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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
    
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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
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1 Answer

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 $B^{-1}>I$.

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I am not sure I understand: with y=STx, T^(-1)y=T^(-1)STx. How can you tell that the norm of this is equal to the norm of Sx ? T is not orthogonal. –  Federico Oct 16 '12 at 15:46
    
@Federico My first attempt, as you pointed out, was wrong. I think this one is correct. –  Davide Giraudo Oct 16 '12 at 18:00
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