Suppose that $\mathbf{A}-\mathbf{B}$ is positive definite. What can we say about $\left( \mathbf{B}^{-1} - \mathbf{A}^{-1} \right)^{-1}$?

Question

In a physical problem related to minimum dissipation, it can be proven from physical considerations that $\mathbf{A}-\mathbf{B}$ is positive definite. According to the definition, we have $$ \forall \mathbf{x} \in \mathbb{R}^n \backslash \{0\}, \quad \mathbf{x}^\top \left( \mathbf{A}-\mathbf{B} \right) \mathbf{x} > 0 \, . $$

Both $\mathbf{A}$ and $\mathbf{B}$ are invertible and are themselves also symmetric positive definite. In addition, the eigenvalues of $\mathbf{A}$ and $\mathbf{B}$ are positive.

To be able to proceed with a mathematical proof, I was wondering whether this imply that $\left( \mathbf{B}^{-1} - \mathbf{A}^{-1} \right)^{-1}$ is positive definite?

Already for diagonal matrices $\mathbf{A}$ and $\mathbf{B}$, it can be shown readily that this is the case indeed.

Can we say anything about the positive definiteness of $\left( \mathbf{B}^{-1} - \mathbf{A}^{-1} \right)^{-1}$ in the general situation of interest?

Answer 1

Yes, you have $$ A(A - B)^{-1}B = \left(B^{-1}(A - B)A^{-1}\right)^{-1} = \left(B^{-1} - A^{-1}\right)^{-1}, $$ which proves that $B^{-1} - A^{-1}$ is invertible (it is not obvious) and its inverse equals $A(A - B)^{-1}B$. This is the matrix version of $\frac{1}{\frac{1}{b} - \frac{1}{a}} = \frac{ab}{a - b}$. Now, let us write $C = A - B$, which is positive definite symmetric (thus $C^{-1}$ is too). And we have, $$ \left(B^{-1} - A^{-1}\right)^{-1} = A(A - B)^{-1}B = (B + C)C^{-1}B = BC^{-1}B + B. $$ And for all $x \neq 0$, $x^\top Bx > 0$ and $x^\top BC^{-1}Bx = (Bx)^\top C^{-1}(Bx) > 0$ which proves that $x^\top\left(B^{-1} - A^{-1}\right)^{-1}x > 0$.

Answer 2

In the below, I use the notation $M \succ 0$ to mean that $M$ is positive definite.

First of all, note that $B^{-1} - A^{-1}$ is positive definite. One way to see this is as follows: $$ A - B \succ 0 \implies\\ B^{-1/2}(A - B)B^{-1/2} \succ 0 \implies\\ B^{-1/2}AB^{-1/2} - I \succ 0 \implies\\ I - [B^{-1/2}AB^{-1/2}]^{-1} \succ 0 \implies\\ I - B^{1/2}A^{-1}B^{1/2} \succ 0\implies\\ B^{-1} - A^{-1} \succ 0 $$ From there, it suffices to note that the inverse of a positive definite matrix is itself positive definite. One simple way to see this is to note that a matrix is positive definite iff it is symmetric with positive eigenvalues, but the inverse of a symmetric matrix is symmetric and the inverse of a matrix with positive eigenvalues has positive eigenvalues.

Thus, the fact that $B^{-1} - A^{-1}$ is positive definite implies that $(B^{-1} - A^{-1})^{-1}$ is positive definite.