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Subject: Re: Solving x'Px = v
Date: Fri, 12 Dec 2008 00:38:03 +0000 (UTC)
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"Benp P" <lightatron@hotmail.com> wrote in message <ghsae8$nlk$1@fred.mathworks.com>...
> Hi, I need to solve the equation
> 
> x'Px = v
> 
> where x is an (n by 1) vector of unknowns and P is a (n by n) square matrix (v is a scalar).
> 
> Any help would be great!
> 
> Thanks, 
> Ben.

  If one were to let P = [1 0;0 1] this would yield the equation

 x^2 + y^2 = v

for the unknown vector [x;y].  In other words you would be giving us the equation of a circle and asking us to find its solution.  That doesn't make much sense to me.  There are infinitely many solutions, and the same would be true in general for your quadratic equation x'*P*x = v.  There are n unknowns and only one equation.

Roger Stafford