Re: least-squares problem: B ~ X.A

*To*: mathgroup at smc.vnet.net*Subject*: [mg57402] Re: least-squares problem: B ~ X.A*From*: dh <dh at metrohm.ch>*Date*: Fri, 27 May 2005 04:56:39 -0400 (EDT)*References*: <d71ipr$46a$1@smc.vnet.net> <d742rb$ipm$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Pascal, If the columns of A are linearly independent PseudoInverse[A] gives the left Inverse. If the rows of A are linearly independent PseudoInverse[A] gives the right Inverse. Now for a 3 x 60 matrix the coulmns can not be linear independent, however the chance is big that the rows are nearly linear independent and you get something close to a right inverse. Further, PseudoInverse[A].A-I and A.PseudoInverse[A]-I with I an identity matrix of corresponding dimension, have minimal sum of squares of elements. Sincerely, Daniel Pascal wrote: > I received the suggestion that what I look for is B.PseudoInverse[A]. > Thanks a lot to Daniel and Ssezi. > I indeed thought of that in the first place but, being a dummy in > maths, I had a doubt. Is such an easy solution rigorously the > bestleast-squares solution of B=X.A ? > > Anyway, I checked on my data that PseudoInverse[A] is a very good quasi > right inverse of A, but a poor quasi left inverse. A priori wouldn't > X.A = B.PseudoInverse[A].A be a poor approximation of B? Dead wrong? > > I also tried Transpose[PseudoInverse[Transpose[A]]] but I again only > obtain an excellent quasi right inverse of A and nearly the same poor > quasi left inverse. > > Thanks for any help. > Pascal. >