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I have an $(n\times n)$ real matrix obtained through the Hadamard product, $H=A\circ B$, of two real $(n\times n)$ symmetric matrices $A,B$. All elements of $A$ are positive, while diagonal elements of $B$ are zero and the rest are non-negative.

My question is if there is any way I can relate (possibly through some inequalities) the Frobenius norm of $H$ to the ones of $A, B$. The Frobenius norm of a real $(n\times n)$ matrix $M$ is defined as $$ \| M \|_F = \sqrt{Tr\left(M^T M\right)} =\sqrt{\sum_{i,j=1}^n M_{i,j}^2} \,. $$

Thanks!

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  • $\begingroup$ Hint: Cauchy Schwarz $\endgroup$ Dec 11, 2018 at 10:55

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You have $\|A\circ B\|_F\leq\operatorname{tr}(AB^T)\le\|A\|_F\|B\|_F$.

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    $\begingroup$ Thanks for the reply. However, I do not see how you get the first inequality. Can you pls explain? $\endgroup$
    – AG1123
    Jan 26, 2017 at 16:30
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    $\begingroup$ @AG1123 Try to write down the squares of both sides in terms of the elements of $A$ and $B$. $\endgroup$
    – user1551
    Jan 27, 2017 at 0:21
  • $\begingroup$ Yep, I see it now. Thanks! $\endgroup$
    – AG1123
    Jan 27, 2017 at 17:58

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