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