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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.


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