Solving Separable Least Square Problem - Isy

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To solve the separable least squares problem min(||Aax-b||2), where A is a known matrix and b is a known vector, isolating the unknown scalar a from the unknown vector x is challenging without specific constraints. The discussion suggests reformulating the problem with a constraint, such as ||x||=1, to facilitate finding a solution. However, this approach introduces ambiguity, as both a and -a can yield valid solutions corresponding to x and -x. Therefore, while a constrained least squares formulation can be applied, it does not eliminate the inherent duality in the solutions. The key takeaway is that additional constraints or properties are necessary to uniquely determine a and x.
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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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