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Oct 28, 2017 at 8:07 history bounty ended Mustafa Eisa
Oct 27, 2017 at 17:59 vote accept Mustafa Eisa
Oct 27, 2017 at 1:37 comment added Mustafa Eisa Sorry I mean Robust PCA is not invariant due to the entry-wise penalty. However replacing that with a sum of row norms makes it so.
Oct 27, 2017 at 1:35 comment added David J. Harris Now I'm confused. PCA is already rotationally invariant, right?
Oct 26, 2017 at 19:59 comment added Mustafa Eisa No it’s just an extension of PCA that incorporates rotational invariance.
Oct 26, 2017 at 19:28 comment added David J. Harris Oh, I see. Would that be a special case of Mahalanobis distance?
Oct 26, 2017 at 18:22 comment added Mustafa Eisa If your answer is, “No” that’s totally fine I’m just wondering.
Oct 26, 2017 at 18:21 comment added Mustafa Eisa In your answer, you distinguish between the two methods by pointing out that the $\ell_1$ penalty in robust PCA is not rotationally-invariant and so is better suited to corruptions in the canonical basis. I’m just asking if you’ve considered or thought about the case in which a sum of (Euclidean) row norms is used in place of the $\ell_1$ coordinate penalties.
Oct 26, 2017 at 18:09 comment added David J. Harris I'm not sure I understand your question. Are you asking me to compare the two approaches you discussed in your question with a different robust PCA approach?
Oct 26, 2017 at 17:27 comment added Mustafa Eisa Thanks for this David, I will take a look at the paper. However, there is a version of robust PCA which imposes a rotationally-invariant penalty on the datum (rows of the data matrix) instead of a penalty on coordinates (such as in the Candes case). Thoughts?
Oct 26, 2017 at 15:21 history answered David J. Harris CC BY-SA 3.0