positive semidefinite}. $$

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Since $M$ is positive semidefinite, we have $$ M = M^T \Leftrightarrow A = A^T, B = C^T, D = D^T. $$

Therefore the difference you are interested in is $$ (A - C^T D^{-1} C)^{-1} - A^{-1}. $$

Using this, we can prove the following claim:

Claim: The matrix $A - C^T D^{-1} C \preceq A$.

Finally, we use the fact that $A_1 \preceq A_2 \implies A_1^{-1} \succeq A_2^{-1}$ to deduce that the difference is PSD.

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Proof of Claim: Note that $$ A - C^T D^{-1} C \preceq A \Leftrightarrow C^T D^{-1} C \succeq 0. $$ Indeed, since $M$ is PSD, so is $D$ (and thus $D^{-1}$), which yields $$ \langle x, C^T D^{-1} C x\rangle = \langle Cx, D^{-1} Cx\rangle \geq 0, \quad \forall x. $$