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I am given three binary vectors v1, v2, v3 represented by unsigned int in my program and a finite field F, which is also a set of binary vectors. I need to check if the vectors are linearly independent that is there are no f1, f2 in F such that f1*v1 +f2*v2 = v3.

The immediate brute force solution is to iterate over the field and check all possible linear combinations.

Does there exist a more efficient algorithm?

I'd like to emphasize two points:

  1. The field elements are vectors, not scalars. Therefor,e a product of a field element f1 and a given vector vi is a dot product. So the Gaussian elimination does not work (if I am not missing something)
  2. the field is finite, so if I find that f1*v1 +f2*v2 = v3 for some f1,f2 it does not mean that f1,f2 belong to F.
double-beep
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user3032925
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  • See [this](http://stackoverflow.com/questions/532029/how-to-check-if-m-n-sized-vectors-are-linearly-independent?rq=1) –  Dec 17 '13 at 11:38
  • for three vectors, check the scalar product between v3 and v2^v1 (cross product) : if it's null, v3 is dependant. – lucasg Dec 17 '13 at 11:39
  • Well, for two vectors you'd use statistical correlation. Don't know offhand if there's an effective way to extend to 3 vectors. Also, are you looking for an exact fit or approximate? (Presumably approximate, if you want to prove "independence".) – Hot Licks Dec 17 '13 at 12:13
  • I am looking for an exact fit, i.e. f1*v1 +f2*v2 = v3 exactly. – user3032925 Dec 17 '13 at 12:58
  • Then, of course, they are not (necessarily) truly "independent". – Hot Licks Dec 17 '13 at 16:16
  • But your problem thus devolves into a "simple" simultaneous equation problem. – Hot Licks Dec 17 '13 at 16:17

3 Answers3

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If vectors are in r^2, then they are automatically dependent because when we make a matrix of them and reduce it to echelon form, there will be atleast one free variable(in this case only one).

If vectors are in R^3, then you can make a matrix from them i. a 2d array and then you can take determinant of that matrix. If determinant is equal to 0 then vectors are linearly dependent otherwise not.

If vectors are in R^4,R^5 and so on the then the appropriate way is to reduce matrix into echelon form.

Ali Raza
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For any finite set of M vectors defined in a space of dimension N, they are linearly independent iff the rank of a MxN matrix constructed by stacking these vectors row by row has rank equal to M.

Regarding numerically stable computation involving linear algebra, the singular value decomposition is usually the way to go and there are plenty of implementations available out there. The key point in this context is to realize the rank of a matrix equals the number of its non zero singular values. One must however note, that due to floating point approximations, a finite precision must be chosen to decide whether a value is effectively zero.

Your question mentions your vectors are defined in the set of integers and that certainly can be taken advantage of to overcome the finite precision of floating point computations, but I would not know how. Maybe somebody out there could help us out?

brunocodutra
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Gaussian elimination does work if you do it inside the finite field. For binary it should be quite simple, because inverse element is trivial. For larger finite fields, you will need somehow to find inverse elements, that may turns into a separate problem.

Askold Ilvento
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