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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.

jub0bs
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Federico
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  • A>B means x'Ax > x'Bx for whatever x (with ' I mean transpose) – Federico Oct 15 '12 at 08:19
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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

2 Answers2

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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 $B^{-1}>I$.

Davide Giraudo
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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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With the identity \begin{aligned} &B^{-1}-A^{-1} \\ =\;&A^{-1}(A-B)A^{-1} + A^{-1}(A-B)B^{-1}(A-B)A^{-1}\\ =\;&((A-B)^{1/2}A^{-1})'((A-B)^{1/2}A^{-1}) + (B^{-1/2}(A-B)A^{-1})'(B^{-1/2}(A-B)A^{-1}), \end{aligned} $B^{-1}-A^{-1}$ can be written as the sum of two positive definite matrices, thereby positive definite.

username
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