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Let $A$ be a matrix in $SL_2(\mathbb R)$. Define the trace norm to be

$$\|A\| = \sqrt{\mathrm{tr}(A^* A)}. $$

Is it true that this norm satisfies some kind of multiplicative property; for example:

$$\|AB\| \leq \|A\|\cdot\|B\|.$$

Can someone give me a brief reference where basic properties of this norm are stated and proved?

Thanks.

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    $\begingroup$ Its true. Here is a reference: matrixanalysis.com/Page279.pdf $\endgroup$ – morgan Sep 17 '11 at 6:34
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    $\begingroup$ It’s also called the Frobenius norm. There’s a brief proof of the inequality that you mention in this short PDF. $\endgroup$ – Brian M. Scott Sep 17 '11 at 6:35
  • $\begingroup$ @Alex: Thank you very much! If you could add that as an answer, and if no other answer comes along, then I could accept it! $\endgroup$ – Derek Sep 17 '11 at 6:37
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As I said in my comment, this is also (and perhaps more commonly) called the Frobenius norm. The argument in the PDF that alex and I mentioned, showing that $\|AB\}\le\|A\|\cdot\|B\|$, also shows that the Frobenius norm of an $m\times n$ matrix $A$ is simply the ordinary Euclidean $2$-norm of $A$ (given by the Pythagorean theorem) if you think of $A$ as a vector in $\mathbb{R}^{mn}$, say be reading it out by rows. Thus, the basic properties of the Frobenius norm follow immediately from those of the $2$-norm.

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