Solving a set of nonlinear quadratic equations

AI Thread Summary
The discussion revolves around solving a system of nonlinear quadratic equations represented by the equation ϒϒ'C – ϒα = B, where ϒ is a column vector and α is a scalar. Participants highlight the challenges in solving this equation, particularly when B is non-zero, as it complicates the relationship between the variables. There is a debate on whether the formulation is correct, with suggestions that B should either be a zero vector or proportional to ϒ for a solution to exist. The conversation also touches on the need for a clear algorithm or code to approach the problem effectively. Overall, the complexity of the equations and the conditions for solvability are central to the discussion.
Tilfani
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I would like to solve this system, which is a sets of non linear quadratic equations, the system needed to be solved can be expressed in general as follow:

ϒϒ'C – ϒα = B

Where ϒ=(ϒ1,ϒ2,...ϒn)’ is a column vector and ϒ’ its transpose

C=(c1,c2,…,cn)’ and B=(b1,b2,…bn)’ are a columns vector

And α is a reel scalar

I would like to solve for ϒ, with approximatively about 30

Can someone propose me an algorihm/method to solve this system.
also a code to do it wil be very useful.
Bests
 
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Tilfani said:
I would like to solve this system, which is a sets of non linear quadratic equations, the system needed to be solved can be expressed in general as follow:

ϒ’ϒC – ϒα = B

Where ϒ=(ϒ1,ϒ2,...ϒn)’ is a column vector and ϒ’ its transpose

C=(c1,c2,…,cn)’ and B=(b1,b2,…bn)’ are a columns vector

And α is a reel scalar

I would like to solve for ϒ, with approximatively about 30

Can someone propose me an algorihm/method to solve this system.
also a code to do it wil be very useful.
Bests
Y' Y is a scalar, ##\vec Y \cdot \vec Y = |\vec Y|^2##. If you know the value of ##\vec Y \cdot \vec Y## (but don't know the components of ##\vec Y##), you can rewrite the equation above as a system of equations:
##\alpha Y_1 = \vec Y \cdot \vec Y - b_1##
##\alpha Y_2 = \vec Y \cdot \vec Y - b_2##
.
.
.
##\alpha Y_n = \vec Y \cdot \vec Y - b_n##

Divide both sides by ##\alpha## to get
##Y_1 = 1/\alpha (\vec Y \cdot \vec Y - b_1)##
##Y_2 = 1/\alpha (\vec Y \cdot \vec Y - b_2)##
.
.
.
##Y_n = 1/\alpha (\vec Y \cdot \vec Y - b_n)##
 
so we're dealing with real scalars here.
- - - -

So your 'equation' is:

ϒϒ'C – ϒα = B
or

ϒϒ'C = ϒα + B

so you have a real symmetric rank one matrix on the Left Hand side (LHS).

The issue is that every possible c you can choose on the LHS gets mapped to zero or is an eigenvector (i.e. ϒ) or a linear combination of the two aforementioned things. So let's hope that B is either a scaled version of ϒ or else the zero vector. If your B is the zero vector, it should be pretty easy. Otherwise you have problems.

More issues: For starters, why write ϒα + B on the Right hand side... why not just write
##\propto ϒ##
 
StoneTemplePython said:
so we're dealing with real scalars here.
- - - -

So your 'equation' is:

ϒϒ'C – ϒα = B
or

ϒϒ'C = ϒα + B

so you have a real symmetric rank one matrix on the Left Hand side (LHS).

The issue is that every possible c you can choose on the LHS gets mapped to zero or is an eigenvector (i.e. ϒ) or a linear combination of the two aforementioned things. So let's hope that B is either a scaled version of ϒ or else the zero vector. If your B is the zero vector, it should be pretty easy. Otherwise you have problems.

More issues: For starters, why write ϒα + B on the Right hand side... why not just write
##\propto ϒ##
B is nonzero column, there is a way to solve that?
 
Tilfani said:
B is nonzero column, there is a way to solve that?

then ##B \propto ϒ## or this is not an equation
 
StoneTemplePython said:
then ##B \propto ϒ## or this is not an equation
B is a constant.
 
Tilfani said:
B is a constant.

I don't know what this means. Your original post, and a quick dimensional check say B is a a column vector.

What I am trying to tell you is your original post is analogous to

## 2 = 3##

or

## 2 = 3 +x##
for real ## x \geq 0##

this is not an equation. It is just wrong.
 
Yes i mean B is a constant column vector. Do you think that is wrong to?
 
  • #10
You're not hearing me. It is one of the 3 options

option a)
## B = \mathbf 0##

option b)
##B \propto ϒ##

option c)
this is not an equation. It is just wrong.
- - - - -
I have nothing more to say on the matter. Good luck.
 
  • #11
Ok, please look at eq (2 19) page 8 on this link, this paper, maybe some thing wrong.
Portfolio Theory: Origins, Markowitz and CAPM Based Selection - Springer
PDFhttps://www.springer.com › document
 
  • #12
Tilfani said:
Ok, please look at eq (2 19) page 8 on this link, this paper, maybe some thing wrong.
Portfolio Theory: Origins, Markowitz and CAPM Based Selection - Springer
PDFhttps://www.springer.com › document
Please provide the actual link to the document. The link you show is just to the Springer site.
 
  • #13
Mark44 said:
Please provide the actual link to the document. The link you show is just to the Springer site.
Please find enclosed the document. Go to page 7 to see the original problem, the resolution of lagrangian (which may be wrong) lead to equation posted which is (2 19)
 

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