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

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

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Is there a motivation for this question? Where does it come from? –  1015 Feb 21 '13 at 21:58
    
@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 ... –  Student Feb 21 '13 at 22:06

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up vote 4 down vote accepted

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 $AB=I$.

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awesome ! Is it something classical or you just come out with that ? –  Student Feb 21 '13 at 23:02
    
@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. –  user1551 Feb 22 '13 at 0:13

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