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Linear equations with errors

asked 2014-09-16 11:04:30 -0600

oren gravatar image

updated 2015-01-14 03:32:37 -0600

FrédéricC gravatar image

I have a system of $n^2$ homogeneous linear equations in $n^2$ variables. Each equation is sparse and only involves $2n$ variables.

I create a list of equations and use solve(). I only get the "all-zero" solution. This is because of inaccuracies in the equations. I know (from theory) that there is be a nonzero kernel.

So, I'd like to find an approximate solution. That is - a solution of norm 1, which "almost fulfills" the equations. Behind the scenes I probably need the SVD decomposition of the matrix describing the equations (or at least, the input vectors corresponding to the small singular values).

  1. Do I have to create a matrix to represent the equations, or can I use my equations directly?
  2. Can it be a sparse matrix?
  3. Do I have to use an SVD routine, or is there some convenient way to solve my problem directly?
  4. Do you have an example of how to do it?
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Indeed, generic matrices are invertible. Before going further, when you speak about the theory, what sort of real numbers are your entries (the coefficients of your equation) ? Integer, rational, algebraic, real symbolic (like pi+sqrt(2), real numeric (like 0.123) ?

tmonteil gravatar imagetmonteil ( 2014-09-16 15:05:33 -0600 )edit

Could you provide your equations or the code you use to create them?

slelievre gravatar imageslelievre ( 2014-09-17 12:33:33 -0600 )edit

@tmontiel: The numbers are real numeric.

oren gravatar imageoren ( 2014-09-18 12:45:25 -0600 )edit

Update: I created a matrix $M$ with rows representing the equations, used M.SVD(), and took the vectors corresponding to the very small singular values. It worked.

oren gravatar imageoren ( 2014-09-18 12:48:41 -0600 )edit

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answered 2014-09-17 04:14:08 -0600

how about the following:

sage: m=matrix(RDF,[[1,1],[2,2]],sparse=True)
sage: m
[1.0 1.0]
[2.0 2.0]
sage: m.kernel()
Vector space of degree 2 and dimension 1 over Real Double Field
Basis matrix:
[ 1.0 -0.5]

Of course this will only work if your close to 0 eigenvalues are close enough to 0...

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Asked: 2014-09-16 11:04:30 -0600

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Last updated: Sep 17 '14