# Angle inequality for inverse of PD diagonal matrix

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...

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