Matrix inequality of inverses of sum of two matrices

Question

$A$ and $B$ are two real symmetric non-singular matrices but not necessarily sign definite. Suppose $A+B$ is non-singular. I want to find a $\lambda \in \Bbb R$ such that $$(A+B)^{-1}\leq(A+\lambda I)^{-1}$$ where $A\leq B$ means $A-B$ is negative semi-definite. Does such $\lambda$ exist? How can I find it?

I have tried the identity $(A+B)^{-1}=A^{-1}-A^{-1}(B^{-1}+A^{-1})^{-1}A^{-1}$. But it led me to

$$(A^{-1}+\lambda^{-1} I)^{-1}\leq(A^{-1}+B^{-1})^{-1}$$

Thanks in advance!

Answer 1

I found a way by Schur's complement identity.

It holds that the following three matrices are congruent. $$\begin{bmatrix}(A+\lambda I)^{-1}&I\\I&A+B\end{bmatrix}\sim\begin{bmatrix}A+B&I\\I&(A+\lambda I)^1-(A+B)^{-1}\end{bmatrix}\sim\begin{bmatrix}(A+\lambda I)^{-1}&I\\I&B-\lambda I\end{bmatrix}$$

Let $n_-(A)$ denote the number of negative eigenvalues of $A$. Then, by Sylvester's law of inertia, $(A+\lambda I)^1\geq(A+B)^{-1}$ iff \begin{equation}n_-(A+B)=n_-(A+\lambda I)+n_-(B-\lambda I)\end{equation} But such $\lambda$ does not always exist, as shown by the excellent counterexample by @Rammus. In that example, $n_-(A+B)=0$ but no $\lambda$ can simultaneously make $n_-(A+\lambda I)=0$ and $n_-(B-\lambda I)$.

Anyways, if such $\lambda$ exists, the identity above provides a way to find it.

Answer 2

No, you cannot always do that. Here's a counterexample. Let $\epsilon > 0$ and $$ A = \begin{pmatrix} 1 & 0 \\ 0 & - \epsilon \end{pmatrix} \qquad B = \begin{pmatrix} - \epsilon & 0 \\ 0 & 1 \end{pmatrix}. $$ Then $$ (A + B)^{-1} = \frac{1}{1-\epsilon}I $$ and $$ (A + \lambda I)^{-1} = \begin{pmatrix} \frac{1}{1+\lambda} & 0 \\ 0 & \frac{1}{\lambda - \epsilon} \end{pmatrix}. $$ Thus $$ \begin{aligned} (A + \lambda I)^{-1} - (A+ B)^{-1} &= \begin{pmatrix} \frac{1}{\epsilon-1} + \frac{1}{1+\lambda} & 0 \\ 0 & \frac{1}{\epsilon-1} + \frac{1}{\lambda-\epsilon} \end{pmatrix} \\ \end{aligned} $$

But then this latter matrix is positive semidefinite iff its diagonal entries are both non-negative. One can then show that if $\epsilon \in (0, 1)$ no such $\lambda$ exists such that they are both non-negative.

Answer 3

For positive Definite symmetric matrices (2/3 down on that page) one has $M\geq N > 0$ iff $N^{-1}\geq M^{-1} >0$ which under your hypotheses translates to $$ 0< (A+B)^{-1} \leq (A +\lambda I)^{-1}$$ iff $$ A+B\geq A +\lambda I > 0$$ or $B\geq \lambda I$. So if $B$ has a non-trivial kernel then you need $\lambda \leq 0$ (and $A$ should be strictly positive for the above to make sense).