Prove that $ 0 \leq B^T(C+B\mathcal{I}_V^{-1}B^T)^{-1}B \leq \mathcal{I}_V $

Question

In Walter Zulehner's article Nonstandard norms and robust estimates for saddle point problems page 546 at the very top, he writes that

$$ 0 \leq B^T(C+B\mathcal{I}_V^{-1}B^T)^{-1}B \leq \mathcal{I}_V \tag{$\ast$}$$

is true for all $ \mathcal{I}_V $. I simply can't figure out how to show this?

It should probably be noted that $ \mathcal{I}_V $ is positive definite, $ C $ is positive semi-definite and $ C+B\mathcal{I}_V^{-1}B^T $ is assumed non-singular. $ C $ and $ \mathcal{I}_V $ are square matrices, say $ \mathcal{I}_V \in \mathbb{R}^{n\times n} $ and $ C \in \mathbb{R}^{m\times m} $, and $ B \in \mathbb{R}^{m\times n} $.

All of this can obviously also be seen in the source article.

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My thoughts so far has been: Since $ \newcommand{\I}{\mathcal{I}} \I_V > 0 $, we have $ \I_V^{-1} > 0 $ and thus $ B\I_V^{-1}B^T \geq 0 $ and $ C + B\I_V^{-1}B^T \geq 0 $. Since this one was assumed non-negative it must be positive definite, $ C + B\I_V^{-1}B^T > 0 $. Thus $ (C + B\I_V^{-1}B^T)^{-1} > 0 $.

So it is clear that the first inequality in $(\ast) $ is true. But I can't figure out how to prove the second one. I feel like it should be simple since he doesn't even mention anything about how it is seen or references a source, but I don't see it. Any hints would be much appreciated.

Answer 1

Let $M = \begin{pmatrix} A & B \\ B^T & C \end{pmatrix}$ be a block matrix with $A,C$ symmetric (not necessarily of the same size) and $C$ positive definite. The Schur complement of $C$ in $M$ is $S_C = A-BC^{-1}B^T$.

The following statement is well known.

Lemma $M$ is positive semi-definite iff $S_C$ is positive semi-definite.

Now consider the matrix $$ M = \begin{pmatrix} D & B \\ B^T & \mathcal I_V \end{pmatrix} $$ with $D = C + B \mathcal I_V^{-1} B^T$. Then the Schur complement of $\mathcal I_V$ in $M$ is $D - B \mathcal I_V^{-1} B^T = C$, which is positive semi-definite. Therefore, $M$ is positive semi-definite. Hence, the Schur complement of $D$ in $M$ must be positive semi-definite too, i.e. $$ S_D = \mathcal I_V - B^T D^{-1} B\ge 0, $$ as required.

Answer 2

Here, you can find a solution for the case $m<n$. Define the inner product in $\mathbb{R}^n$ as $\langle v,w \rangle_1= v^t I_V^{-1} w$.

Let $B_k\in \mathbb{R}^{m\times n}$ be a sequence of rank $m$ matrices converging to $B\in \mathbb{R}^{m\times n}$.

Let $R_k$ be the row space of $B$ and $R_k^{\perp}$ be the orthogonal complement of $R_k$ with respect to $\langle \cdot,\cdot \rangle_1$.

Let $\widetilde{B_k}$ be a real matrix of order $n$ whose first $m$ rows are ocuppied by the matrix $B_k$ and the others are occupied by an orthonormal basis of $V_k^{\perp}$. Therefore $\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t$ is the direct sum $B_k^tI_V^{-1}B_k\oplus I_{n-m\times n-m}$ and $(\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t)^{-1}=(B_k^tI_V^{-1}B_k)^{-1}\oplus I_{n-m\times n-m}$.

Let $\widetilde{C}=C\oplus 0_{n-m\times n-m}$ and notice that $0<\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t\leq \widetilde{C} +\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t$.

Next, since $0<D\leq E$ implies $0<E^{-1}\leq D^{-1}$ (left as an exercise).

Therefore, $0<\widetilde{B_k}^t(\widetilde{C}+\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t)^{-1}\widetilde{B_k}\leq \widetilde{B_k}^t(\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t)^{-1}\widetilde{B_k}=I_V$.

Since $\widetilde{C}+\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t=(C+(B_k^tI_V^{-1}B_k)^{-1})\oplus I_{n-m\times n-m}$ then $ B_k^t(C+(B_k^tI_V^{-1}B_k)^{-1})B_k\leq \widetilde{B_k}^t(\widetilde{C}+\widetilde{B_k}I_V^{-1}\widetilde{B_k}^t)^{-1}\widetilde{B_k}\leq I_V$.

As $k$ aprroach infinity we get $0\leq B^t(C+BI_V^{-1}B^t)^{-1}B\leq I_V$.