Timeline for What're the differences between PCA and autoencoder?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| S Apr 25, 2020 at 13:16 | history | suggested | brazofuerte | CC BY-SA 4.0 |
Slight correction.
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| Apr 24, 2020 at 16:00 | review | Suggested edits | |||
| S Apr 25, 2020 at 13:16 | |||||
| Feb 22, 2018 at 11:45 | comment | added | elliotp | The paper you cite uses a linear autoencoder, i.e., no non-linear activation function. That is why its weights span the same subspace spanned by PCA exactly. | |
| S Jan 23, 2017 at 20:21 | history | suggested | salvador | CC BY-SA 3.0 |
This is confusing since there is always a transfer function. What is meant here is linear TF.
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| Jan 23, 2017 at 17:45 | review | Suggested edits | |||
| S Jan 23, 2017 at 20:21 | |||||
| May 17, 2016 at 9:12 | comment | added | johnblund | This is better suited as a comment but as I lack the reputation for that it will be given as an answer. I somewhat confused with notion of nearly in bayerj:s answer. Reading Neural Networks and Principal Component Analysis: Learning from Examples Without Local Minima where the proof is given. > ''In the auto-associative case ... and therefore the unique locally and globally optimal map W is the orthogonal projection onto the space spanned by the first $p$ eigenvectors of $\Sigma_{XX}$'' Is this then not the exactly the corres | |
| Apr 2, 2016 at 12:48 | comment | added | bayerj | It is the same objective function, which is convex. All solutions will find equivalent minima, but only the PCA solver is constrained to find an orthogonal subspace. | |
| Apr 2, 2016 at 1:29 | comment | added | DikobrAz | Can anybody provide a link to the explanation why linear auto encoder finds the same space as PCA? | |
| Oct 17, 2014 at 9:21 | comment | added | bayerj | Or any other transfer function used. | |
| Oct 16, 2014 at 17:26 | comment | added | RockTheStar | "when the hidden units are activated in the near linear regions", you mean the linear part in the sigmoid function, right? | |
| Oct 16, 2014 at 6:07 | comment | added | bayerj | Yes. (Also it might still be linear in some cases, e.g. when the hidden units are activated in near linear regions.) | |
| Oct 15, 2014 at 22:40 | comment | added | RockTheStar | So, with non-linear transformation, even there is only 1 layer of hidden unit. The solution is still non-linear? | |
| Oct 15, 2014 at 21:13 | comment | added | amoeba | @RockTheStar: it's not the number of layers that matters, but the activation function [transfer function]. With linear transfer function, no number of layers will lead to a non-linear autoencoder. | |
| Oct 15, 2014 at 16:49 | comment | added | RockTheStar | I see! So i need to have two layers for non-linear transformation. So multiple layers means very complex non-linear? | |
| Oct 15, 2014 at 16:48 | vote | accept | RockTheStar | ||
| Oct 15, 2014 at 8:55 | history | answered | bayerj | CC BY-SA 3.0 |