3
$\begingroup$

Given A and B two real symmetric positive definite matrices is it true that, for some norm $\|.\|$, this inequality holds

$$ \|AB-I\| \leq \|A^2B^2-I\| \qquad ? $$

$\endgroup$
  • $\begingroup$ Is there a motivation for this question? Where does it come from? $\endgroup$ – Julien Feb 21 '13 at 21:58
  • $\begingroup$ @julien : A friend which is working on some statistic estimation project asked me. I have to admit that I don't know the background, but I tried to solve it and got stuck ... $\endgroup$ – Student Feb 21 '13 at 22:06
4
$\begingroup$

Yes, the Frobenius norm $\|M\|_F=\|\operatorname{vec}(M)\|_2$ will do. Since $A,B$ are positive definite, so is $B\otimes A$. Therefore all eigenvalues of the symmetric matrix $B\otimes A + I$ are larger than $1$ and hence \begin{align*} \|A^2B^2-I\|_F &=\|A(AB-I)B + AB-I\|_F\\ &=\|(B\otimes A + I)\operatorname{vec}(AB-I)\|_2\\ &\ge\|\operatorname{vec}(AB-I)\|_2\\ &=\|AB-I\|_F. \end{align*} Equality holds if and only if $AB=I$.

$\endgroup$
  • 1
    $\begingroup$ awesome ! Is it something classical or you just come out with that ? $\endgroup$ – Student Feb 21 '13 at 23:02
  • $\begingroup$ @Student I just tried a few norms. Numerical experiements showed that the inequality holds for some other matrix norms, too, such as the spectral norm, but I simply picked the one that makes the inequality the easiest to prove. So I don't know if this is a known result or not, or whether the inequality holds for a broader class of norms. $\endgroup$ – user1551 Feb 22 '13 at 0:13

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.