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I have two diagonal, positive definite matrices $A$ and $B$, and I have that the angle between $Ax$ and $x$ is smaller than that of $Bx$ and $x$:

$$\frac{x^T A x}{\|Ax\|_2} \geq \frac{x^T B x}{\|Bx\|_2}.$$

Intuitively, this means (to me) that $A$ is somehow "closer to identity" than $B$, and so the same statement should also hold for the inverse:

$$\frac{x^T A^{-1} x}{\|A^{-1}x\|_2} \geq \frac{x^T B^{-1} x}{\|B^{-1}x\|_2},$$

as if $A$ is closer to identity than $B$, so should $A^{-1}$ be closer to identity than $B^{-1}$.

Somehow though, I cannot prove this, or find a counterexample...

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  • $\begingroup$ If you replace in the first inequality $x$ by $(AB)^(-1/2)x$ you end up with $x^TB^(-1)x/\|A^(1/2)B^(-1/2)x\|\geq x^TA^(-1)x/\|A^(-1/2)B^(1/2)x\|$ $\endgroup$ – user35593 Apr 23 at 7:01

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