# Inequality concerning inverses of positive definite matrices

I don't find a way to prove this: given $A$, $B$, symmetric and positive definite:

$$A>B \Rightarrow A^{-1} < B^{-1},$$ where $A>B$ means that $A-B$ is positive definite.

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A>B means x'Ax > x'Bx for whatever x (with ' I mean transpose) – Federico Oct 15 '12 at 8:19
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Oct 15 '12 at 20:59
Ok, thanks. I am self studying linear algebra in the context of control theory. So this is not strictly homework. – Federico Oct 16 '12 at 15:36

First, assume we have solved it when $A=I$. We have, as $A>0$, that $A$ admit a positive define square root $A^{1/2}$. We have
$I-A^{-1/2}BA^{-1/2}>0$. Let $B':=A^{-1/2}BA^{-1/2}>0$. Then $B'^{—1}>I$, hence $A^{1/2}B^{-1}A^{1/2}>I$ and we are done.
Now we solve this case: write $B:=C^2$, where $C>0$. Then for $x\neq 0$, $\lVert Cx\rVert^2<\lVert x\rVert^2$, which gives $\lVert y\rVert^2<\lVert C^{—1}y\rVert^2$ for $y\neq 0$. This gives $C^{—2}>I$ hence $B^{-1}>I$.