Farkas' lemma

Farkas' lemma is a result in mathematics stating that a vector is either in a given convex cone or that there exists a (hyper)plane separating the vector from the cone, but not both. It was originally proved by the Hungarian mathematician Gyula Farkas  (1894, 1902). It is used amongst other things in the proof of the Karush–Kuhn–Tucker theorem in nonlinear programming.

Farkas' lemma is an example of a theorem of the alternative; a theorem stating that of two systems, one or the other has a solution, but not both or none.

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Statement of the lemma

Let A be an m × n matrix and b an m-dimensional vector. Then, exactly one of the following two statements is true:

  1. There exists an xRn such that Ax = b and x ≥ 0.
  2. There exists a yRm such that ATy ≥ 0 and bTy < 0.

Here, the notation x ≥ 0 means that all components of the vector x are nonnegative.

There are a number of slightly different (but equivalent) formulations of the Lemma in the literature. The one given here is due to Gale, Kuhn and Tucker in 1951.

Geometric interpretation

Let a1, …, anRm denote the columns of A. In terms of these vectors, Farkas' lemma states that exactly one of the following two statements is true:

  1. There exist non-negative coefficients x1, …, xnR such that b = x1a1 + ··· + xnan.
  2. There exists a vector yRm such that ai · y ≥ 0 for i = 1, …, n and b · y < 0.

The vectors x1a1 + ··· + xnan with nonnegative coefficients constitute the convex cone of the set {a1, …, an} so the first statement says that b is in this cone.

The second statement says that there exists a vector y such that the angle of y with the vectors ai is at most 90° while the angle of y with the vector b is more than 90°. The hyperplane normal to this vector has the vectors ai on one side and the vector b on the other side. Hence, this hyperplane separates the vectors in the cone of {a1, …, an} and the vector b.

For example, let n,m=2 and a1 = (1,0)T and a2 = (1,1)T. The convex cone spanned by a1 and a2 can be seen as a wedge-shaped slice of the first quadrant in the x-y plane. Now, suppose b = (0,1). Certainly, b is not in the convex cone a1x1+a2x2. Hence, there must be a separating hyperplane. Let y = (1,−1)T. We can see that a1 · y = 1, a2 · y = 0, and b · y = −1. Hence, the hyperplane with normal y indeed separates the convex cone a1x1+a2x2 from b.

Farkas' lemma can thus be interpreted geometrically as follows: Given a convex cone and a vector, either the vector is in the cone or there is a hyperplane separating the vector from the cone, but not both.

Further implications

Farkas' lemma can be varied to many further theorems of alternative by simple modifications, such as Gordan's theorem: Either Ax < 0 has a solution x, or A^T y = 0 has a nonzero solution y with y ≥ 0.

A particularly suggestive and easy-to-remember version is the following: if a set of inequalities has no solution, then a contradiction can be produced from it by linear combination with nonnegative coefficients. In formulas: if Axb is unsolvable then y^T A = 0, y^T b = -1, y0 has a solution[1]. (Note that y^T A is a combination of the left hand sides, y^T b a combination of the right hand side of the inequalities. Since the positive combination produces a zero vector on the left and a −1 on the right, the contradiction is apparent.)

References

Notes

  1. ^ Boyd, Stephen P.; Vandenberghe, Lieven (2004) (pdf). Convex Optimization. Cambridge University Press. ISBN 9780521833783. http://www.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf. Retrieved October 15, 2011.