Solving Separable Least Square Problem - Isy

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Discussion Overview

The discussion revolves around solving a separable least squares problem involving a known matrix A, a known vector b, an unknown vector x, and an unknown scalar a. Participants explore methods to isolate the scalar a from the vector x within the context of least squares optimization.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using the pseudo-inverse method to solve for x but questions how to isolate a from x.
  • Another participant suggests that without specific properties of x, such as its magnitude being 1, it is not possible to separate a from x, indicating that the solution is inherently a vector.
  • A later reply proposes reformulating the problem as a constrained least squares problem, suggesting the minimization of ||Aax-b||2 subject to the constraint ||x||=1.
  • It is noted that this reformulation would lead to two possible solutions for a, corresponding to both a and -a, which would be associated with x and -x.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of isolating a from x, with some suggesting constraints may allow for a reformulation while others maintain that separation is not possible without specific conditions.

Contextual Notes

The discussion highlights limitations regarding the assumptions needed for isolating a from x and the implications of the constraints on the solution space.

isolde_isy
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Hi I have to solve the following LSQ problem:
min(||Aax-b||2)

where
A is a known matrix,
b is a know vector
x is an unknows vector
a is an unknown scalar

I can solve directly via pseudo inverse
ax=inv(A'A)A'b
but how can I isolate a from x?

Could Separable least square a scheme to follow...
Thnaks in advance
Isy :smile:
 
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Unless you have some particularly property (for example the magnitude of x = 1), you can't "separate" a from x. Your solution is a vector.
 
So it can be reformulate the problem in lsq constrained
min(||Aax-b||2)
subject to ||x||=1

Right?
Thanks Isy
 
isolde_isy said:
So it can be reformulate the problem in lsq constrained
min(||Aax-b||2)
subject to ||x||=1

Right?
Thanks Isy
Yes - but note that there would (for real vector spaces) be two possibilities, since a and -a, associated with x and -x would satisfy.
 

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