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If $A$ and $B$ are positive definite matrices, and $A-B$ is also positive definite, is $B^{-1}-A^{-1}$ necessarily positive definite?

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2 Answers 2

A more conventional proof would be something like this: Let $A^{1/2}$ be a Hermitian square root of $A$ (such a square root always exists: since $A$ is positive definite, it can be unitarily diagonalized to $UDU^\ast$; now, take $A^{1/2}=UD^{1/2}U^\ast$ where $D^{1/2}$ is the entrywise square root of the positive diagonal matrix $D$). Let let $\sim$ denotes the $\phantom{}^\ast$congruence ("star-congruence") relation, i.e. $X\sim Y$ iff $X=P^\ast Y P$ for some invertible (but not necessarily unitary) matrix $P$. Note that positive definiteness is preserved under $\phantom{}^\ast$congruence. That is, if $X\sim Y$, then $X$ is positive definite if and only if $Y$ is positive definite.

Let $C = A^{-1/2}BA^{-1/2}$. Then $C(\sim B)$ is positive definite. Hence $C=W^\ast\Lambda W$ for some unitary matrix $W$ and some diagonal matrix $\Lambda$. Now, \begin{align} A-B &= A^{1/2} (I - A^{-1/2}BA^{-1/2}) A^{1/2}\\ &\sim I - C\\ &= W^\ast(I-\Lambda)W\\ &\sim I - \Lambda,\\ B^{-1} - A^{-1} &= A^{-1/2} (A^{1/2}B^{-1} A^{1/2} - I) A^{-1/2}\\ &\sim C^{-1} - I\\ &= W^\ast(\Lambda^{-1}-I)W\\ &\sim \Lambda^{-1} - I. \end{align} So the problem boils down to the special and trivial case where $A=I$ and $B=\Lambda$.

In the above, actually we have also prove the following proposition, which is included in some textbooks:

Proposition. If $A$ is positive definite and $B$ is Hermitian, then there exists a nonsingular matrix $P$ such that $P^\ast AP=I$ and $P^\ast BP$ is diagonal.

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It's true. Let me write $B\ge0$ if $B$ is positive definite. First, let $B, C\ge0$ and consider $$\begin{aligned} (B+C)^{-1}&=\bigl(B^{1/2}(I+B^{-1/2}CB^{-1/2})B^{1/2}\bigr)^{-1}\\ &=B^{-1/2}(I+B^{-1/2}CB^{-1/2})^{-1}B^{-1/2}\\ &=B^{-1/2}\sum_{n=0}^\infty(-B^{-1/2}CB^{-1/2})^nB^{-1/2}, \end{aligned}$$ which holds if $C$ is small enough. So replace $C$ by $tC$ and take the derivative at $t=0$ (only the $n=1$ term takes part): $$\left.\frac{d}{dt}(B+tC)^{-1}\right|_{t=0}=-B^{-1}CB^{-1}\le0.$$ I could have done this calculation at any $t>0$, so in fact $$\frac{d}{dt}(B+tC)^{-1}\le0.$$ Integrating this from $t=0$ to $t=1$ yields $(B+C)^{-1}\le B^{-1}$. Now let $C=B-A$.

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How do you know that $C=B-A$ is small enough for series expansion of $(I+B^{-1/2}CB^{-1/2})^{-1}$? –  user1551 Nov 14 '12 at 8:26
    
@user1551: I don't. But $tC$ is, for small enough $t$, and that is all I need. That said, I must admit that the proof in the question noted in the comments above as a duplicate of this one is better than mine. (I also could have polished mine a bit, making the structure clearer, but in light of that other proof, I didn't want to.) –  Harald Hanche-Olsen Nov 14 '12 at 8:31

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