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

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

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.

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