I am trying to get a feel for which DR scheme is most suitable for my problem, with the stipulation that the scheme is bijectiv; i.e. there is a two-way map between the low-dimensional manifold and the original space.


For clarity: is it possible to select your dimensional reduction so that you can project UP from the low dimensional space that you have found, such that information loss is acceptable. E.g. say I have a high-dimensional joint configuration of some character, and I want to find its low-dimensional joint projection and then traverse that manifold, or sample it, and visualise those samples in the original space.

  • 1
    $\begingroup$ I do not see how a dimension reduction scheme could be bijective in this sense. (In essence this is part of the definition of dimensionality of a space) However in principle any dimension reduction scheme could be bijective in the sense of preserving the data. For example if the data lie exactly on a hyperplane, then PCA would be able to preserve the data while dropping some eigenvectors. $\endgroup$ – GeoMatt22 Sep 9 '16 at 16:48
  • 2
    $\begingroup$ (My comment was posted before your edit.) Note that if PCA is used as a dimension reduction scheme, then the reduced basis-set will no longer map to the entire original space. If you keep all the eigenvectors, then PCA is just a rotation of the coordinate axes in the original space, but does not reduce the number of dimensions! $\endgroup$ – GeoMatt22 Sep 9 '16 at 16:50
  • $\begingroup$ Well no they can be, you will have information loss of course, but you should still just be able to map it back. $\endgroup$ – Astrid Sep 9 '16 at 16:51
  • 2
    $\begingroup$ I am not sure what you are asking. It sounds like it may be more "what alternative coordinate systems for $\mathbb{R}^n$ are bijective?". For that question, the answer is: those with an invertible Jacobian. $\endgroup$ – GeoMatt22 Sep 9 '16 at 16:58
  • 2
    $\begingroup$ You might find this question helpful. BTW I found that by searching on the manifold learning tag, which you might consider adding to your question. $\endgroup$ – GeoMatt22 Sep 9 '16 at 17:12

So in general your question is a bit weird, dimensionality reduction is related to projection which, as the name says, puts something onto something but many things can be put on the same thing. (Think about a point in 3D that is projected on a surface then think about the perpendicular line to the surface passing through that point. every point on that line is projected onto the same point)

Ok let's consider PCA which you claim is bijective (it really isn't), actually consider SVD since PCA is really just SVD.

Full SVD is:

$ M = U\Lambda V^T$

when you consider dimensionality reduction, you only consider the first largest elements of the $\Lambda$, a diagonal matrix, hence the representation

$\hat M = U\hat\Lambda V^T$

now this representation can correspond to an infinite number of matrices $M$. In other words, you can build an infinite number of matrices $M'$ such that, when applying the same process of SVD truncation, you get $\hat M$ (so it's really not bijective)

Indeed: let

$M' = U \Lambda' V^T$

where $\Lambda'$ has the same first few singular values than $\Lambda$ and all the others are different (but smaller). The SVD truncation of all these $M'$ will give you $\hat M$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.