[Date Index]
[Thread Index]
[Author Index]
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.
>
Prev by Date:
** a dangerous feature of Module[]?**
Next by Date:
**Re: Re: Applying a list of functions to a list of arguments**
Previous by thread:
**Re: least-squares problem: B ~ X.A**
Next by thread:
** Fourier Integrals**
| |