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I have read many of the questions already here in regards to the Frobenius norm, but they do not help me too much.

My question is, why is the Frobenius norm not considered a 'proper' norm?

In a book I am reading, (p.143), we have a square real matrix $\mathbf W$, that needs to be (iteratively) symmetrically orthogonalized. To start off, we first create a random matrix, and then normalize it as such:

$$ \mathbf W = \mathbf W / ||(\mathbf W)||_{\; 1,\; or\; 2\; ,\; or \; \infty,\;\text{but not frobenius}} $$

As the matrix becomes symmetrically orthogonalized during loops of the iteration, we will approach the condition that:

$$ \mathbf {W^TW \approx I} $$

Anyway, my book states the following:

...the norm in the normalization of $\mathbf W$ must be chosen appropriately. That is, it must be a proper norm in the space of matricies; most conventional norms have this property, except for the Frobenius norm.

Why is this the case? What makes it a 'proper' norm, vs the matrix $l_1$, $l_2$, or $l_\infty$ matrix norms? Why is $l_\text{Fro}$ a bad guy?

Thanks!

EDIT:

Here is the full text: Question is labelled in red:

enter image description here enter image description here

(Start reading from symmetric orthoginalization at near top of first page).

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    Where in what book?2012-08-30
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    @JonasMeyer Thanks, I have edited with context and book to make it clearer.2012-08-30
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    I'm not sure what orthogonalized means unless you are in a Hilbert space, which restricts the choice of norm?2012-08-30
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    @copper.hat Oh, basically what we want to do, is make all the columns of the matrix $\mathbf W$ orthogonal to each other.2012-08-30
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    I realize that. I think without more details, it is hard to guess what the issue is here. The Frobenius norm is not an induced norm, if that matters.2012-08-30
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    @copper.hat Ok, I have added full context. :-)2012-08-30

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