LSA is better way for extracted new concepts from large text documents collections .. in the following example :

i have spend lot of time in Google to get explanation about the following

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My questions are:

The result matrix Vtk .. similarity matrix can be build based on cosine similarity for all documents .. now how do we use this matrix for document clustering ?

can some one give a brief explanation how can we get Ak matrix .. is this matrix contain the original Term-Document matrix with reduced version?

thanks for any suggestion


You never compute Ak. Too expensive. It will usually be less sparse than A, so even worse. Ak is a reconstruction of the original term-document matrix.

For clustering, you could use V_k, the documents x topics matrix.

But usually you wouldn't even do clustering, you would hope the topics (factors) already are what you are looking for.

As a matter of fact, you can formalize k-means the very same way, as a matrix decomposition. Uk is the matrix storing your centroids, sigma is identity, and Vk is a binary matrix with exactly one 1 for every document. k-means is a matrix factorization.

  • $\begingroup$ regard this note : For clustering, you could use V_k, the documents x topics matrix. .. so is k-means working on this matrix .. is this right? $\endgroup$ – Ray ben Nov 16 '16 at 18:38
  • $\begingroup$ Do you mean im using NMF (non-negative matrix factorization) where NMF give two matrices are U and V and applying k-means. is this right or do you mean some thing else? $\endgroup$ – Ray ben Nov 16 '16 at 18:45
  • $\begingroup$ No. I am saying that k-means is also a variant of NMF. And why would you apply NMF twice in a row? You certainly can, but it lacks any good reason to do so. $\endgroup$ – Has QUIT--Anony-Mousse Nov 17 '16 at 0:27
  • $\begingroup$ ok, so the result matrix Vk for documents x Topics ... how do we clustering similar documents .. ? can i use cosine to find ( documents x documents ) matrix similarity based in cosine .. then applying such algorithm ? what do you think about that ? thanks $\endgroup$ – Ray ben Nov 17 '16 at 7:49
  • $\begingroup$ You have the "clusters" right there in the matrix. You are done. $\endgroup$ – Has QUIT--Anony-Mousse Nov 17 '16 at 7:52

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