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...)

  • $\begingroup$ All norms are equivalent, so the phenomenon you mention in your second paragraph cannot happen. $\endgroup$ Commented Jan 12, 2011 at 18:28
  • $\begingroup$ ...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 $n$? $\endgroup$
    – shabbychef
    Commented Jan 12, 2011 at 18:31
  • $\begingroup$ See this. According to the article, if a set of data contains outliers, then an $L_1$ norm might be better to use. $\endgroup$ Commented Jan 12, 2011 at 18:33

2 Answers 2


The Frobenius norm is actually quite nice, and also natural. it is defined by merely $$\|A\|_F^2 = \mbox{trace}(A'A)$$

and since it is naturally an inner-product norm, it makes optimization, etc. with it much easier (think quadratic programs, instead of semidefinite programs)

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).

Otherwise, what norm you use, should be governed by the application where you are trying to use it.


As Joel Tropp puts it:

Frobenius norm error bounds are typically vacuous.

"An Introduction to Matrix Concentration Inequalities," (arxiv), page 84

The reason is explained there, and also in the Appendix of this paper. Essentially, noise shows up as a long tail of singular values that are individually much smaller than the leading singular value, but when summed up, may exceed the leading singular value. The Frobenius norm is the sum of squares of the singular values, and hence, you are just measuring noise--the signal has little effect. The spectral norm, on the other hand, is just the leading singular value--and hence it is measuring the actual signal.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .