On Wed, Aug 13, 2014 at 11:57 PM, Peter Brady <subscripti...@simonplace.net> wrote: > Hi All, > > I've inherited some R code that I can't work out what they've done. It > appears to work and give sort of reasonable answers, I'm just trying to > work out why they've done what they have. I suspect that this is a > simple vector identity that I've just been staring at too long and have > forgotten... > > The code: > > GGt <- M0 - M1 %*% M0inv %*% t(M1) > svdGG <- svd(GGt) > Gmat <- svdGG$u %*% diag(sqrt(svdGG$d)) > > It is supposed to solve: > > G*G^T = M0 - M1*M0^-1*M1^T > > for G, where G^T is the transpose of G. It is designed to reproduce a > numerical method described in two papers: > > Srikanthan and Pegram, Journal of Hydrology, 371 (2009) 142-153, > Equation A13, who suggest the SVD method but don't describe the > specifics, eg: "...G is found by singular value decomposition..." > > Alternatively, Matalas (1967) Water Resources Research 3 (4) 937-945, > Equation 17, say that the above can be solved using Principle Component > Analysis (PCA). > > I use PCA (specifically POD) and SVD to look at the components after > decomposition, so I'm a bit lost as to how the original matrix G can be > constructed in this case from only the singular values and the left > singular vectors.
GG' is a symmetric matrix, so left- and right-singular vectors are the same. If I recall right, in general it is impossible to find G from GG' (I denote the transpose by ') since, given an orthogonal transformation U (that is, UU'=1), GUU'G' = GG', so you can only find G up to multiplication with an orthogonal transformation matrix. Since SVD decomposes a matrix X = UDV', the decomposition for GG' is GG' = UDU'; setting S = sqrt(D) (i.e., diagonal matrix with elements that are sqrt of those in D), GG' = USSU' = USS'U', so one solution is G = US which is the solution used. You could use PCA on G, which is roughly equivalent to doing SVD on GG' (up to centering and scaling of the columns of G). I am not very familiar with PCA in R since I always use SVD, but here's what the help file for prcomp (PCA in R) says: The calculation is done by a singular value decomposition of the (centered and possibly scaled) data matrix, not by using ‘eigen’ on the covariance matrix. This is generally the preferred method for numerical accuracy. HTH, Peter > Like I said earlier, I suspect that this is a simple > array identity that I've forgotten. My Google Fu is letting me down at > this point. > > My questions: > 1) What is the proof, or where can I better find it to satisfy myself, > that the above works? > > 2) Alternatively, can anyone suggest how I could apply PCA in R to > compute the same? > > Thanks in advance, > -pete > > -- > Peter Brady > Email: pdbr...@ans.com.au > Skype: pbrady77 > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.