# Why should I avoid the Frobenius Norm?

I vaguely remember the Frobenius matrix norm ( ${||A||}_F = \sqrt{\sum_{i,j} a_{i,j}^2}$ ) was somehow considered unsuitable for numerical analysis applications. I only remember, however, that it was not a subordinate matrix norm, but only because it did not take the identity matrix to $1$. It seems this latter problem could be solved with a rescaling, though. I don't remember my numerical analysis text considering this norm any further after introducing this fact, which seemed to be its death-knell for some reason.

The question, then: for fixed $n$, when looking at $n \times n$ matrices, are there any weird gotchas, deficiencies, oddities, etc, when using the (possibly rescaled) Frobenius norm? For example, is there some weird series of matrices $A_i$ such that the Frobenius norm of the $A_i$ approaches zero while the $\ell_2$-subordinate norm does not converge to zero? (It seems like that can not happen because the $\ell_2$ norm is the square root of the largest eigenvalue of $A^*A$, and thus bounded from above by the Frobenius norm...)

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All norms are equivalent, so the phenomenon you mention in your second paragraph cannot happen. – Mariano Suárez-Alvarez Jan 12 '11 at 18:28
...for finite dimensional vector spaces. Perhaps the argument against Frobenius norms relied on a sequence where $n$ diverged to infinity, and is irrelevant for fixed $n$? – shabbychef Jan 12 '11 at 18:31
See this. According to the article, if a set of data contains outliers, then an $L_1$ norm might be better to use. – PEV Jan 12 '11 at 18:33

The Frobenius norm is actually quite nice, and also natural. it is defined by merely $$\|A\|_F^2 = \mbox{trace}(A'A)$$
Numerical analysis probably like the operator norm perhaps because they often exploit $\|Ax\| \le \|A\| \|x\|$, and if you use the operator-2 norm, you get a tighter inequality (in general).