4
$\begingroup$

Let $A$ be a $n\times n$ real (symmetric) positive definite matrix with spectrum contained in $[m, M]$ and also let $X$ be an $n\times p$ matrix such that $X'X=I_p$. It is known that $$\frac{(M+m)^2}{4Mm}(X'AX)^{-1}-X'A^{-1}X \tag{$\star$}$$ is positive semidefinite. Is it true that $$\frac{(M+m)^4}{16M^2m^2}(X'AX)^{-2}-(X'A^{-1}X)^2 \tag{$\dagger$}$$ is also positive semidefinite?

$\endgroup$
  • $\begingroup$ Yes, you comment is right. We know generally, for psd of $S,T$, psd of $S-T$ does not imply psd of $S^2-T^2$. $\endgroup$ – Sunni Aug 24 '11 at 20:04
2
$\begingroup$

From maximizing the given condition, denote it by $(\star)$, we obtain, $$ \left( \frac{(M+m)^2}{4Mm}\frac{1}{m}-\frac{1}{M}\right) I \succeq (\star)\succeq 0 $$ and thus, $(M+m)^2-4m^2\geq0$ which gives us nothing other than $M\geq m$. On the other hand, minimizing the second expression, say $(\dagger)$, we obtain $$ \frac{(M+m)^4}{16M^2m^2}\frac{1}{M^2}-\frac{1}{m^2} = \frac{(M+m)^4-(2M)^4}{16M^4m^2} $$ and this expression is positive semi-definite if and only if $m\geq M$. Therefore combining both, if $M=m$ then $(\dagger)$ is positive semi-definite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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