I have an n x n similarity matrix which I'd like to reduce to a smaller square matrix. I am aware of this answer:

How to use SVD for dimensionality reduction to reduce the number of columns (features) of the data matrix?

The answer shows how you can use SVD to reduce the number of columns of a matrix. Essentially, if $A= USV^T$, then taking $US$, gives a matrix with fewer columns than the original.

Therefore, I know how to reduce the number of columns using SVD, but how about the number of rows and columns?

Can I apply the procedure twice, first to reduce the columns, then to the transpose of the result to reduce the rows?

If my similarity matrix is symmetrical; does it make the problem simpler?

  • 1
    $\begingroup$ svd does not reduce the number of "columns" or "rows", it helps to reduce dimensionality of the space which is spanned by those columns, as well as rows, as vectors. So your question is unclear - what do you want? $\endgroup$
    – ttnphns
    Dec 16, 2018 at 13:02
  • $\begingroup$ @ttnphns see the link the question. You can use SVD to reduce the number of columns. I.e. if $A= USV^T$, then the column reduced matrix is given by $U_kS_k$ where k is the top k columns from U, and the top k rows from the diagonal of S. $\endgroup$
    – usgroup
    Dec 17, 2018 at 13:40
  • 1
    $\begingroup$ Reduction of dimensionality is often conceptualized in terms of latent variables. People typically have data matrices w/ columns as variables & rows as observations. Assuming your setup is like that, cluster analysis is generally used for observations, & factor analysis is often used for features. $\endgroup$ Dec 17, 2018 at 15:03
  • 2
    $\begingroup$ However, you can cluster variables. (This typically leads to worse fitting solutions, but does enforce simple structure.) From there, note that there are biclustering methods that cluster rows and columns simultaneously. That isn't SVD, & isn't specific to the special case of symmetrical matrices, but may get you where you want to go $\endgroup$ Dec 17, 2018 at 15:03
  • $\begingroup$ thanks @gung, that's good advise. I'll have a look at biclustering. I actually wanted to do some dimension reduction to make a similarity matrix smaller so that I could cluster it, but the biclustering stuff seems fascinating anyway. $\endgroup$
    – usgroup
    Dec 18, 2018 at 9:26


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