-1
$\begingroup$

This question already has an answer here:

Let $X,Y$ be some invertible positive definite matrices such that $X>Y>0$, how do we prove that $Y^{-1}>X^{-1}>0$?

Any hint is greatly appreciated! I just need some initial directions.

Thanks a lot!!

$\endgroup$

marked as duplicate by Igor Rivin, Did, Xander Henderson, The Phenotype, Saad Mar 20 '18 at 0:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Does the inequality mean that the difference is positive definite? $\endgroup$ – Igor Rivin Mar 19 '18 at 16:45
  • $\begingroup$ Yes yes! Sorry for not clearing it up. $X>Y$ means that $X-Y>0$. $\endgroup$ – Penthrite Mar 19 '18 at 16:48
0
$\begingroup$

Can you adapt the basic fact about real numbers $$\dfrac{1}{y} - \dfrac{1}{x} = \dfrac{x-y}{xy}$$ to apply to matrices? (I haven't thought any further on this to see if it works out, but it's what I'd start with.)

Edit: Thinking about it some more, you'd immediately hit a problem with matrix multiplication not being commutative. I imagine that there's a way to work around that, but I haven't got it yet.

$\endgroup$
-1
$\begingroup$

Let $\Delta Y= X-Y$. Then $$\frac{\mathrm{d}}{\mathrm{d}t}\frac{1}{Y+t\Delta Y}=-\frac{1}{Y+t\Delta Y}\Delta Y\frac{1}{Y+t\Delta Y}$$ so that $$\frac{1}{Y}-\frac{1}{X}=\int_0^1\mathrm{d} t\frac{1}{Y+t\Delta Y}\Delta Y\frac{1}{Y+t\Delta Y}\text{.}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.