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I have a question about the properties of the Frobenius and L$_{2,1}$ norms. Why is the L$_{2,1}$ norm more robust to the outlier than the Frobenius norm?

PS: For a matrix $A\in\mathbb{R}^{n\times d}$, it can be easily seen that $$ \text{Frobenius norm:}\qquad\Vert A\Vert_F= \left(\sum_{i=1}^{n}\sum_{j=1}^{d}\vert a_{i,j}\vert^2 \right)^{\frac{1}{2}}=\sum_{i=1}^{n}\Vert A(i,:)\Vert_2^2, $$ and $$ L_{2,1}\,\,norm: \qquad\Vert A\Vert_{2,1}=\sum_{i=1}^{n}\left(\sum_{j=1}^{d}\vert a_{i,j}\vert^2 \right)^{\frac{1}{2}}=\sum_{i=1}^{n}\Vert A(i,:)\Vert_2,$$ where $A(i,:)$ is the $i$-th row of $A$.

I would be very grateful if some could answer my question.

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I have only a couple of hints:

  1. Frobenius norm, by definition, takes equal account of all data in matrix (all rows and columns).
  2. Whereas $L_{2,1}$ norm is Frobenius norm but per row instead, so outliers in other rows do not affect (equally) norm of current row.
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  • $\begingroup$ Thank you so much. $\endgroup$ – Math-Data Feb 21 at 14:22

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