# The robustness of the Frobenius and L2,1 norms to the outlier

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.

## 1 Answer

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.
• Thank you so much. – Math-Data Feb 21 at 14:22