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I am currently reading a scientific paper about clustering of brain signals, which consist on long time series across many channels (each signal is a matrix of C channels by T time samples). In the preprocessing of their datas, the authors normalize each signal matrix by dividing it with its Frobenius norm. My problem is that they don't even say why they do so... is this so obvious that I can't see it?

Any thought?


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This is difficult to answer without knowing more about what they do with the matrix once they normalize it. – Rahul Apr 13 '12 at 18:08
Clustering ! :) Precisely, ascendant hierarchical clustering using Ward's method. – CTZStef Apr 13 '12 at 18:15
up vote 3 down vote accepted

My educated guess is that they want to compare relative values, not absolute ones, for purposes of clustering. Perhaps they chose the Frobenius norm because it is easy to calculate.

edit: So the Frobenius norm was chosen out of computational considerations. It is easy to compute because it does not necessitate SVD, as required of the spectral norm. Moreover, it is easy to update when expanding or reducing the matrix.

Once you normalize something by its norm, it is obviously going to have a unit norm (i.e., lie on the unit sphere).

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I quote the paper : "Finally, each signal matrix was normalized through division by the Frobenius norm of the matrix. This operation transform each matrix to a point on the surface of the unit sphere in vector space R^CxT" Frobenius is only for easiness of computation? Why is it easier? – CTZStef Apr 13 '12 at 18:20
@CTZStef, Emre: Another good reason to use the Frobenius norm is that operator norms like the spectral norm make sense for linear transformations, but the matrix in question doesn't really represent a linear transformation. It's just a collection of $C\times T$ samples. So it makes sense to compute its norm as a vector in $\mathbb R^{C\times T}$, which happens to be the Frobenius norm of the corresponding matrix. – Rahul Apr 13 '12 at 22:55

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