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The Frobenius, or Hilbert-Schmidt, norm of an $n$ by $n$ matrix $A$ is defined as $\|A\|_2 = \sqrt{\sum_{i,j=1}^n |A_{ij}|^2}$. The absolute value of $A$ is the unique positive matrix $|A|$ satisfying $|A|^2 = A^* A$. Are there any known relations between $\| |A| \|_2$ and $\|A\|_2$?

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They are equal, since the Hilbert-Schmidt norm of A is the square root of the trace of $A^*A$ –  Yemon Choi Oct 5 '10 at 10:32
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closed as off-topic by Ricardo Andrade, Daniel Moskovich, Chris Godsil, Jack Huizenga, Suvrit Dec 1 '13 at 6:13

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1 Answer

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It holds that

|||A|||_2 =||A||_2

Proof: Let the singular value decomposition of A be given by $A=U\Sigma V^H$. Since the HS-norm is invariant under unitary transformations, $||A||_2= ||\Sigma||_2$ holds. As for $|A|$, we obtain $|A|= |A^HA|=|V\Sigma U^HU\Sigma V^H|=|V\Sigma^2 V^H|= V\Sigma V^H$. Again, unitary invariance yields $|| |A| ||_2=||\Sigma ||_2=||A||_2$, qed.

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See my comment above –  Yemon Choi Oct 5 '10 at 18:27
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